3.142 \(\int f^{a+\frac{b}{x^2}} x^8 \, dx\)

Optimal. Leaf size=34 \[ \frac{1}{2} x^9 f^a \left (-\frac{b \log (f)}{x^2}\right )^{9/2} \text{Gamma}\left (-\frac{9}{2},-\frac{b \log (f)}{x^2}\right ) \]

[Out]

(f^a*x^9*Gamma[-9/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(9/2))/2

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Rubi [A]  time = 0.0236792, antiderivative size = 34, 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} \[ \frac{1}{2} x^9 f^a \left (-\frac{b \log (f)}{x^2}\right )^{9/2} \text{Gamma}\left (-\frac{9}{2},-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^8,x]

[Out]

(f^a*x^9*Gamma[-9/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(9/2))/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^8 \, dx &=\frac{1}{2} f^a x^9 \Gamma \left (-\frac{9}{2},-\frac{b \log (f)}{x^2}\right ) \left (-\frac{b \log (f)}{x^2}\right )^{9/2}\\ \end{align*}

Mathematica [A]  time = 0.004138, size = 34, normalized size = 1. \[ \frac{1}{2} x^9 f^a \left (-\frac{b \log (f)}{x^2}\right )^{9/2} \text{Gamma}\left (-\frac{9}{2},-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^8,x]

[Out]

(f^a*x^9*Gamma[-9/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(9/2))/2

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Maple [A]  time = 0.054, size = 133, normalized size = 3.9 \begin{align*}{\frac{{f}^{a}{x}^{9}}{9}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{2\,{f}^{a}\ln \left ( f \right ) b{x}^{7}}{63}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{x}^{5}}{315}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{8\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}{x}^{3}}{945}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{16\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{4}{b}^{4}x}{945}{f}^{{\frac{b}{{x}^{2}}}}}-{\frac{16\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{5}{b}^{5}\sqrt{\pi }}{945}{\it Erf} \left ({\frac{1}{x}\sqrt{-b\ln \left ( f \right ) }} \right ){\frac{1}{\sqrt{-b\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^8,x)

[Out]

1/9*f^a*x^9*f^(b/x^2)+2/63*f^a*ln(f)*b*x^7*f^(b/x^2)+4/315*f^a*ln(f)^2*b^2*x^5*f^(b/x^2)+8/945*f^a*ln(f)^3*b^3
*x^3*f^(b/x^2)+16/945*f^a*ln(f)^4*b^4*x*f^(b/x^2)-16/945*f^a*ln(f)^5*b^5*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(
f))^(1/2)/x)

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Maxima [A]  time = 1.20365, size = 38, normalized size = 1.12 \begin{align*} \frac{1}{2} \, f^{a} x^{9} \left (-\frac{b \log \left (f\right )}{x^{2}}\right )^{\frac{9}{2}} \Gamma \left (-\frac{9}{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^8,x, algorithm="maxima")

[Out]

1/2*f^a*x^9*(-b*log(f)/x^2)^(9/2)*gamma(-9/2, -b*log(f)/x^2)

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Fricas [A]  time = 1.75018, size = 259, normalized size = 7.62 \begin{align*} \frac{16}{945} \, \sqrt{\pi } \sqrt{-b \log \left (f\right )} b^{4} f^{a} \operatorname{erf}\left (\frac{\sqrt{-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{4} + \frac{1}{945} \,{\left (105 \, x^{9} + 30 \, b x^{7} \log \left (f\right ) + 12 \, b^{2} x^{5} \log \left (f\right )^{2} + 8 \, b^{3} x^{3} \log \left (f\right )^{3} + 16 \, b^{4} x \log \left (f\right )^{4}\right )} f^{\frac{a x^{2} + b}{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^8,x, algorithm="fricas")

[Out]

16/945*sqrt(pi)*sqrt(-b*log(f))*b^4*f^a*erf(sqrt(-b*log(f))/x)*log(f)^4 + 1/945*(105*x^9 + 30*b*x^7*log(f) + 1
2*b^2*x^5*log(f)^2 + 8*b^3*x^3*log(f)^3 + 16*b^4*x*log(f)^4)*f^((a*x^2 + b)/x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)*x**8,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{2}}} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^8,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^8, x)