3.131 \(\int f^{a+\frac {b}{x^2}} x^5 \, dx\)

Optimal. Leaf size=81 \[ -\frac {1}{12} b^3 f^a \log ^3(f) \text {Ei}\left (\frac {b \log (f)}{x^2}\right )+\frac {1}{12} b^2 x^2 \log ^2(f) f^{a+\frac {b}{x^2}}+\frac {1}{6} x^6 f^{a+\frac {b}{x^2}}+\frac {1}{12} b x^4 \log (f) f^{a+\frac {b}{x^2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2214, 2210} \[ -\frac {1}{12} b^3 f^a \log ^3(f) \text {Ei}\left (\frac {b \log (f)}{x^2}\right )+\frac {1}{12} b^2 x^2 \log ^2(f) f^{a+\frac {b}{x^2}}+\frac {1}{6} x^6 f^{a+\frac {b}{x^2}}+\frac {1}{12} b x^4 \log (f) f^{a+\frac {b}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^5,x]

[Out]

(f^(a + b/x^2)*x^6)/6 + (b*f^(a + b/x^2)*x^4*Log[f])/12 + (b^2*f^(a + b/x^2)*x^2*Log[f]^2)/12 - (b^3*f^a*ExpIn
tegralEi[(b*Log[f])/x^2]*Log[f]^3)/12

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^2}} x^5 \, dx &=\frac {1}{6} f^{a+\frac {b}{x^2}} x^6+\frac {1}{3} (b \log (f)) \int f^{a+\frac {b}{x^2}} x^3 \, dx\\ &=\frac {1}{6} f^{a+\frac {b}{x^2}} x^6+\frac {1}{12} b f^{a+\frac {b}{x^2}} x^4 \log (f)+\frac {1}{6} \left (b^2 \log ^2(f)\right ) \int f^{a+\frac {b}{x^2}} x \, dx\\ &=\frac {1}{6} f^{a+\frac {b}{x^2}} x^6+\frac {1}{12} b f^{a+\frac {b}{x^2}} x^4 \log (f)+\frac {1}{12} b^2 f^{a+\frac {b}{x^2}} x^2 \log ^2(f)+\frac {1}{6} \left (b^3 \log ^3(f)\right ) \int \frac {f^{a+\frac {b}{x^2}}}{x} \, dx\\ &=\frac {1}{6} f^{a+\frac {b}{x^2}} x^6+\frac {1}{12} b f^{a+\frac {b}{x^2}} x^4 \log (f)+\frac {1}{12} b^2 f^{a+\frac {b}{x^2}} x^2 \log ^2(f)-\frac {1}{12} b^3 f^a \text {Ei}\left (\frac {b \log (f)}{x^2}\right ) \log ^3(f)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.70 \[ \frac {1}{12} f^a \left (x^2 f^{\frac {b}{x^2}} \left (b^2 \log ^2(f)+b x^2 \log (f)+2 x^4\right )-b^3 \log ^3(f) \text {Ei}\left (\frac {b \log (f)}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^5,x]

[Out]

(f^a*(-(b^3*ExpIntegralEi[(b*Log[f])/x^2]*Log[f]^3) + f^(b/x^2)*x^2*(2*x^4 + b*x^2*Log[f] + b^2*Log[f]^2)))/12

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fricas [A]  time = 0.42, size = 60, normalized size = 0.74 \[ -\frac {1}{12} \, b^{3} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{2}}\right ) \log \relax (f)^{3} + \frac {1}{12} \, {\left (2 \, x^{6} + b x^{4} \log \relax (f) + b^{2} x^{2} \log \relax (f)^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^5,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^5,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^5, x)

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maple [A]  time = 0.06, size = 79, normalized size = 0.98 \[ \frac {b^{3} f^{a} \Ei \left (1, -\frac {b \ln \relax (f )}{x^{2}}\right ) \ln \relax (f )^{3}}{12}+\frac {b^{2} x^{2} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )^{2}}{12}+\frac {b \,x^{4} f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )}{12}+\frac {x^{6} f^{a} f^{\frac {b}{x^{2}}}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^5,x)

[Out]

1/6*f^a*x^6*f^(b/x^2)+1/12*f^a*ln(f)*b*x^4*f^(b/x^2)+1/12*f^a*ln(f)^2*b^2*x^2*f^(b/x^2)+1/12*f^a*ln(f)^3*b^3*E
i(1,-b/x^2*ln(f))

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maxima [A]  time = 1.48, size = 22, normalized size = 0.27 \[ -\frac {1}{2} \, b^{3} f^{a} \Gamma \left (-3, -\frac {b \log \relax (f)}{x^{2}}\right ) \log \relax (f)^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.74, size = 69, normalized size = 0.85 \[ \frac {b^3\,f^a\,{\ln \relax (f)}^3\,\left (f^{\frac {b}{x^2}}\,\left (\frac {x^2}{6\,b\,\ln \relax (f)}+\frac {x^4}{6\,b^2\,{\ln \relax (f)}^2}+\frac {x^6}{3\,b^3\,{\ln \relax (f)}^3}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^2}\right )}{6}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^5,x)

[Out]

(b^3*f^a*log(f)^3*(f^(b/x^2)*(x^2/(6*b*log(f)) + x^4/(6*b^2*log(f)^2) + x^6/(3*b^3*log(f)^3)) + expint(-(b*log
(f))/x^2)/6))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)*x**5,x)

[Out]

Integral(f**(a + b/x**2)*x**5, x)

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