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

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

[Out]

1/2*f^a*x^9*(-32/945*Pi^(1/2)*erfc((-b*ln(f)/x^2)^(1/2))+32/945/(-b*ln(f)/x^2)^(1/2)*exp(b*ln(f)/x^2)-16/945/(
-b*ln(f)/x^2)^(3/2)*exp(b*ln(f)/x^2)+8/315/(-b*ln(f)/x^2)^(5/2)*exp(b*ln(f)/x^2)-4/63/(-b*ln(f)/x^2)^(7/2)*exp
(b*ln(f)/x^2)+2/9/(-b*ln(f)/x^2)^(9/2)*exp(b*ln(f)/x^2))*(-b*ln(f)/x^2)^(9/2)

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Rubi [A]  time = 0.02, antiderivative size = 34, 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} \[ \frac {1}{2} x^9 f^a \left (-\frac {b \log (f)}{x^2}\right )^{9/2} \text {Gamma}\left (-\frac {9}{2},-\frac {b \log (f)}{x^2}\right ) \]

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*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.00 \[ \frac {1}{2} x^9 f^a \left (-\frac {b \log (f)}{x^2}\right )^{9/2} \Gamma \left (-\frac {9}{2},-\frac {b \log (f)}{x^2}\right ) \]

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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fricas [A]  time = 0.42, size = 98, normalized size = 2.88 \[ \frac {16}{945} \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b^{4} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right ) \log \relax (f)^{4} + \frac {1}{945} \, {\left (105 \, x^{9} + 30 \, b x^{7} \log \relax (f) + 12 \, b^{2} x^{5} \log \relax (f)^{2} + 8 \, b^{3} x^{3} \log \relax (f)^{3} + 16 \, b^{4} x \log \relax (f)^{4}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{8}\,{d x} \]

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)

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maple [A]  time = 0.08, size = 133, normalized size = 3.91 \[ -\frac {16 \sqrt {\pi }\, b^{5} f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right ) \ln \relax (f )^{5}}{945 \sqrt {-b \ln \relax (f )}}+\frac {16 b^{4} x \,f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{4}}{945}+\frac {8 b^{3} x^{3} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{3}}{945}+\frac {4 b^{2} x^{5} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{2}}{315}+\frac {2 b \,x^{7} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )}{63}+\frac {x^{9} f^{a} f^{\frac {b}{x^{2}}}}{9} \]

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 = 1.34, size = 28, normalized size = 0.82 \[ \frac {1}{2} \, f^{a} x^{9} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {9}{2}} \Gamma \left (-\frac {9}{2}, -\frac {b \log \relax (f)}{x^{2}}\right ) \]

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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mupad [B]  time = 3.66, size = 151, normalized size = 4.44 \[ \frac {f^a\,f^{\frac {b}{x^2}}\,x^9}{9}+\frac {16\,f^a\,x^9\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}^{9/2}}{945}-\frac {16\,f^a\,x^9\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \relax (f)}{x^2}}\right )\,{\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}^{9/2}}{945}+\frac {16\,b^4\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \relax (f)}^4}{945}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x^5\,{\ln \relax (f)}^2}{315}+\frac {8\,b^3\,f^a\,f^{\frac {b}{x^2}}\,x^3\,{\ln \relax (f)}^3}{945}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^7\,\ln \relax (f)}{63} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^8,x)

[Out]

(f^a*f^(b/x^2)*x^9)/9 + (16*f^a*x^9*pi^(1/2)*(-(b*log(f))/x^2)^(9/2))/945 - (16*f^a*x^9*pi^(1/2)*erfc((-(b*log
(f))/x^2)^(1/2))*(-(b*log(f))/x^2)^(9/2))/945 + (16*b^4*f^a*f^(b/x^2)*x*log(f)^4)/945 + (4*b^2*f^a*f^(b/x^2)*x
^5*log(f)^2)/315 + (8*b^3*f^a*f^(b/x^2)*x^3*log(f)^3)/945 + (2*b*f^a*f^(b/x^2)*x^7*log(f))/63

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{8}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**8, x)

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