Integrand size = 13, antiderivative size = 34 \[ \int f^{a+b x^2} x^{10} \, dx=-\frac {f^a x^{11} \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}} \]
-1/2*f^a*x^11*(1048576/61836869254970658257624840625*GAMMA(51/2,-b*x^2*ln( f))-1048576/61836869254970658257624840625*(-b*x^2*ln(f))^(49/2)*exp(b*x^2* ln(f))-524288/1261976923570829760359690625*(-b*x^2*ln(f))^(47/2)*exp(b*x^2 *ln(f))-262144/26850572841932548092759375*(-b*x^2*ln(f))^(45/2)*exp(b*x^2* ln(f))-131072/596679396487389957616875*(-b*x^2*ln(f))^(43/2)*exp(b*x^2*ln( f))-65536/13876265034590464130625*(-b*x^2*ln(f))^(41/2)*exp(b*x^2*ln(f))-3 2768/338445488648547905625*(-b*x^2*ln(f))^(39/2)*exp(b*x^2*ln(f))-16384/86 78089452526869375*(-b*x^2*ln(f))^(37/2)*exp(b*x^2*ln(f))-8192/234542958176 401875*(-b*x^2*ln(f))^(35/2)*exp(b*x^2*ln(f))-4096/6701227376468625*(-b*x^ 2*ln(f))^(33/2)*exp(b*x^2*ln(f))-2048/203067496256625*(-b*x^2*ln(f))^(31/2 )*exp(b*x^2*ln(f))-1024/6550564395375*(-b*x^2*ln(f))^(29/2)*exp(b*x^2*ln(f ))-512/225881530875*(-b*x^2*ln(f))^(27/2)*exp(b*x^2*ln(f))-256/8365982625* (-b*x^2*ln(f))^(25/2)*exp(b*x^2*ln(f))-128/334639305*(-b*x^2*ln(f))^(23/2) *exp(b*x^2*ln(f))-64/14549535*(-b*x^2*ln(f))^(21/2)*exp(b*x^2*ln(f))-32/69 2835*(-b*x^2*ln(f))^(19/2)*exp(b*x^2*ln(f))-16/36465*(-b*x^2*ln(f))^(17/2) *exp(b*x^2*ln(f))-8/2145*(-b*x^2*ln(f))^(15/2)*exp(b*x^2*ln(f))-4/143*(-b* x^2*ln(f))^(13/2)*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)
Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^2} x^{10} \, dx=-\frac {f^a x^{11} \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}} \]
Time = 0.17 (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.
\(\displaystyle \int x^{10} f^{a+b x^2} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle -\frac {x^{11} f^a \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}}\) |
3.1.83.3.1 Defintions of rubi rules used
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 = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.26
method | result | size |
meijerg | \(-\frac {f^{a} \left (\frac {x \left (-b \right )^{\frac {11}{2}} \sqrt {\ln \left (f \right )}\, \left (176 b^{4} x^{8} \ln \left (f \right )^{4}-792 b^{3} x^{6} \ln \left (f \right )^{3}+2772 b^{2} x^{4} \ln \left (f \right )^{2}-6930 b \,x^{2} \ln \left (f \right )+10395\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{176 b^{5}}-\frac {945 \left (-b \right )^{\frac {11}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{32 b^{\frac {11}{2}}}\right )}{2 b^{5} \ln \left (f \right )^{\frac {11}{2}} \sqrt {-b}}\) | \(111\) |
risch | \(\frac {f^{a} x^{9} f^{b \,x^{2}}}{2 \ln \left (f \right ) b}-\frac {9 f^{a} x^{7} f^{b \,x^{2}}}{4 \ln \left (f \right )^{2} b^{2}}+\frac {63 f^{a} x^{5} f^{b \,x^{2}}}{8 \ln \left (f \right )^{3} b^{3}}-\frac {315 f^{a} x^{3} f^{b \,x^{2}}}{16 \ln \left (f \right )^{4} b^{4}}+\frac {945 f^{a} x \,f^{b \,x^{2}}}{32 \ln \left (f \right )^{5} b^{5}}-\frac {945 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{64 \ln \left (f \right )^{5} b^{5} \sqrt {-b \ln \left (f \right )}}\) | \(142\) |
