Integrand size = 13, antiderivative size = 34 \[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {1}{2} f^a x^{11} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \] Output:
1/2*f^a*x^11*(64/10395*Pi^(1/2)*erfc((-b*ln(f)/x^2)^(1/2))-64/10395/(-b*ln (f)/x^2)^(1/2)*exp(b*ln(f)/x^2)+32/10395/(-b*ln(f)/x^2)^(3/2)*exp(b*ln(f)/ x^2)-16/3465/(-b*ln(f)/x^2)^(5/2)*exp(b*ln(f)/x^2)+8/693/(-b*ln(f)/x^2)^(7 /2)*exp(b*ln(f)/x^2)-4/99/(-b*ln(f)/x^2)^(9/2)*exp(b*ln(f)/x^2)+2/11/(-b*l n(f)/x^2)^(11/2)*exp(b*ln(f)/x^2))*(-b*ln(f)/x^2)^(11/2)
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {1}{2} f^a x^{11} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right ) \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \] Input:
Integrate[f^(a + b/x^2)*x^10,x]
Output:
(f^a*x^11*Gamma[-11/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(11/2))/2
Time = 0.27 (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.
\(\displaystyle \int x^{10} f^{a+\frac {b}{x^2}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \frac {1}{2} x^{11} f^a \left (-\frac {b \log (f)}{x^2}\right )^{11/2} \Gamma \left (-\frac {11}{2},-\frac {b \log (f)}{x^2}\right )\) |
Input:
Int[f^(a + b/x^2)*x^10,x]
Output:
(f^a*x^11*Gamma[-11/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^(11/2))/2
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 = 0.38 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.65
method | result | size |
meijerg | \(\frac {f^{a} b^{5} \ln \left (f \right )^{\frac {11}{2}} \sqrt {-b}\, \left (-\frac {2 x^{11} \left (\frac {32 b^{5} \ln \left (f \right )^{5}}{945 x^{10}}+\frac {16 b^{4} \ln \left (f \right )^{4}}{945 x^{8}}+\frac {8 b^{3} \ln \left (f \right )^{3}}{315 x^{6}}+\frac {4 b^{2} \ln \left (f \right )^{2}}{63 x^{4}}+\frac {2 b \ln \left (f \right )}{9 x^{2}}+1\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{11 \left (-b \right )^{\frac {11}{2}} \ln \left (f \right )^{\frac {11}{2}}}+\frac {64 b^{\frac {11}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{10395 \left (-b \right )^{\frac {11}{2}}}\right )}{2}\) | \(124\) |
risch | \(\frac {f^{a} x^{11} f^{\frac {b}{x^{2}}}}{11}+\frac {2 f^{a} \ln \left (f \right ) b \,x^{9} f^{\frac {b}{x^{2}}}}{99}+\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} x^{7} f^{\frac {b}{x^{2}}}}{693}+\frac {8 f^{a} \ln \left (f \right )^{3} b^{3} x^{5} f^{\frac {b}{x^{2}}}}{3465}+\frac {16 f^{a} \ln \left (f \right )^{4} b^{4} x^{3} f^{\frac {b}{x^{2}}}}{10395}+\frac {32 f^{a} \ln \left (f \right )^{5} b^{5} x \,f^{\frac {b}{x^{2}}}}{10395}-\frac {32 f^{a} \ln \left (f \right )^{6} b^{6} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{10395 \sqrt {-b \ln \left (f \right )}}\) | \(155\) |
Input:
int(f^(a+b/x^2)*x^10,x,method=_RETURNVERBOSE)
Output:
1/2*f^a*b^5*ln(f)^(11/2)*(-b)^(1/2)*(-2/11*x^11/(-b)^(11/2)/ln(f)^(11/2)*( 32/945*b^5*ln(f)^5/x^10+16/945*b^4*ln(f)^4/x^8+8/315*b^3*ln(f)^3/x^6+4/63* b^2*ln(f)^2/x^4+2/9*b*ln(f)/x^2+1)*exp(b*ln(f)/x^2)+64/10395/(-b)^(11/2)*b ^(11/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.24 \[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {32}{10395} \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{5} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{5} + \frac {1}{10395} \, {\left (945 \, x^{11} + 210 \, b x^{9} \log \left (f\right ) + 60 \, b^{2} x^{7} \log \left (f\right )^{2} + 24 \, b^{3} x^{5} \log \left (f\right )^{3} + 16 \, b^{4} x^{3} \log \left (f\right )^{4} + 32 \, b^{5} x \log \left (f\right )^{5}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \] Input:
