\(\int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx\) [152]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\frac {f^a \Gamma \left (\frac {11}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{11} \left (-\frac {b \log (f)}{x^2}\right )^{11/2}} \]

[Out]

1/2*f^a*(1048576/61836869254970658257624840625*GAMMA(51/2,-b*ln(f)/x^2)-1048576/61836869254970658257624840625*
(-b*ln(f)/x^2)^(49/2)*exp(b*ln(f)/x^2)-524288/1261976923570829760359690625*(-b*ln(f)/x^2)^(47/2)*exp(b*ln(f)/x
^2)-262144/26850572841932548092759375*(-b*ln(f)/x^2)^(45/2)*exp(b*ln(f)/x^2)-131072/596679396487389957616875*(
-b*ln(f)/x^2)^(43/2)*exp(b*ln(f)/x^2)-65536/13876265034590464130625*(-b*ln(f)/x^2)^(41/2)*exp(b*ln(f)/x^2)-327
68/338445488648547905625*(-b*ln(f)/x^2)^(39/2)*exp(b*ln(f)/x^2)-16384/8678089452526869375*(-b*ln(f)/x^2)^(37/2
)*exp(b*ln(f)/x^2)-8192/234542958176401875*(-b*ln(f)/x^2)^(35/2)*exp(b*ln(f)/x^2)-4096/6701227376468625*(-b*ln
(f)/x^2)^(33/2)*exp(b*ln(f)/x^2)-2048/203067496256625*(-b*ln(f)/x^2)^(31/2)*exp(b*ln(f)/x^2)-1024/655056439537
5*(-b*ln(f)/x^2)^(29/2)*exp(b*ln(f)/x^2)-512/225881530875*(-b*ln(f)/x^2)^(27/2)*exp(b*ln(f)/x^2)-256/836598262
5*(-b*ln(f)/x^2)^(25/2)*exp(b*ln(f)/x^2)-128/334639305*(-b*ln(f)/x^2)^(23/2)*exp(b*ln(f)/x^2)-64/14549535*(-b*
ln(f)/x^2)^(21/2)*exp(b*ln(f)/x^2)-32/692835*(-b*ln(f)/x^2)^(19/2)*exp(b*ln(f)/x^2)-16/36465*(-b*ln(f)/x^2)^(1
7/2)*exp(b*ln(f)/x^2)-8/2145*(-b*ln(f)/x^2)^(15/2)*exp(b*ln(f)/x^2)-4/143*(-b*ln(f)/x^2)^(13/2)*exp(b*ln(f)/x^
2)-2/11*(-b*ln(f)/x^2)^(11/2)*exp(b*ln(f)/x^2))/x^11/(-b*ln(f)/x^2)^(11/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\frac {f^a \Gamma \left (\frac {11}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{11} \left (-\frac {b \log (f)}{x^2}\right )^{11/2}} \]

[In]

Int[f^(a + b/x^2)/x^12,x]

[Out]

(f^a*Gamma[11/2, -((b*Log[f])/x^2)])/(2*x^11*(-((b*Log[f])/x^2))^(11/2))

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 {f^a \Gamma \left (\frac {11}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{11} \left (-\frac {b \log (f)}{x^2}\right )^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\frac {f^a \Gamma \left (\frac {11}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{11} \left (-\frac {b \log (f)}{x^2}\right )^{11/2}} \]

[In]

Integrate[f^(a + b/x^2)/x^12,x]

[Out]

(f^a*Gamma[11/2, -((b*Log[f])/x^2)])/(2*x^11*(-((b*Log[f])/x^2))^(11/2))

