\(\int \frac {f^{a+b x^2}}{x^{12}} \, dx\) [94]

Optimal result

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

[Out]

-1/2*f^a*(64/10395*Pi^(1/2)*erfc((-b*x^2*ln(f))^(1/2))-64/10395/(-b*x^2*ln(f))^(1/2)*exp(b*x^2*ln(f))+32/10395
/(-b*x^2*ln(f))^(3/2)*exp(b*x^2*ln(f))-16/3465/(-b*x^2*ln(f))^(5/2)*exp(b*x^2*ln(f))+8/693/(-b*x^2*ln(f))^(7/2
)*exp(b*x^2*ln(f))-4/99/(-b*x^2*ln(f))^(9/2)*exp(b*x^2*ln(f))+2/11/(-b*x^2*ln(f))^(11/2)*exp(b*x^2*ln(f)))*(-b
*x^2*ln(f))^(11/2)/x^11

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

[In]

Int[f^(a + b*x^2)/x^12,x]

[Out]

-1/2*(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/x^11

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {f^a \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{11/2}}{2 x^{11}} \]

[In]

Integrate[f^(a + b*x^2)/x^12,x]

[Out]

-1/2*(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/x^11

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.59

method	result	size
meijerg	\(\frac {f^{a} b^{6} \ln \left (f \right )^{\frac {11}{2}} \left (-\frac {2 \left (\frac {32 b^{5} x^{10} \ln \left (f \right )^{5}}{945}+\frac {16 b^{4} x^{8} \ln \left (f \right )^{4}}{945}+\frac {8 b^{3} x^{6} \ln \left (f \right )^{3}}{315}+\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{63}+\frac {2 b \,x^{2} \ln \left (f \right )}{9}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{11 x^{11} \left (-b \right )^{\frac {11}{2}} \ln \left (f \right )^{\frac {11}{2}}}+\frac {64 b^{\frac {11}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{10395 \left (-b \right )^{\frac {11}{2}}}\right )}{2 \sqrt {-b}}\)	\(122\)
risch	\(-\frac {f^{a} f^{b \,x^{2}}}{11 x^{11}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{99 x^{9}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{693 x^{7}}-\frac {8 f^{a} \ln \left (f \right )^{3} b^{3} f^{b \,x^{2}}}{3465 x^{5}}-\frac {16 f^{a} \ln \left (f \right )^{4} b^{4} f^{b \,x^{2}}}{10395 x^{3}}-\frac {32 f^{a} \ln \left (f \right )^{5} b^{5} f^{b \,x^{2}}}{10395 x}+\frac {32 f^{a} \ln \left (f \right )^{6} b^{6} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{10395 \sqrt {-b \ln \left (f \right )}}\)	\(155\)

[In]

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

[Out]

1/2*f^a*b^6*ln(f)^(11/2)/(-b)^(1/2)*(-2/11/x^11/(-b)^(11/2)/ln(f)^(11/2)*(32/945*b^5*x^10*ln(f)^5+16/945*b^4*x
^8*ln(f)^4+8/315*b^3*x^6*ln(f)^3+4/63*b^2*x^4*ln(f)^2+2/9*b*x^2*ln(f)+1)*exp(b*x^2*ln(f))+64/10395/(-b)^(11/2)
*b^(11/2)*Pi^(1/2)*erfi(x*b^(1/2)*ln(f)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.21 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {32 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{5} f^{a} x^{11} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right )^{5} + {\left (32 \, b^{5} x^{10} \log \left (f\right )^{5} + 16 \, b^{4} x^{8} \log \left (f\right )^{4} + 24 \, b^{3} x^{6} \log \left (f\right )^{3} + 60 \, b^{2} x^{4} \log \left (f\right )^{2} + 210 \, b x^{2} \log \left (f\right ) + 945\right )} f^{b x^{2} + a}}{10395 \, x^{11}} \]

[In]

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

[Out]

-1/10395*(32*sqrt(pi)*sqrt(-b*log(f))*b^5*f^a*x^11*erf(sqrt(-b*log(f))*x)*log(f)^5 + (32*b^5*x^10*log(f)^5 + 1
6*b^4*x^8*log(f)^4 + 24*b^3*x^6*log(f)^3 + 60*b^2*x^4*log(f)^2 + 210*b*x^2*log(f) + 945)*f^(b*x^2 + a))/x^11

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**12, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {11}{2}} f^{a} \Gamma \left (-\frac {11}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{11}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^12, x)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 175, normalized size of antiderivative = 5.15 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=\frac {32\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{11/2}}{10395\,x^{11}}-\frac {f^a\,f^{b\,x^2}}{11\,x^{11}}-\frac {32\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \left (f\right )}\right )\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{11/2}}{10395\,x^{11}}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^2}{693\,x^7}-\frac {8\,b^3\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^3}{3465\,x^5}-\frac {16\,b^4\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^4}{10395\,x^3}-\frac {32\,b^5\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^5}{10395\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \left (f\right )}{99\,x^9} \]

[In]

int(f^(a + b*x^2)/x^12,x)

[Out]

(32*f^a*pi^(1/2)*(-b*x^2*log(f))^(11/2))/(10395*x^11) - (f^a*f^(b*x^2))/(11*x^11) - (32*f^a*pi^(1/2)*erfc((-b*
x^2*log(f))^(1/2))*(-b*x^2*log(f))^(11/2))/(10395*x^11) - (4*b^2*f^a*f^(b*x^2)*log(f)^2)/(693*x^7) - (8*b^3*f^
a*f^(b*x^2)*log(f)^3)/(3465*x^5) - (16*b^4*f^a*f^(b*x^2)*log(f)^4)/(10395*x^3) - (32*b^5*f^a*f^(b*x^2)*log(f)^
5)/(10395*x) - (2*b*f^a*f^(b*x^2)*log(f))/(99*x^9)