3.141 \(\int f^{a+\frac {b}{x^2}} x^{10} \, dx\)

Optimal. Leaf size=34 \[ \frac {1}{2} x^{11} f^a \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \]

[Out]

1/2*f^a*x^11*(64/10395*Pi^(1/2)*erfc((-b*ln(f)/x^2)^(1/2))-64/10395/(-b*ln(f)/x^2)^(1/2)*exp(b*ln(f)/x^2)+32/1
0395/(-b*ln(f)/x^2)^(3/2)*exp(b*ln(f)/x^2)-16/3465/(-b*ln(f)/x^2)^(5/2)*exp(b*ln(f)/x^2)+8/693/(-b*ln(f)/x^2)^
(7/2)*exp(b*ln(f)/x^2)-4/99/(-b*ln(f)/x^2)^(9/2)*exp(b*ln(f)/x^2)+2/11/(-b*ln(f)/x^2)^(11/2)*exp(b*ln(f)/x^2))
*(-b*ln(f)/x^2)^(11/2)

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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} \[ \frac {1}{2} x^{11} f^a \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \text {Gamma}\left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^10,x]

[Out]

(f^a*x^11*Gamma[-11/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(11/2))/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^{10} \, dx &=\frac {1}{2} f^a x^{11} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \left (-\frac {b \log (f)}{x^2}\right )^{11/2}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.00 \[ \frac {1}{2} x^{11} f^a \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^10,x]

[Out]

(f^a*x^11*Gamma[-11/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(11/2))/2

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fricas [A]  time = 0.42, size = 110, normalized size = 3.24 \[ \frac {32}{10395} \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b^{5} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right ) \log \relax (f)^{5} + \frac {1}{10395} \, {\left (945 \, x^{11} + 210 \, b x^{9} \log \relax (f) + 60 \, b^{2} x^{7} \log \relax (f)^{2} + 24 \, b^{3} x^{5} \log \relax (f)^{3} + 16 \, b^{4} x^{3} \log \relax (f)^{4} + 32 \, b^{5} x \log \relax (f)^{5}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^10,x, algorithm="fricas")

[Out]

32/10395*sqrt(pi)*sqrt(-b*log(f))*b^5*f^a*erf(sqrt(-b*log(f))/x)*log(f)^5 + 1/10395*(945*x^11 + 210*b*x^9*log(
f) + 60*b^2*x^7*log(f)^2 + 24*b^3*x^5*log(f)^3 + 16*b^4*x^3*log(f)^4 + 32*b^5*x*log(f)^5)*f^((a*x^2 + b)/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{10}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^10,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^10, x)

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maple [A]  time = 0.13, size = 155, normalized size = 4.56 \[ -\frac {32 \sqrt {\pi }\, b^{6} f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right ) \ln \relax (f )^{6}}{10395 \sqrt {-b \ln \relax (f )}}+\frac {32 b^{5} x \,f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{5}}{10395}+\frac {16 b^{4} x^{3} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{4}}{10395}+\frac {8 b^{3} x^{5} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{3}}{3465}+\frac {4 b^{2} x^{7} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{2}}{693}+\frac {2 b \,x^{9} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )}{99}+\frac {x^{11} f^{a} f^{\frac {b}{x^{2}}}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^10,x)

[Out]

1/11*f^a*x^11*f^(b/x^2)+2/99*f^a*ln(f)*b*x^9*f^(b/x^2)+4/693*f^a*ln(f)^2*b^2*x^7*f^(b/x^2)+8/3465*f^a*ln(f)^3*
b^3*x^5*f^(b/x^2)+16/10395*f^a*ln(f)^4*b^4*x^3*f^(b/x^2)+32/10395*f^a*ln(f)^5*b^5*x*f^(b/x^2)-32/10395*f^a*ln(
f)^6*b^6*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)/x)

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maxima [A]  time = 1.22, size = 28, normalized size = 0.82 \[ \frac {1}{2} \, f^{a} x^{11} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {11}{2}} \Gamma \left (-\frac {11}{2}, -\frac {b \log \relax (f)}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^10,x, algorithm="maxima")

[Out]

1/2*f^a*x^11*(-b*log(f)/x^2)^(11/2)*gamma(-11/2, -b*log(f)/x^2)

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mupad [B]  time = 3.66, size = 173, normalized size = 5.09 \[ \frac {f^a\,f^{\frac {b}{x^2}}\,x^{11}}{11}-\frac {32\,f^a\,x^{11}\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}^{11/2}}{10395}+\frac {32\,f^a\,x^{11}\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \relax (f)}{x^2}}\right )\,{\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}^{11/2}}{10395}+\frac {32\,b^5\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \relax (f)}^5}{10395}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x^7\,{\ln \relax (f)}^2}{693}+\frac {8\,b^3\,f^a\,f^{\frac {b}{x^2}}\,x^5\,{\ln \relax (f)}^3}{3465}+\frac {16\,b^4\,f^a\,f^{\frac {b}{x^2}}\,x^3\,{\ln \relax (f)}^4}{10395}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^9\,\ln \relax (f)}{99} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^10,x)

[Out]

(f^a*f^(b/x^2)*x^11)/11 - (32*f^a*x^11*pi^(1/2)*(-(b*log(f))/x^2)^(11/2))/10395 + (32*f^a*x^11*pi^(1/2)*erfc((
-(b*log(f))/x^2)^(1/2))*(-(b*log(f))/x^2)^(11/2))/10395 + (32*b^5*f^a*f^(b/x^2)*x*log(f)^5)/10395 + (4*b^2*f^a
*f^(b/x^2)*x^7*log(f)^2)/693 + (8*b^3*f^a*f^(b/x^2)*x^5*log(f)^3)/3465 + (16*b^4*f^a*f^(b/x^2)*x^3*log(f)^4)/1
0395 + (2*b*f^a*f^(b/x^2)*x^9*log(f))/99

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{10}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)*x**10,x)

[Out]

Integral(f**(a + b/x**2)*x**10, x)

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