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.38

method	result	size
meijerg	\(-\frac {f^{a} \sqrt {-b}\, \left (\frac {\left (-b \right )^{\frac {11}{2}} \sqrt {\ln \left (f \right )}\, \left (\frac {176 b^{4} \ln \left (f \right )^{4}}{x^{8}}-\frac {792 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {2772 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {6930 b \ln \left (f \right )}{x^{2}}+10395\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{176 x \,b^{5}}-\frac {945 \left (-b \right )^{\frac {11}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{32 b^{\frac {11}{2}}}\right )}{2 b^{6} \ln \left (f \right )^{\frac {11}{2}}}\)	\(115\)
risch	\(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{9} b \ln \left (f \right )}+\frac {9 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{7}}-\frac {63 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x^{5}}+\frac {315 f^{a} f^{\frac {b}{x^{2}}}}{16 \ln \left (f \right )^{4} b^{4} x^{3}}-\frac {945 f^{a} f^{\frac {b}{x^{2}}}}{32 \ln \left (f \right )^{5} b^{5} x}+\frac {945 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{64 \ln \left (f \right )^{5} b^{5} \sqrt {-b \ln \left (f \right )}}\)	\(146\)

[In]

int(f^(a+b/x^2)/x^12,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a/b^6/ln(f)^(11/2)*(-b)^(1/2)*(1/176/x*(-b)^(11/2)*ln(f)^(1/2)*(176*b^4*ln(f)^4/x^8-792*b^3*ln(f)^3/x^6
+2772*b^2*ln(f)^2/x^4-6930*b*ln(f)/x^2+10395)/b^5*exp(b*ln(f)/x^2)-945/32*(-b)^(11/2)/b^(11/2)*Pi^(1/2)*erfi(b
^(1/2)*ln(f)^(1/2)/x))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.29 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=-\frac {945 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{9} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (945 \, b x^{8} \log \left (f\right ) - 630 \, b^{2} x^{6} \log \left (f\right )^{2} + 252 \, b^{3} x^{4} \log \left (f\right )^{3} - 72 \, b^{4} x^{2} \log \left (f\right )^{4} + 16 \, b^{5} \log \left (f\right )^{5}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{64 \, b^{6} x^{9} \log \left (f\right )^{6}} \]

[In]

integrate(f^(a+b/x^2)/x^12,x, algorithm="fricas")

[Out]

-1/64*(945*sqrt(pi)*sqrt(-b*log(f))*f^a*x^9*erf(sqrt(-b*log(f))/x) + 2*(945*b*x^8*log(f) - 630*b^2*x^6*log(f)^
2 + 252*b^3*x^4*log(f)^3 - 72*b^4*x^2*log(f)^4 + 16*b^5*log(f)^5)*f^((a*x^2 + b)/x^2))/(b^6*x^9*log(f)^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\text {Timed out} \]

[In]

integrate(f**(a+b/x**2)/x**12,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\frac {f^{a} \Gamma \left (\frac {11}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{11} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {11}{2}}} \]

[In]

integrate(f^(a+b/x^2)/x^12,x, algorithm="maxima")

[Out]

1/2*f^a*gamma(11/2, -b*log(f)/x^2)/(x^11*(-b*log(f)/x^2)^(11/2))

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{12}} \,d x } \]

[In]

integrate(f^(a+b/x^2)/x^12,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^12, x)

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 142, normalized size of antiderivative = 4.18 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{12}} \, dx=\frac {\frac {f^a\,\left (945\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )-\frac {1890\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{x}\right )}{64\,\sqrt {b\,\ln \left (f\right )}}-\frac {63\,b^2\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^2}{8\,x^5}+\frac {9\,b^3\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^3}{4\,x^7}-\frac {b^4\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^4}{2\,x^9}+\frac {315\,b\,f^a\,f^{\frac {b}{x^2}}\,\ln \left (f\right )}{16\,x^3}}{b^5\,{\ln \left (f\right )}^5} \]

[In]

int(f^(a + b/x^2)/x^12,x)

[Out]

((f^a*(945*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))) - (1890*f^(b/x^2)*(b*log(f))^(1/2))/x))/(64*(b*log(f
))^(1/2)) - (63*b^2*f^a*f^(b/x^2)*log(f)^2)/(8*x^5) + (9*b^3*f^a*f^(b/x^2)*log(f)^3)/(4*x^7) - (b^4*f^a*f^(b/x
^2)*log(f)^4)/(2*x^9) + (315*b*f^a*f^(b/x^2)*log(f))/(16*x^3))/(b^5*log(f)^5)