3.129 \(\int f^{a+\frac{b}{x^2}} x^9 \, dx\)

Optimal. Leaf size=24 \[ -\frac{1}{2} b^5 f^a \log ^5(f) \text{Gamma}\left (-5,-\frac{b \log (f)}{x^2}\right ) \]

[Out]

-(b^5*f^a*Gamma[-5, -((b*Log[f])/x^2)]*Log[f]^5)/2

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Rubi [A]  time = 0.0247295, antiderivative size = 24, 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} \[ -\frac{1}{2} b^5 f^a \log ^5(f) \text{Gamma}\left (-5,-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^9,x]

[Out]

-(b^5*f^a*Gamma[-5, -((b*Log[f])/x^2)]*Log[f]^5)/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^9 \, dx &=-\frac{1}{2} b^5 f^a \Gamma \left (-5,-\frac{b \log (f)}{x^2}\right ) \log ^5(f)\\ \end{align*}

Mathematica [A]  time = 0.0023841, size = 24, normalized size = 1. \[ -\frac{1}{2} b^5 f^a \log ^5(f) \text{Gamma}\left (-5,-\frac{b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^9,x]

[Out]

-(b^5*f^a*Gamma[-5, -((b*Log[f])/x^2)]*Log[f]^5)/2

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Maple [B]  time = 0.063, size = 123, normalized size = 5.1 \begin{align*}{\frac{{f}^{a}{x}^{10}}{10}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{{f}^{a}\ln \left ( f \right ) b{x}^{8}}{40}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{x}^{6}}{120}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}{x}^{4}}{240}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{4}{b}^{4}{x}^{2}}{240}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{5}{b}^{5}}{240}{\it Ei} \left ( 1,-{\frac{b\ln \left ( f \right ) }{{x}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^9,x)

[Out]

1/10*f^a*x^10*f^(b/x^2)+1/40*f^a*ln(f)*b*x^8*f^(b/x^2)+1/120*f^a*ln(f)^2*b^2*x^6*f^(b/x^2)+1/240*f^a*ln(f)^3*b
^3*x^4*f^(b/x^2)+1/240*f^a*ln(f)^4*b^4*x^2*f^(b/x^2)+1/240*f^a*ln(f)^5*b^5*Ei(1,-b*ln(f)/x^2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^9,x, algorithm="maxima")

[Out]

Exception raised: TypeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^9,x, algorithm="fricas")

[Out]

Exception raised: TypeError

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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**9,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^{9}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^9,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^9, x)