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

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

[Out]

1/2*f^a*x^10*Ei(6,-b*ln(f)/x^2)

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ -\frac {1}{2} b^5 f^a \log ^5(f) \Gamma \left (-5,-\frac {b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^9,x]

[Out]

-1/2*(b^5*f^a*Gamma[-5, -((b*Log[f])/x^2)]*Log[f]^5)

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fricas [B]  time = 0.42, size = 84, normalized size = 3.50 \[ -\frac {1}{240} \, b^{5} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{2}}\right ) \log \relax (f)^{5} + \frac {1}{240} \, {\left (24 \, x^{10} + 6 \, b x^{8} \log \relax (f) + 2 \, b^{2} x^{6} \log \relax (f)^{2} + b^{3} x^{4} \log \relax (f)^{3} + b^{4} x^{2} \log \relax (f)^{4}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*b^5*f^a*Ei(b*log(f)/x^2)*log(f)^5 + 1/240*(24*x^10 + 6*b*x^8*log(f) + 2*b^2*x^6*log(f)^2 + b^3*x^4*log(
f)^3 + b^4*x^2*log(f)^4)*f^((a*x^2 + b)/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{9}\,{d x} \]

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)

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maple [B]  time = 0.08, size = 123, normalized size = 5.12 \[ \frac {b^{5} f^{a} \Ei \left (1, -\frac {b \ln \relax (f )}{x^{2}}\right ) \ln \relax (f )^{5}}{240}+\frac {b^{4} x^{2} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{4}}{240}+\frac {b^{3} x^{4} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{3}}{240}+\frac {b^{2} x^{6} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{2}}{120}+\frac {b \,x^{8} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )}{40}+\frac {x^{10} f^{a} f^{\frac {b}{x^{2}}}}{10} \]

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/x^2*ln(f))

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maxima [B]  time = 1.32, size = 22, normalized size = 0.92 \[ -\frac {1}{2} \, b^{5} f^{a} \Gamma \left (-5, -\frac {b \log \relax (f)}{x^{2}}\right ) \log \relax (f)^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^5*f^a*gamma(-5, -b*log(f)/x^2)*log(f)^5

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mupad [B]  time = 3.79, size = 102, normalized size = 4.25 \[ \frac {b^5\,f^a\,{\ln \relax (f)}^5\,\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}{240}+\frac {b^5\,f^a\,f^{\frac {b}{x^2}}\,{\ln \relax (f)}^5\,\left (\frac {x^2}{120\,b\,\ln \relax (f)}+\frac {x^4}{120\,b^2\,{\ln \relax (f)}^2}+\frac {x^6}{60\,b^3\,{\ln \relax (f)}^3}+\frac {x^8}{20\,b^4\,{\ln \relax (f)}^4}+\frac {x^{10}}{5\,b^5\,{\ln \relax (f)}^5}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^9,x)

[Out]

(b^5*f^a*log(f)^5*expint(-(b*log(f))/x^2))/240 + (b^5*f^a*f^(b/x^2)*log(f)^5*(x^2/(120*b*log(f)) + x^4/(120*b^
2*log(f)^2) + x^6/(60*b^3*log(f)^3) + x^8/(20*b^4*log(f)^4) + x^10/(5*b^5*log(f)^5)))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{9}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**9, x)

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