\(\int f^{a+\frac {b}{x^2}} x^5 \, dx\) [131]

Optimal result

Integrand size = 13, antiderivative size = 81 \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=\frac {1}{6} f^{a+\frac {b}{x^2}} x^6+\frac {1}{12} b f^{a+\frac {b}{x^2}} x^4 \log (f)+\frac {1}{12} b^2 f^{a+\frac {b}{x^2}} x^2 \log ^2(f)-\frac {1}{12} b^3 f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log ^3(f) \]

[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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=-\frac {1}{12} b^3 f^a \log ^3(f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right )+\frac {1}{12} b^2 x^2 \log ^2(f) f^{a+\frac {b}{x^2}}+\frac {1}{6} x^6 f^{a+\frac {b}{x^2}}+\frac {1}{12} b x^4 \log (f) f^{a+\frac {b}{x^2}} \]

[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 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^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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=\frac {1}{12} f^a \left (-b^3 \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log ^3(f)+f^{\frac {b}{x^2}} x^2 \left (2 x^4+b x^2 \log (f)+b^2 \log ^2(f)\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {f^{a} x^{6} f^{\frac {b}{x^{2}}}}{6}+\frac {f^{a} \ln \left (f \right ) b \,x^{4} f^{\frac {b}{x^{2}}}}{12}+\frac {f^{a} \ln \left (f \right )^{2} b^{2} x^{2} f^{\frac {b}{x^{2}}}}{12}+\frac {f^{a} \ln \left (f \right )^{3} b^{3} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{12}\)	\(79\)
meijerg	\(\frac {f^{a} b^{3} \ln \left (f \right )^{3} \left (\frac {x^{6}}{3 b^{3} \ln \left (f \right )^{3}}+\frac {x^{4}}{2 b^{2} \ln \left (f \right )^{2}}+\frac {x^{2}}{2 b \ln \left (f \right )}+\frac {11}{36}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (-b \right )}{6}-\frac {\ln \left (\ln \left (f \right )\right )}{6}-\frac {x^{6} \left (\frac {22 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {36 b^{2} \ln \left (f \right )^{2}}{x^{4}}+\frac {36 b \ln \left (f \right )}{x^{2}}+24\right )}{72 b^{3} \ln \left (f \right )^{3}}+\frac {x^{6} \left (\frac {4 b^{2} \ln \left (f \right )^{2}}{x^{4}}+\frac {4 b \ln \left (f \right )}{x^{2}}+8\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{24 b^{3} \ln \left (f \right )^{3}}+\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{6}+\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{6}\right )}{2}\)	\(177\)

[In]

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

[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*ln(f)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=-\frac {1}{12} \, b^{3} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{3} + \frac {1}{12} \, {\left (2 \, x^{6} + b x^{4} \log \left (f\right ) + b^{2} x^{2} \log \left (f\right )^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

[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)

Sympy [F]

\[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{5}\, dx \]

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.27 \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=-\frac {1}{2} \, b^{3} f^{a} \Gamma \left (-3, -\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{3} \]

[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

Giac [F]

\[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{5} \,d x } \]

[In]

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

[Out]

integrate(f^(a + b/x^2)*x^5, x)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int f^{a+\frac {b}{x^2}} x^5 \, dx=\frac {b^3\,f^a\,{\ln \left (f\right )}^3\,\left (f^{\frac {b}{x^2}}\,\left (\frac {x^2}{6\,b\,\ln \left (f\right )}+\frac {x^4}{6\,b^2\,{\ln \left (f\right )}^2}+\frac {x^6}{3\,b^3\,{\ln \left (f\right )}^3}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}{6}\right )}{2} \]

[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