3.158 \(\int f^{a+\frac {b}{x^3}} x^5 \, dx\)

Optimal. Leaf size=58 \[ -\frac {1}{6} b^2 f^a \log ^2(f) \text {Ei}\left (\frac {b \log (f)}{x^3}\right )+\frac {1}{6} b x^3 \log (f) f^{a+\frac {b}{x^3}}+\frac {1}{6} x^6 f^{a+\frac {b}{x^3}} \]

[Out]

1/6*f^(a+b/x^3)*x^6+1/6*b*f^(a+b/x^3)*x^3*ln(f)-1/6*b^2*f^a*Ei(b*ln(f)/x^3)*ln(f)^2

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Rubi [A]  time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2214, 2210} \[ -\frac {1}{6} b^2 f^a \log ^2(f) \text {Ei}\left (\frac {b \log (f)}{x^3}\right )+\frac {1}{6} x^6 f^{a+\frac {b}{x^3}}+\frac {1}{6} b x^3 \log (f) f^{a+\frac {b}{x^3}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^3)*x^5,x]

[Out]

(f^(a + b/x^3)*x^6)/6 + (b*f^(a + b/x^3)*x^3*Log[f])/6 - (b^2*f^a*ExpIntegralEi[(b*Log[f])/x^3]*Log[f]^2)/6

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

\begin {align*} \int f^{a+\frac {b}{x^3}} x^5 \, dx &=\frac {1}{6} f^{a+\frac {b}{x^3}} x^6+\frac {1}{2} (b \log (f)) \int f^{a+\frac {b}{x^3}} x^2 \, dx\\ &=\frac {1}{6} f^{a+\frac {b}{x^3}} x^6+\frac {1}{6} b f^{a+\frac {b}{x^3}} x^3 \log (f)+\frac {1}{2} \left (b^2 \log ^2(f)\right ) \int \frac {f^{a+\frac {b}{x^3}}}{x} \, dx\\ &=\frac {1}{6} f^{a+\frac {b}{x^3}} x^6+\frac {1}{6} b f^{a+\frac {b}{x^3}} x^3 \log (f)-\frac {1}{6} b^2 f^a \text {Ei}\left (\frac {b \log (f)}{x^3}\right ) \log ^2(f)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.76 \[ \frac {1}{6} f^a \left (x^3 f^{\frac {b}{x^3}} \left (b \log (f)+x^3\right )-b^2 \log ^2(f) \text {Ei}\left (\frac {b \log (f)}{x^3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^3)*x^5,x]

[Out]

(f^a*(-(b^2*ExpIntegralEi[(b*Log[f])/x^3]*Log[f]^2) + f^(b/x^3)*x^3*(x^3 + b*Log[f])))/6

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fricas [A]  time = 0.46, size = 47, normalized size = 0.81 \[ -\frac {1}{6} \, b^{2} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{2} + \frac {1}{6} \, {\left (x^{6} + b x^{3} \log \relax (f)\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^5,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^5,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)*x^5, x)

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maple [B]  time = 0.07, size = 141, normalized size = 2.43 \[ -\frac {\left (-\frac {\left (\frac {3 b \ln \relax (f )}{x^{3}}+3\right ) x^{6} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{3}}}}{6 b^{2} \ln \relax (f )^{2}}+\frac {\left (\frac {12 b \ln \relax (f )}{x^{3}}+\frac {9 b^{2} \ln \relax (f )^{2}}{x^{6}}+6\right ) x^{6}}{12 b^{2} \ln \relax (f )^{2}}-\frac {x^{6}}{2 b^{2} \ln \relax (f )^{2}}-\frac {x^{3}}{b \ln \relax (f )}-\frac {\Ei \left (1, -\frac {b \ln \relax (f )}{x^{3}}\right )}{2}-\frac {3 \ln \relax (x )}{2}+\frac {\ln \left (-b \right )}{2}-\frac {\ln \left (-\frac {b \ln \relax (f )}{x^{3}}\right )}{2}+\frac {\ln \left (\ln \relax (f )\right )}{2}-\frac {3}{4}\right ) b^{2} f^{a} \ln \relax (f )^{2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^3)*x^5,x)

[Out]

-1/3*f^a*b^2*ln(f)^2*(1/12/b^2*x^6/ln(f)^2*(9*b^2/x^6*ln(f)^2+12*b/x^3*ln(f)+6)-1/6/b^2*x^6/ln(f)^2*(3+3*b/x^3
*ln(f))*exp(b/x^3*ln(f))-1/2*ln(-b/x^3*ln(f))-1/2*Ei(1,-b/x^3*ln(f))-3/4-3/2*ln(x)+1/2*ln(-b)+1/2*ln(ln(f))-1/
2/b^2*x^6/ln(f)^2-1/b*x^3/ln(f))

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maxima [A]  time = 1.35, size = 22, normalized size = 0.38 \[ \frac {1}{3} \, b^{2} f^{a} \Gamma \left (-2, -\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^5,x, algorithm="maxima")

[Out]

1/3*b^2*f^a*gamma(-2, -b*log(f)/x^3)*log(f)^2

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mupad [B]  time = 3.65, size = 57, normalized size = 0.98 \[ \frac {b^2\,f^a\,{\ln \relax (f)}^2\,\left (f^{\frac {b}{x^3}}\,\left (\frac {x^3}{2\,b\,\ln \relax (f)}+\frac {x^6}{2\,b^2\,{\ln \relax (f)}^2}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}{2}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^3)*x^5,x)

[Out]

(b^2*f^a*log(f)^2*(f^(b/x^3)*(x^3/(2*b*log(f)) + x^6/(2*b^2*log(f)^2)) + expint(-(b*log(f))/x^3)/2))/3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**3)*x**5,x)

[Out]

Integral(f**(a + b/x**3)*x**5, x)

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