Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {f^a \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{11/2}}{2 x^{11}} \] Output:
-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*l n(f))^(11/2)*exp(b*x^2*ln(f)))*(-b*x^2*ln(f))^(11/2)/x^11
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {f^a \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{11/2}}{2 x^{11}} \] Input:
Integrate[f^(a + b*x^2)/x^12,x]
Output:
-1/2*(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/x^11
Time = 0.26 (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 = {2648}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {f^{a+b x^2}}{x^{12}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle -\frac {f^a \left (-b x^2 \log (f)\right )^{11/2} \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right )}{2 x^{11}}\) |
Input:
Int[f^(a + b*x^2)/x^12,x]
Output:
-1/2*(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/x^11
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] /; FreeQ[{F , a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.59
method | result | size |
meijerg | \(\frac {f^{a} b^{6} \ln \left (f \right )^{\frac {11}{2}} \left (-\frac {2 \left (\frac {32 b^{5} x^{10} \ln \left (f \right )^{5}}{945}+\frac {16 b^{4} x^{8} \ln \left (f \right )^{4}}{945}+\frac {8 b^{3} x^{6} \ln \left (f \right )^{3}}{315}+\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{63}+\frac {2 b \,x^{2} \ln \left (f \right )}{9}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{11 x^{11} \left (-b \right )^{\frac {11}{2}} \ln \left (f \right )^{\frac {11}{2}}}+\frac {64 b^{\frac {11}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, x \sqrt {\ln \left (f \right )}\right )}{10395 \left (-b \right )^{\frac {11}{2}}}\right )}{2 \sqrt {-b}}\) | \(122\) |
risch | \(-\frac {f^{b \,x^{2}} f^{a}}{11 x^{11}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{99 x^{9}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{693 x^{7}}-\frac {8 f^{a} \ln \left (f \right )^{3} b^{3} f^{b \,x^{2}}}{3465 x^{5}}-\frac {16 f^{a} \ln \left (f \right )^{4} b^{4} f^{b \,x^{2}}}{10395 x^{3}}-\frac {32 f^{a} \ln \left (f \right )^{5} b^{5} f^{b \,x^{2}}}{10395 x}+\frac {32 f^{a} \ln \left (f \right )^{6} b^{6} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{10395 \sqrt {-b \ln \left (f \right )}}\) | \(155\) |
Input:
int(f^(b*x^2+a)/x^12,x,method=_RETURNVERBOSE)
Output:
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(b^(1/2)*x*ln(f)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.21 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {32 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{5} f^{a} x^{11} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right )^{5} + {\left (32 \, b^{5} x^{10} \log \left (f\right )^{5} + 16 \, b^{4} x^{8} \log \left (f\right )^{4} + 24 \, b^{3} x^{6} \log \left (f\right )^{3} + 60 \, b^{2} x^{4} \log \left (f\right )^{2} + 210 \, b x^{2} \log \left (f\right ) + 945\right )} f^{b x^{2} + a}}{10395 \, x^{11}} \] Input:
integrate(f^(b*x^2+a)/x^12,x, algorithm="fricas")
Output:
-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 + 16*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
\[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=\int \frac {f^{a + b x^{2}}}{x^{12}}\, dx \] Input:
integrate(f**(b*x**2+a)/x**12,x)
Output:
Integral(f**(a + b*x**2)/x**12, x)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {11}{2}} f^{a} \Gamma \left (-\frac {11}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{11}} \] Input:
integrate(f^(b*x^2+a)/x^12,x, algorithm="maxima")
Output:
-1/2*(-b*x^2*log(f))^(11/2)*f^a*gamma(-11/2, -b*x^2*log(f))/x^11
\[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{12}} \,d x } \] Input:
integrate(f^(b*x^2+a)/x^12,x, algorithm="giac")
Output:
integrate(f^(b*x^2 + a)/x^12, x)
Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 5.15 \[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=\frac {32\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{11/2}}{10395\,x^{11}}-\frac {f^a\,f^{b\,x^2}}{11\,x^{11}}-\frac {32\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \left (f\right )}\right )\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{11/2}}{10395\,x^{11}}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^2}{693\,x^7}-\frac {8\,b^3\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^3}{3465\,x^5}-\frac {16\,b^4\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^4}{10395\,x^3}-\frac {32\,b^5\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^5}{10395\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \left (f\right )}{99\,x^9} \] Input:
int(f^(a + b*x^2)/x^12,x)
Output:
(32*f^a*pi^(1/2)*(-b*x^2*log(f))^(11/2))/(10395*x^11) - (f^a*f^(b*x^2))/(1 1*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)
\[ \int \frac {f^{a+b x^2}}{x^{12}} \, dx=f^{a} \left (\int \frac {f^{b \,x^{2}}}{x^{12}}d x \right ) \] Input:
int(f^(b*x^2+a)/x^12,x)
Output:
f**a*int(f**(b*x**2)/x**12,x)