integrate(f^(a+b/x^2)*x^10,x, algorithm="fricas")
Output:
32/10395*sqrt(pi)*sqrt(-b*log(f))*b^5*f^a*erf(sqrt(-b*log(f))/x)*log(f)^5 + 1/10395*(945*x^11 + 210*b*x^9*log(f) + 60*b^2*x^7*log(f)^2 + 24*b^3*x^5* log(f)^3 + 16*b^4*x^3*log(f)^4 + 32*b^5*x*log(f)^5)*f^((a*x^2 + b)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\int f^{a + \frac {b}{x^{2}}} x^{10}\, dx \] Input:
integrate(f**(a+b/x**2)*x**10,x)
Output:
Integral(f**(a + b/x**2)*x**10, x)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {1}{2} \, f^{a} x^{11} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {11}{2}} \Gamma \left (-\frac {11}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \] Input:
integrate(f^(a+b/x^2)*x^10,x, algorithm="maxima")
Output:
1/2*f^a*x^11*(-b*log(f)/x^2)^(11/2)*gamma(-11/2, -b*log(f)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{10} \,d x } \] Input:
integrate(f^(a+b/x^2)*x^10,x, algorithm="giac")
Output:
integrate(f^(a + b/x^2)*x^10, x)
Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 5.09 \[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {f^a\,f^{\frac {b}{x^2}}\,x^{11}}{11}-\frac {32\,f^a\,x^{11}\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{11/2}}{10395}+\frac {32\,f^a\,x^{11}\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (f\right )}{x^2}}\right )\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{11/2}}{10395}+\frac {32\,b^5\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \left (f\right )}^5}{10395}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x^7\,{\ln \left (f\right )}^2}{693}+\frac {8\,b^3\,f^a\,f^{\frac {b}{x^2}}\,x^5\,{\ln \left (f\right )}^3}{3465}+\frac {16\,b^4\,f^a\,f^{\frac {b}{x^2}}\,x^3\,{\ln \left (f\right )}^4}{10395}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^9\,\ln \left (f\right )}{99} \] Input:
int(f^(a + b/x^2)*x^10,x)
Output:
(f^a*f^(b/x^2)*x^11)/11 - (32*f^a*x^11*pi^(1/2)*(-(b*log(f))/x^2)^(11/2))/ 10395 + (32*f^a*x^11*pi^(1/2)*erfc((-(b*log(f))/x^2)^(1/2))*(-(b*log(f))/x ^2)^(11/2))/10395 + (32*b^5*f^a*f^(b/x^2)*x*log(f)^5)/10395 + (4*b^2*f^a*f ^(b/x^2)*x^7*log(f)^2)/693 + (8*b^3*f^a*f^(b/x^2)*x^5*log(f)^3)/3465 + (16 *b^4*f^a*f^(b/x^2)*x^3*log(f)^4)/10395 + (2*b*f^a*f^(b/x^2)*x^9*log(f))/99
\[ \int f^{a+\frac {b}{x^2}} x^{10} \, dx=\frac {-64 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{6} b^{6}+32 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{5} b^{5} x^{2}+16 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{4} b^{4} x^{4}+24 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{3} b^{3} x^{6}+60 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{2} b^{2} x^{8}+210 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right ) b \,x^{10}+945 f^{\frac {a \,x^{2}+b}{x^{2}}} x^{12}-128 \left (\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{4}}d x \right ) \mathrm {log}\left (f \right )^{7} b^{7} x}{10395 x} \] Input:
int(f^(a+b/x^2)*x^10,x)
Output:
( - 64*f**((a*x**2 + b)/x**2)*log(f)**6*b**6 + 32*f**((a*x**2 + b)/x**2)*l og(f)**5*b**5*x**2 + 16*f**((a*x**2 + b)/x**2)*log(f)**4*b**4*x**4 + 24*f* *((a*x**2 + b)/x**2)*log(f)**3*b**3*x**6 + 60*f**((a*x**2 + b)/x**2)*log(f )**2*b**2*x**8 + 210*f**((a*x**2 + b)/x**2)*log(f)*b*x**10 + 945*f**((a*x* *2 + b)/x**2)*x**12 - 128*int(f**((a*x**2 + b)/x**2)/x**4,x)*log(f)**7*b** 7*x)/(10395*x)