\(\int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx\) [139]

Optimal result

Integrand size = 13, antiderivative size = 69 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {f^{a+\frac {b}{x^2}} \left (24 x^8-24 b x^6 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^2 \log ^3(f)+b^4 \log ^4(f)\right )}{2 b^5 x^8 \log ^5(f)} \]

[Out]

-1/2*f^(a+b/x^2)*(24*x^8-24*b*x^6*ln(f)+12*b^2*x^4*ln(f)^2-4*b^3*x^2*ln(f)^3+b^4*ln(f)^4)/b^5/x^8/ln(f)^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, 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 = {2249} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {f^{a+\frac {b}{x^2}} \left (b^4 \log ^4(f)-4 b^3 x^2 \log ^3(f)+12 b^2 x^4 \log ^2(f)-24 b x^6 \log (f)+24 x^8\right )}{2 b^5 x^8 \log ^5(f)} \]

[In]

Int[f^(a + b/x^2)/x^11,x]

[Out]

-1/2*(f^(a + b/x^2)*(24*x^8 - 24*b*x^6*Log[f] + 12*b^2*x^4*Log[f]^2 - 4*b^3*x^2*Log[f]^3 + b^4*Log[f]^4))/(b^5
*x^8*Log[f]^5)

Rule 2249

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{p = Simplify
[(m + 1)/n]}, Simp[(-F^a)*((f/d)^m/(d*n*((-b)*Log[F])^p))*Simplify[FunctionExpand[Gamma[p, (-b)*(c + d*x)^n*Lo
g[F]]]], x] /; IGtQ[p, 0]] /; FreeQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = -\frac {f^{a+\frac {b}{x^2}} \left (24 x^8-24 b x^6 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^2 \log ^3(f)+b^4 \log ^4(f)\right )}{2 b^5 x^8 \log ^5(f)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.35 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {f^a \Gamma \left (5,-\frac {b \log (f)}{x^2}\right )}{2 b^5 \log ^5(f)} \]

[In]

Integrate[f^(a + b/x^2)/x^11,x]

[Out]

-1/2*(f^a*Gamma[5, -((b*Log[f])/x^2)])/(b^5*Log[f]^5)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03

method	result	size
meijerg	\(\frac {f^{a} \left (24-\frac {\left (\frac {5 b^{4} \ln \left (f \right )^{4}}{x^{8}}-\frac {20 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {60 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {120 b \ln \left (f \right )}{x^{2}}+120\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{5}\right )}{2 b^{5} \ln \left (f \right )^{5}}\)	\(71\)
risch	\(-\frac {\left (24 x^{8}-24 b \,x^{6} \ln \left (f \right )+12 b^{2} x^{4} \ln \left (f \right )^{2}-4 b^{3} x^{2} \ln \left (f \right )^{3}+b^{4} \ln \left (f \right )^{4}\right ) f^{\frac {a \,x^{2}+b}{x^{2}}}}{2 b^{5} \ln \left (f \right )^{5} x^{8}}\)	\(72\)
parallelrisch	\(\frac {-f^{a +\frac {b}{x^{2}}} b^{4} \ln \left (f \right )^{4}+4 f^{a +\frac {b}{x^{2}}} x^{2} b^{3} \ln \left (f \right )^{3}-12 f^{a +\frac {b}{x^{2}}} x^{4} b^{2} \ln \left (f \right )^{2}+24 f^{a +\frac {b}{x^{2}}} x^{6} b \ln \left (f \right )-24 f^{a +\frac {b}{x^{2}}} x^{8}}{2 x^{8} b^{5} \ln \left (f \right )^{5}}\)	\(105\)
norman	\(\frac {-\frac {x^{2} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}+\frac {2 x^{4} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{2} \ln \left (f \right )^{2}}-\frac {6 x^{6} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{3} \ln \left (f \right )^{3}}+\frac {12 x^{8} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{4} \ln \left (f \right )^{4}}-\frac {12 x^{10} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{5} \ln \left (f \right )^{5}}}{x^{10}}\)	\(121\)

[In]

int(f^(a+b/x^2)/x^11,x,method=_RETURNVERBOSE)

[Out]

1/2*f^a/b^5/ln(f)^5*(24-1/5*(5*b^4*ln(f)^4/x^8-20*b^3*ln(f)^3/x^6+60*b^2*ln(f)^2/x^4-120*b*ln(f)/x^2+120)*exp(
b*ln(f)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {{\left (24 \, x^{8} - 24 \, b x^{6} \log \left (f\right ) + 12 \, b^{2} x^{4} \log \left (f\right )^{2} - 4 \, b^{3} x^{2} \log \left (f\right )^{3} + b^{4} \log \left (f\right )^{4}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{2 \, b^{5} x^{8} \log \left (f\right )^{5}} \]

[In]

integrate(f^(a+b/x^2)/x^11,x, algorithm="fricas")

[Out]

-1/2*(24*x^8 - 24*b*x^6*log(f) + 12*b^2*x^4*log(f)^2 - 4*b^3*x^2*log(f)^3 + b^4*log(f)^4)*f^((a*x^2 + b)/x^2)/
(b^5*x^8*log(f)^5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=\frac {f^{a + \frac {b}{x^{2}}} \left (- b^{4} \log {\left (f \right )}^{4} + 4 b^{3} x^{2} \log {\left (f \right )}^{3} - 12 b^{2} x^{4} \log {\left (f \right )}^{2} + 24 b x^{6} \log {\left (f \right )} - 24 x^{8}\right )}{2 b^{5} x^{8} \log {\left (f \right )}^{5}} \]

[In]

integrate(f**(a+b/x**2)/x**11,x)

[Out]

f**(a + b/x**2)*(-b**4*log(f)**4 + 4*b**3*x**2*log(f)**3 - 12*b**2*x**4*log(f)**2 + 24*b*x**6*log(f) - 24*x**8
)/(2*b**5*x**8*log(f)**5)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.32 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {f^{a} \Gamma \left (5, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(a+b/x^2)/x^11,x, algorithm="maxima")

[Out]

-1/2*f^a*gamma(5, -b*log(f)/x^2)/(b^5*log(f)^5)

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{11}} \,d x } \]

[In]

integrate(f^(a+b/x^2)/x^11,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^11, x)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{11}} \, dx=-\frac {f^{a+\frac {b}{x^2}}\,\left (\frac {1}{2\,b\,\ln \left (f\right )}-\frac {2\,x^2}{b^2\,{\ln \left (f\right )}^2}+\frac {6\,x^4}{b^3\,{\ln \left (f\right )}^3}-\frac {12\,x^6}{b^4\,{\ln \left (f\right )}^4}+\frac {12\,x^8}{b^5\,{\ln \left (f\right )}^5}\right )}{x^8} \]

[In]

int(f^(a + b/x^2)/x^11,x)

[Out]

-(f^(a + b/x^2)*(1/(2*b*log(f)) - (2*x^2)/(b^2*log(f)^2) + (6*x^4)/(b^3*log(f)^3) - (12*x^6)/(b^4*log(f)^4) +
(12*x^8)/(b^5*log(f)^5)))/x^8