\(\int f^{a+\frac {b}{x^3}} x^5 \, dx\) [158]

Optimal result

Integrand size = 13, antiderivative size = 58 \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\frac {1}{6} f^{a+\frac {b}{x^3}} x^6+\frac {1}{6} b f^{a+\frac {b}{x^3}} x^3 \log (f)-\frac {1}{6} b^2 f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right ) \log ^2(f) \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=-\frac {1}{6} b^2 f^a \log ^2(f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right )+\frac {1}{6} b x^3 \log (f) f^{a+\frac {b}{x^3}}+\frac {1}{6} x^6 f^{a+\frac {b}{x^3}} \]

[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 2241

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 2245

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\frac {1}{6} f^a \left (-b^2 \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right ) \log ^2(f)+f^{\frac {b}{x^3}} x^3 \left (x^3+b \log (f)\right )\right ) \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(140\) vs. \(2(52)=104\).

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.43

method	result	size
meijerg	\(-\frac {f^{a} b^{2} \ln \left (f \right )^{2} \left (-\frac {x^{6}}{2 b^{2} \ln \left (f \right )^{2}}-\frac {x^{3}}{b \ln \left (f \right )}-\frac {3}{4}-\frac {3 \ln \left (x \right )}{2}+\frac {\ln \left (-b \right )}{2}+\frac {\ln \left (\ln \left (f \right )\right )}{2}+\frac {x^{6} \left (\frac {9 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {12 b \ln \left (f \right )}{x^{3}}+6\right )}{12 b^{2} \ln \left (f \right )^{2}}-\frac {x^{6} \left (3+\frac {3 b \ln \left (f \right )}{x^{3}}\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{6 b^{2} \ln \left (f \right )^{2}}-\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{2}\right )}{3}\)	\(141\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81 \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=-\frac {1}{6} \, b^{2} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{2} + \frac {1}{6} \, {\left (x^{6} + b x^{3} \log \left (f\right )\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

[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)

Sympy [F]

\[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\int f^{a + \frac {b}{x^{3}}} x^{5}\, dx \]

[In]

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

[Out]

Integral(f**(a + b/x**3)*x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.38 \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\frac {1}{3} \, b^{2} f^{a} \Gamma \left (-2, -\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{2} \]

[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

Giac [F]

\[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\int { f^{a + \frac {b}{x^{3}}} x^{5} \,d x } \]

[In]

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

[Out]

integrate(f^(a + b/x^3)*x^5, x)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int f^{a+\frac {b}{x^3}} x^5 \, dx=\frac {b^2\,f^a\,{\ln \left (f\right )}^2\,\left (f^{\frac {b}{x^3}}\,\left (\frac {x^3}{2\,b\,\ln \left (f\right )}+\frac {x^6}{2\,b^2\,{\ln \left (f\right )}^2}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^3}\right )}{2}\right )}{3} \]

[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