3.156 \(\int f^{a+\frac {b}{x^3}} x^{11} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{3} b^4 f^a \log ^4(f) \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \]

[Out]

1/3*f^a*x^12*Ei(5,-b*ln(f)/x^3)

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Rubi [A]  time = 0.03, 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} \[ \frac {1}{3} b^4 f^a \log ^4(f) \text {Gamma}\left (-4,-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^3)*x^11,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^3)]*Log[f]^4)/3

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^3}} x^{11} \, dx &=\frac {1}{3} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \log ^4(f)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ \frac {1}{3} b^4 f^a \log ^4(f) \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^3)*x^11,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^3)]*Log[f]^4)/3

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fricas [B]  time = 0.43, size = 72, normalized size = 3.00 \[ -\frac {1}{72} \, b^{4} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{4} + \frac {1}{72} \, {\left (6 \, x^{12} + 2 \, b x^{9} \log \relax (f) + b^{2} x^{6} \log \relax (f)^{2} + b^{3} x^{3} \log \relax (f)^{3}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="fricas")

[Out]

-1/72*b^4*f^a*Ei(b*log(f)/x^3)*log(f)^4 + 1/72*(6*x^12 + 2*b*x^9*log(f) + b^2*x^6*log(f)^2 + b^3*x^3*log(f)^3)
*f^((a*x^3 + b)/x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{11}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)*x^11, x)

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maple [B]  time = 0.08, size = 213, normalized size = 8.88 \[ -\frac {\left (-\frac {\left (\frac {10 b \ln \relax (f )}{x^{3}}+\frac {5 b^{2} \ln \relax (f )^{2}}{x^{6}}+\frac {5 b^{3} \ln \relax (f )^{3}}{x^{9}}+30\right ) x^{12} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{3}}}}{120 b^{4} \ln \relax (f )^{4}}+\frac {\left (\frac {480 b \ln \relax (f )}{x^{3}}+\frac {360 b^{2} \ln \relax (f )^{2}}{x^{6}}+\frac {240 b^{3} \ln \relax (f )^{3}}{x^{9}}+\frac {125 b^{4} \ln \relax (f )^{4}}{x^{12}}+360\right ) x^{12}}{1440 b^{4} \ln \relax (f )^{4}}-\frac {x^{12}}{4 b^{4} \ln \relax (f )^{4}}-\frac {x^{9}}{3 b^{3} \ln \relax (f )^{3}}-\frac {x^{6}}{4 b^{2} \ln \relax (f )^{2}}-\frac {x^{3}}{6 b \ln \relax (f )}-\frac {\Ei \left (1, -\frac {b \ln \relax (f )}{x^{3}}\right )}{24}-\frac {\ln \relax (x )}{8}+\frac {\ln \left (-b \right )}{24}-\frac {\ln \left (-\frac {b \ln \relax (f )}{x^{3}}\right )}{24}+\frac {\ln \left (\ln \relax (f )\right )}{24}-\frac {25}{288}\right ) b^{4} f^{a} \ln \relax (f )^{4}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^3)*x^11,x)

[Out]

-1/3*f^a*b^4*ln(f)^4*(1/1440/b^4*x^12/ln(f)^4*(125*b^4/x^12*ln(f)^4+240*b^3/x^9*ln(f)^3+360*b^2/x^6*ln(f)^2+48
0*b/x^3*ln(f)+360)-1/120/b^4*x^12/ln(f)^4*(5*b^3/x^9*ln(f)^3+5*b^2/x^6*ln(f)^2+10*b/x^3*ln(f)+30)*exp(b/x^3*ln
(f))-1/24*ln(-b/x^3*ln(f))-1/24*Ei(1,-b/x^3*ln(f))-25/288-1/8*ln(x)+1/24*ln(-b)+1/24*ln(ln(f))-1/4/b^4*x^12/ln
(f)^4-1/3/b^3*x^9/ln(f)^3-1/4/b^2*x^6/ln(f)^2-1/6/b*x^3/ln(f))

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maxima [B]  time = 1.29, size = 22, normalized size = 0.92 \[ \frac {1}{3} \, b^{4} f^{a} \Gamma \left (-4, -\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="maxima")

[Out]

1/3*b^4*f^a*gamma(-4, -b*log(f)/x^3)*log(f)^4

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mupad [B]  time = 3.78, size = 90, normalized size = 3.75 \[ \frac {b^4\,f^a\,{\ln \relax (f)}^4\,\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}{72}+\frac {b^4\,f^a\,f^{\frac {b}{x^3}}\,{\ln \relax (f)}^4\,\left (\frac {x^3}{24\,b\,\ln \relax (f)}+\frac {x^6}{24\,b^2\,{\ln \relax (f)}^2}+\frac {x^9}{12\,b^3\,{\ln \relax (f)}^3}+\frac {x^{12}}{4\,b^4\,{\ln \relax (f)}^4}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^3)*x^11,x)

[Out]

(b^4*f^a*log(f)^4*expint(-(b*log(f))/x^3))/72 + (b^4*f^a*f^(b/x^3)*log(f)^4*(x^3/(24*b*log(f)) + x^6/(24*b^2*l
og(f)^2) + x^9/(12*b^3*log(f)^3) + x^12/(4*b^4*log(f)^4)))/3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{11}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**3)*x**11,x)

[Out]

Integral(f**(a + b/x**3)*x**11, x)

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