\(\int f^{a+\frac {b}{x^2}} x^3 \, dx\) [132]

Optimal result

Integrand size = 13, antiderivative size = 58 \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\frac {1}{4} f^{a+\frac {b}{x^2}} x^4+\frac {1}{4} b f^{a+\frac {b}{x^2}} x^2 \log (f)-\frac {1}{4} b^2 f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log ^2(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (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} \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=-\frac {1}{4} b^2 f^a \log ^2(f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right )+\frac {1}{4} b x^2 \log (f) f^{a+\frac {b}{x^2}}+\frac {1}{4} x^4 f^{a+\frac {b}{x^2}} \]

[In]

Int[f^(a + b/x^2)*x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\frac {1}{4} f^a \left (-b^2 \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log ^2(f)+f^{\frac {b}{x^2}} x^2 \left (x^2+b \log (f)\right )\right ) \]

[In]

Integrate[f^(a + b/x^2)*x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {f^{a} x^{4} f^{\frac {b}{x^{2}}}}{4}+\frac {f^{a} \ln \left (f \right ) b \,x^{2} f^{\frac {b}{x^{2}}}}{4}+\frac {f^{a} \ln \left (f \right )^{2} b^{2} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{4}\)	\(57\)
meijerg	\(-\frac {f^{a} b^{2} \ln \left (f \right )^{2} \left (-\frac {x^{4}}{2 b^{2} \ln \left (f \right )^{2}}-\frac {x^{2}}{b \ln \left (f \right )}-\frac {3}{4}-\ln \left (x \right )+\frac {\ln \left (-b \right )}{2}+\frac {\ln \left (\ln \left (f \right )\right )}{2}+\frac {x^{4} \left (\frac {9 b^{2} \ln \left (f \right )^{2}}{x^{4}}+\frac {12 b \ln \left (f \right )}{x^{2}}+6\right )}{12 b^{2} \ln \left (f \right )^{2}}-\frac {x^{4} \left (3+\frac {3 b \ln \left (f \right )}{x^{2}}\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{6 b^{2} \ln \left (f \right )^{2}}-\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{2}\right )}{2}\)	\(141\)

[In]

int(f^(a+b/x^2)*x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81 \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=-\frac {1}{4} \, b^{2} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{2} + \frac {1}{4} \, {\left (x^{4} + b x^{2} \log \left (f\right )\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

[In]

integrate(f^(a+b/x^2)*x^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{3}\, dx \]

[In]

integrate(f**(a+b/x**2)*x**3,x)

[Out]

Integral(f**(a + b/x**2)*x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.38 \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\frac {1}{2} \, b^{2} f^{a} \Gamma \left (-2, -\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{2} \]

[In]

integrate(f^(a+b/x^2)*x^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{3} \,d x } \]

[In]

integrate(f^(a+b/x^2)*x^3,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^3, x)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int f^{a+\frac {b}{x^2}} x^3 \, dx=\frac {b^2\,f^a\,{\ln \left (f\right )}^2\,\left (f^{\frac {b}{x^2}}\,\left (\frac {x^2}{2\,b\,\ln \left (f\right )}+\frac {x^4}{2\,b^2\,{\ln \left (f\right )}^2}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}{2}\right )}{2} \]

[In]

int(f^(a + b/x^2)*x^3,x)

[Out]

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