3.128 \(\int f^{a+\frac{b}{x^2}} x^m \, dx\)

Optimal. Leaf size=46 \[ \frac{1}{2} f^a x^{m+1} \left (-\frac{b \log (f)}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (f)}{x^2}\right ) \]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^((1 + m)/2))/2

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Rubi [A]  time = 0.0236666, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2218} \[ \frac{1}{2} f^a x^{m+1} \left (-\frac{b \log (f)}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^m,x]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^((1 + m)/2))/2

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int f^{a+\frac{b}{x^2}} x^m \, dx &=\frac{1}{2} f^a x^{1+m} \Gamma \left (\frac{1}{2} (-1-m),-\frac{b \log (f)}{x^2}\right ) \left (-\frac{b \log (f)}{x^2}\right )^{\frac{1+m}{2}}\\ \end{align*}

Mathematica [A]  time = 0.0102872, size = 46, normalized size = 1. \[ \frac{1}{2} f^a x^{m+1} \left (-\frac{b \log (f)}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^m,x]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^((1 + m)/2))/2

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Maple [B]  time = 0.038, size = 169, normalized size = 3.7 \begin{align*} -{\frac{{f}^{a}}{2} \left ( -b \right ) ^{{\frac{m}{2}}+{\frac{1}{2}}} \left ( \ln \left ( f \right ) \right ) ^{{\frac{m}{2}}+{\frac{1}{2}}} \left ( 2\,{\frac{{x}^{-1+m} \left ( -b \right ) ^{-m/2-1/2} \left ( \ln \left ( f \right ) \right ) ^{1/2-m/2}b\Gamma \left ( 1/2-m/2 \right ) }{1+m} \left ( -{\frac{b\ln \left ( f \right ) }{{x}^{2}}} \right ) ^{-1/2+m/2}}-2\,{\frac{{x}^{1+m} \left ( -b \right ) ^{-m/2-1/2} \left ( \ln \left ( f \right ) \right ) ^{-m/2-1/2}}{1+m}{{\rm e}^{{\frac{b\ln \left ( f \right ) }{{x}^{2}}}}}}-2\,{\frac{{x}^{-1+m} \left ( -b \right ) ^{-m/2-1/2} \left ( \ln \left ( f \right ) \right ) ^{1/2-m/2}b}{1+m} \left ( -{\frac{b\ln \left ( f \right ) }{{x}^{2}}} \right ) ^{-1/2+m/2}\Gamma \left ( 1/2-m/2,-{\frac{b\ln \left ( f \right ) }{{x}^{2}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^m,x)

[Out]

-1/2*f^a*(-b)^(1/2*m+1/2)*ln(f)^(1/2*m+1/2)*(2/(1+m)*x^(-1+m)*(-b)^(-1/2*m-1/2)*ln(f)^(1/2-1/2*m)*b*(-b*ln(f)/
x^2)^(-1/2+1/2*m)*GAMMA(1/2-1/2*m)-2/(1+m)*x^(1+m)*(-b)^(-1/2*m-1/2)*ln(f)^(-1/2*m-1/2)*exp(b*ln(f)/x^2)-2/(1+
m)*x^(-1+m)*(-b)^(-1/2*m-1/2)*ln(f)^(1/2-1/2*m)*b*(-b*ln(f)/x^2)^(-1/2+1/2*m)*GAMMA(1/2-1/2*m,-b*ln(f)/x^2))

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Maxima [A]  time = 1.3925, size = 51, normalized size = 1.11 \begin{align*} \frac{1}{2} \, f^{a} x^{m + 1} \left (-\frac{b \log \left (f\right )}{x^{2}}\right )^{\frac{1}{2} \, m + \frac{1}{2}} \Gamma \left (-\frac{1}{2} \, m - \frac{1}{2}, -\frac{b \log \left (f\right )}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^a*x^(m + 1)*(-b*log(f)/x^2)^(1/2*m + 1/2)*gamma(-1/2*m - 1/2, -b*log(f)/x^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (f^{\frac{a x^{2} + b}{x^{2}}} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^((a*x^2 + b)/x^2)*x^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{2}}} x^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{2}}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^2)*x^m, x)