\(\int f^{a+\frac {b}{x^2}} x^7 \, dx\) [130]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\frac {1}{2} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^2}\right ) \log ^4(f) \]

[Out]

1/2*f^a*x^8*Ei(5,-b*ln(f)/x^2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2250} \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\frac {1}{2} b^4 f^a \log ^4(f) \Gamma \left (-4,-\frac {b \log (f)}{x^2}\right ) \]

[In]

Int[f^(a + b/x^2)*x^7,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^2)]*Log[f]^4)/2

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^2}\right ) \log ^4(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\frac {1}{2} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^2}\right ) \log ^4(f) \]

[In]

Integrate[f^(a + b/x^2)*x^7,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^2)]*Log[f]^4)/2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(100\) vs. \(2(18)=36\).

Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 4.21

method	result	size
risch	\(\frac {f^{a} x^{8} f^{\frac {b}{x^{2}}}}{8}+\frac {f^{a} \ln \left (f \right ) b \,x^{6} f^{\frac {b}{x^{2}}}}{24}+\frac {f^{a} \ln \left (f \right )^{2} b^{2} x^{4} f^{\frac {b}{x^{2}}}}{48}+\frac {f^{a} \ln \left (f \right )^{3} b^{3} x^{2} f^{\frac {b}{x^{2}}}}{48}+\frac {f^{a} \ln \left (f \right )^{4} b^{4} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{48}\)	\(101\)
meijerg	\(-\frac {f^{a} b^{4} \ln \left (f \right )^{4} \left (-\frac {x^{8}}{4 b^{4} \ln \left (f \right )^{4}}-\frac {x^{6}}{3 b^{3} \ln \left (f \right )^{3}}-\frac {x^{4}}{4 b^{2} \ln \left (f \right )^{2}}-\frac {x^{2}}{6 b \ln \left (f \right )}-\frac {25}{288}-\frac {\ln \left (x \right )}{12}+\frac {\ln \left (-b \right )}{24}+\frac {\ln \left (\ln \left (f \right )\right )}{24}+\frac {x^{8} \left (\frac {125 b^{4} \ln \left (f \right )^{4}}{x^{8}}+\frac {240 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {360 b^{2} \ln \left (f \right )^{2}}{x^{4}}+\frac {480 b \ln \left (f \right )}{x^{2}}+360\right )}{1440 b^{4} \ln \left (f \right )^{4}}-\frac {x^{8} \left (\frac {5 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {5 b^{2} \ln \left (f \right )^{2}}{x^{4}}+\frac {10 b \ln \left (f \right )}{x^{2}}+30\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{120 b^{4} \ln \left (f \right )^{4}}-\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{24}-\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{24}\right )}{2}\)	\(213\)

[In]

int(f^(a+b/x^2)*x^7,x,method=_RETURNVERBOSE)

[Out]

1/8*f^a*x^8*f^(b/x^2)+1/24*f^a*ln(f)*b*x^6*f^(b/x^2)+1/48*f^a*ln(f)^2*b^2*x^4*f^(b/x^2)+1/48*f^a*ln(f)^3*b^3*x
^2*f^(b/x^2)+1/48*f^a*ln(f)^4*b^4*Ei(1,-b*ln(f)/x^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (18) = 36\).

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=-\frac {1}{48} \, b^{4} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{4} + \frac {1}{48} \, {\left (6 \, x^{8} + 2 \, b x^{6} \log \left (f\right ) + b^{2} x^{4} \log \left (f\right )^{2} + b^{3} x^{2} \log \left (f\right )^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

[In]

integrate(f^(a+b/x^2)*x^7,x, algorithm="fricas")

[Out]

-1/48*b^4*f^a*Ei(b*log(f)/x^2)*log(f)^4 + 1/48*(6*x^8 + 2*b*x^6*log(f) + b^2*x^4*log(f)^2 + b^3*x^2*log(f)^3)*
f^((a*x^2 + b)/x^2)

Sympy [F]

\[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{7}\, dx \]

[In]

integrate(f**(a+b/x**2)*x**7,x)

[Out]

Integral(f**(a + b/x**2)*x**7, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\frac {1}{2} \, b^{4} f^{a} \Gamma \left (-4, -\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right )^{4} \]

[In]

integrate(f^(a+b/x^2)*x^7,x, algorithm="maxima")

[Out]

1/2*b^4*f^a*gamma(-4, -b*log(f)/x^2)*log(f)^4

Giac [F]

\[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{7} \,d x } \]

[In]

integrate(f^(a+b/x^2)*x^7,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^7, x)

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75 \[ \int f^{a+\frac {b}{x^2}} x^7 \, dx=\frac {b^4\,f^a\,{\ln \left (f\right )}^4\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}{48}+\frac {b^4\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^4\,\left (\frac {x^2}{24\,b\,\ln \left (f\right )}+\frac {x^4}{24\,b^2\,{\ln \left (f\right )}^2}+\frac {x^6}{12\,b^3\,{\ln \left (f\right )}^3}+\frac {x^8}{4\,b^4\,{\ln \left (f\right )}^4}\right )}{2} \]

[In]

int(f^(a + b/x^2)*x^7,x)

[Out]

(b^4*f^a*log(f)^4*expint(-(b*log(f))/x^2))/48 + (b^4*f^a*f^(b/x^2)*log(f)^4*(x^2/(24*b*log(f)) + x^4/(24*b^2*l
og(f)^2) + x^6/(12*b^3*log(f)^3) + x^8/(4*b^4*log(f)^4)))/2