-1/2*f^a/b^5/ln(f)^(11/2)/(-b)^(1/2)*(1/176*x*(-b)^(11/2)*ln(f)^(1/2)*(176 *b^4*x^8*ln(f)^4-792*b^3*x^6*ln(f)^3+2772*b^2*x^4*ln(f)^2-6930*b*x^2*ln(f) +10395)/b^5*exp(b*x^2*ln(f))-945/32*(-b)^(11/2)/b^(11/2)*Pi^(1/2)*erfi(x*b ^(1/2)*ln(f)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.97 \[ \int f^{a+b x^2} x^{10} \, dx=\frac {945 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) + 2 \, {\left (16 \, b^{5} x^{9} \log \left (f\right )^{5} - 72 \, b^{4} x^{7} \log \left (f\right )^{4} + 252 \, b^{3} x^{5} \log \left (f\right )^{3} - 630 \, b^{2} x^{3} \log \left (f\right )^{2} + 945 \, b x \log \left (f\right )\right )} f^{b x^{2} + a}}{64 \, b^{6} \log \left (f\right )^{6}} \]
1/64*(945*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x) + 2*(16*b^5* x^9*log(f)^5 - 72*b^4*x^7*log(f)^4 + 252*b^3*x^5*log(f)^3 - 630*b^2*x^3*lo g(f)^2 + 945*b*x*log(f))*f^(b*x^2 + a))/(b^6*log(f)^6)
\[ \int f^{a+b x^2} x^{10} \, dx=\int f^{a + b x^{2}} x^{10}\, dx \]
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.29 \[ \int f^{a+b x^2} x^{10} \, dx=\frac {{\left (16 \, b^{4} f^{a} x^{9} \log \left (f\right )^{4} - 72 \, b^{3} f^{a} x^{7} \log \left (f\right )^{3} + 252 \, b^{2} f^{a} x^{5} \log \left (f\right )^{2} - 630 \, b f^{a} x^{3} \log \left (f\right ) + 945 \, f^{a} x\right )} f^{b x^{2}}}{32 \, b^{5} \log \left (f\right )^{5}} - \frac {945 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{64 \, \sqrt {-b \log \left (f\right )} b^{5} \log \left (f\right )^{5}} \]
1/32*(16*b^4*f^a*x^9*log(f)^4 - 72*b^3*f^a*x^7*log(f)^3 + 252*b^2*f^a*x^5* log(f)^2 - 630*b*f^a*x^3*log(f) + 945*f^a*x)*f^(b*x^2)/(b^5*log(f)^5) - 94 5/64*sqrt(pi)*f^a*erf(sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^5*log(f)^5)
Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.06 \[ \int f^{a+b x^2} x^{10} \, dx=\frac {945 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (f\right )} x\right )}{64 \, \sqrt {-b \log \left (f\right )} b^{5} \log \left (f\right )^{5}} + \frac {{\left (16 \, b^{4} x^{9} \log \left (f\right )^{4} - 72 \, b^{3} x^{7} \log \left (f\right )^{3} + 252 \, b^{2} x^{5} \log \left (f\right )^{2} - 630 \, b x^{3} \log \left (f\right ) + 945 \, x\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{32 \, b^{5} \log \left (f\right )^{5}} \]
945/64*sqrt(pi)*f^a*erf(-sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^5*log(f)^5) + 1/32*(16*b^4*x^9*log(f)^4 - 72*b^3*x^7*log(f)^3 + 252*b^2*x^5*log(f)^2 - 630*b*x^3*log(f) + 945*x)*e^(b*x^2*log(f) + a*log(f))/(b^5*log(f)^5)
Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 4.09 \[ \int f^{a+b x^2} x^{10} \, dx=-\frac {\frac {f^a\,\left (945\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )-1890\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \left (f\right )}\right )}{64\,\sqrt {b\,\ln \left (f\right )}}-\frac {63\,b^2\,f^a\,f^{b\,x^2}\,x^5\,{\ln \left (f\right )}^2}{8}+\frac {9\,b^3\,f^a\,f^{b\,x^2}\,x^7\,{\ln \left (f\right )}^3}{4}-\frac {b^4\,f^a\,f^{b\,x^2}\,x^9\,{\ln \left (f\right )}^4}{2}+\frac {315\,b\,f^a\,f^{b\,x^2}\,x^3\,\ln \left (f\right )}{16}}{b^5\,{\ln \left (f\right )}^5} \]