\(\int \frac {f^{a+b x^2}}{x} \, dx\) [76]

Optimal result

Integrand size = 13, antiderivative size = 15 \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {1}{2} f^a \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \]

[Out]

1/2*f^a*Ei(b*x^2*ln(f))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2241} \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {1}{2} f^a \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \]

[In]

Int[f^(a + b*x^2)/x,x]

[Out]

(f^a*ExpIntegralEi[b*x^2*Log[f]])/2

Rule 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} f^a \text {Ei}\left (b x^2 \log (f)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {1}{2} f^a \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \]

[In]

Integrate[f^(a + b*x^2)/x,x]

[Out]

(f^a*ExpIntegralEi[b*x^2*Log[f]])/2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {f^{a} \operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )}{2}\)	\(16\)
meijerg	\(\frac {f^{a} \left (2 \ln \left (x \right )+\ln \left (-b \right )+\ln \left (\ln \left (f \right )\right )-\ln \left (-b \,x^{2} \ln \left (f \right )\right )-\operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )\right )}{2}\)	\(41\)

[In]

int(f^(b*x^2+a)/x,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a*Ei(1,-b*x^2*ln(f))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {1}{2} \, f^{a} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \]

[In]

integrate(f^(b*x^2+a)/x,x, algorithm="fricas")

[Out]

1/2*f^a*Ei(b*x^2*log(f))

Sympy [F]

\[ \int \frac {f^{a+b x^2}}{x} \, dx=\int \frac {f^{a + b x^{2}}}{x}\, dx \]

[In]

integrate(f**(b*x**2+a)/x,x)

[Out]

Integral(f**(a + b*x**2)/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {1}{2} \, f^{a} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \]

[In]

integrate(f^(b*x^2+a)/x,x, algorithm="maxima")

[Out]

1/2*f^a*Ei(b*x^2*log(f))

Giac [F]

\[ \int \frac {f^{a+b x^2}}{x} \, dx=\int { \frac {f^{b x^{2} + a}}{x} \,d x } \]

[In]

integrate(f^(b*x^2+a)/x,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x, x)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {f^{a+b x^2}}{x} \, dx=\frac {f^a\,\mathrm {ei}\left (b\,x^2\,\ln \left (f\right )\right )}{2} \]

[In]

int(f^(a + b*x^2)/x,x)

[Out]

(f^a*ei(b*x^2*log(f)))/2