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

Optimal. Leaf size=24 \[ -\frac{1}{2} b^4 f^a \log ^4(f) \text{Gamma}\left (-4,-b x^2 \log (f)\right ) \]

[Out]

-(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)/2

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Rubi [A]  time = 0.0221914, 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} \[ -\frac{1}{2} b^4 f^a \log ^4(f) \text{Gamma}\left (-4,-b x^2 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^9,x]

[Out]

-(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)/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 \frac{f^{a+b x^2}}{x^9} \, dx &=-\frac{1}{2} b^4 f^a \Gamma \left (-4,-b x^2 \log (f)\right ) \log ^4(f)\\ \end{align*}

Mathematica [A]  time = 0.002278, size = 24, normalized size = 1. \[ -\frac{1}{2} b^4 f^a \log ^4(f) \text{Gamma}\left (-4,-b x^2 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^9,x]

[Out]

-(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)/2

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Maple [B]  time = 0.044, size = 101, normalized size = 4.2 \begin{align*} -{\frac{{f}^{a}{f}^{b{x}^{2}}}{8\,{x}^{8}}}-{\frac{{f}^{a}\ln \left ( f \right ) b{f}^{b{x}^{2}}}{24\,{x}^{6}}}-{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{f}^{b{x}^{2}}}{48\,{x}^{4}}}-{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}{f}^{b{x}^{2}}}{48\,{x}^{2}}}-{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{4}{b}^{4}{\it Ei} \left ( 1,-b{x}^{2}\ln \left ( f \right ) \right ) }{48}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*f^a/x^8*f^(b*x^2)-1/24*f^a*ln(f)*b/x^6*f^(b*x^2)-1/48*f^a*ln(f)^2*b^2/x^4*f^(b*x^2)-1/48*f^a*ln(f)^3*b^3/
x^2*f^(b*x^2)-1/48*f^a*ln(f)^4*b^4*Ei(1,-b*x^2*ln(f))

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

[Out]

Exception raised: TypeError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x^{2}}}{x^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**9, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x^{2} + a}}{x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^9, x)