Equivalent codes, different results (Python, Mathematica)

These are two codes, one written with Python 3 and the other with Wolfram Mathematica. The codes are equivalent, so the results (graphs) should be the same. But the codes give different graphs. Here are the codes.

Python code:

import numpy as np
import matplotlib.pyplot as plt 
from scipy.special import k0, k1, i0, i1

k=100.0
x = 0.0103406
B = 80.0

def fdens(f):
    return (1/2*(1-f**2)**2+f **4/2
            +1/2*B*k*x**2*f**2*(1-f**2)*np.log(1+2/(B*k*x**2))
            +(B*f**2*(1+B*k*x**2))/((k*(2+B*k*x**2))**2)
            -f**4/(2+B*k*x**2)
            +(B*f)/(k*x)*
            (k0(f*x)*i1(f *np.sqrt(2/(k*B)+x**2))
            +i0(f*x)*k1(f *np.sqrt(2/(k*B)+x**2)))/
            (k1(f*x)*i1(f *np.sqrt(2/(k*B)+x**2))
            -i1(f*x)*k1(f *np.sqrt(2/(k*B)+x**2)))
            )

plt.figure(figsize=(10, 8), dpi=70)
X = np.linspace(0, 1, 100, endpoint=True)
C = fdens(X)
plt.plot(X, C, color="blue", linewidth=2.0, linestyle="-")
plt.show()

      

        
        
        
      

    

python result

Mathematica code:

k=100.;B=80.;
x=0.0103406;
func[f_]:=1/2*(1-f^2)^2+1/2*B*k*x^2*f^2*(1-f^2)*Log[1+2/(B*k*x^2)]+f^4/2-f^4/(2+B*k*x^2)+B*f^2*(1+B*k*x^2)/(k*(2+B*k*x^2)^2)+(B*f)/(k*x)*(BesselI[1, (f*Sqrt[2/(B*k) + x^2])]*BesselK[0, f*x] + BesselI[0, f*x]*BesselK[1, (f*Sqrt[2/(B*k) + x^2])])/(BesselI[1, (f*Sqrt[2/(B*k) + x^2])]*BesselK[1,f*x] - BesselI[1,f*x]*BesselK[1, (f*Sqrt[2/(B*k) + x^2])]);

Plot[func[f],{f,0,1}]

      

        
        
        
      

    

Mathematica result (fix)

The results are different. Does anyone know why?