Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=-\frac {F^a \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{11/2}}{2 d (c+d x)^{11}} \]
-1/2*F^a*(64/10395*Pi^(1/2)*erfc((-b*(d*x+c)^2*ln(F))^(1/2))-64/10395/(-b* (d*x+c)^2*ln(F))^(1/2)*exp(b*(d*x+c)^2*ln(F))+32/10395/(-b*(d*x+c)^2*ln(F) )^(3/2)*exp(b*(d*x+c)^2*ln(F))-16/3465/(-b*(d*x+c)^2*ln(F))^(5/2)*exp(b*(d *x+c)^2*ln(F))+8/693/(-b*(d*x+c)^2*ln(F))^(7/2)*exp(b*(d*x+c)^2*ln(F))-4/9 9/(-b*(d*x+c)^2*ln(F))^(9/2)*exp(b*(d*x+c)^2*ln(F))+2/11/(-b*(d*x+c)^2*ln( F))^(11/2)*exp(b*(d*x+c)^2*ln(F)))*(-b*(d*x+c)^2*ln(F))^(11/2)/d/(d*x+c)^1 1
Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=-\frac {F^a \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{11/2}}{2 d (c+d x)^{11}} \]
-1/2*(F^a*Gamma[-11/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^ (11/2))/(d*(c + d*x)^11)
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, 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.
| \(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx\) |
| \(\Big \downarrow \) 2648 |
| \(\displaystyle -\frac {F^a \left (-b \log (F) (c+d x)^2\right )^{11/2} \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right )}{2 d (c+d x)^{11}}\) |
-1/2*(F^a*Gamma[-11/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^ (11/2))/(d*(c + d*x)^11)
3.3.79.3.1 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 = 1.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 4.65
| method | result | size |
| risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{11 d \left (d x +c \right )^{11}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{99 d \left (d x +c \right )^{9}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{693 d \left (d x +c \right )^{7}}-\frac {8 b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{3465 d \left (d x +c \right )^{5}}-\frac {16 b^{4} \ln \left (F \right )^{4} F^{b \left (d x +c \right )^{2}} F^{a}}{10395 d \left (d x +c \right )^{3}}-\frac {32 b^{5} \ln \left (F \right )^{5} F^{b \left (d x +c \right )^{2}} F^{a}}{10395 d \left (d x +c \right )}+\frac {32 b^{6} \ln \left (F \right )^{6} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{10395 d \sqrt {-b \ln \left (F \right )}}\) | \(228\) |
-1/11/d/(d*x+c)^11*F^(b*(d*x+c)^2)*F^a-2/99/d*b*ln(F)/(d*x+c)^9*F^(b*(d*x+ c)^2)*F^a-4/693/d*b^2*ln(F)^2/(d*x+c)^7*F^(b*(d*x+c)^2)*F^a-8/3465/d*b^3*l n(F)^3/(d*x+c)^5*F^(b*(d*x+c)^2)*F^a-16/10395/d*b^4*ln(F)^4/(d*x+c)^3*F^(b *(d*x+c)^2)*F^a-32/10395/d*b^5*ln(F)^5/(d*x+c)*F^(b*(d*x+c)^2)*F^a+32/1039 5/d*b^6*ln(F)^6*Pi^(1/2)*F^a/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)*(d*x+c) )
Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (212) = 424\).
Time = 0.27 (sec) , antiderivative size = 795, normalized size of antiderivative = 16.22 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=-\frac {32 \, \sqrt {\pi } {\left (b^{5} d^{11} x^{11} + 11 \, b^{5} c d^{10} x^{10} + 55 \, b^{5} c^{2} d^{9} x^{9} + 165 \, b^{5} c^{3} d^{8} x^{8} + 330 \, b^{5} c^{4} d^{7} x^{7} + 462 \, b^{5} c^{5} d^{6} x^{6} + 462 \, b^{5} c^{6} d^{5} x^{5} + 330 \, b^{5} c^{7} d^{4} x^{4} + 165 \, b^{5} c^{8} d^{3} x^{3} + 55 \, b^{5} c^{9} d^{2} x^{2} + 11 \, b^{5} c^{10} d x + b^{5} c^{11}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right )^{5} + {\left (32 \, {\left (b^{5} d^{11} x^{10} + 10 \, b^{5} c d^{10} x^{9} + 45 \, b^{5} c^{2} d^{9} x^{8} + 120 \, b^{5} c^{3} d^{8} x^{7} + 210 \, b^{5} c^{4} d^{7} x^{6} + 252 \, b^{5} c^{5} d^{6} x^{5} + 210 \, b^{5} c^{6} d^{5} x^{4} + 120 \, b^{5} c^{7} d^{4} x^{3} + 45 \, b^{5} c^{8} d^{3} x^{2} + 10 \, b^{5} c^{9} d^{2} x + b^{5} c^{10} d\right )} \log \left (F\right )^{5} + 16 \, {\left (b^{4} d^{9} x^{8} + 8 \, b^{4} c d^{8} x^{7} + 28 \, b^{4} c^{2} d^{7} x^{6} + 56 \, b^{4} c^{3} d^{6} x^{5} + 70 \, b^{4} c^{4} d^{5} x^{4} + 56 \, b^{4} c^{5} d^{4} x^{3} + 28 \, b^{4} c^{6} d^{3} x^{2} + 8 \, b^{4} c^{7} d^{2} x + b^{4} c^{8} d\right )} \log \left (F\right )^{4} + 24 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \left (F\right )^{3} + 60 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 210 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 945 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{10395 \, {\left (d^{13} x^{11} + 11 \, c d^{12} x^{10} + 55 \, c^{2} d^{11} x^{9} + 165 \, c^{3} d^{10} x^{8} + 330 \, c^{4} d^{9} x^{7} + 462 \, c^{5} d^{8} x^{6} + 462 \, c^{6} d^{7} x^{5} + 330 \, c^{7} d^{6} x^{4} + 165 \, c^{8} d^{5} x^{3} + 55 \, c^{9} d^{4} x^{2} + 11 \, c^{10} d^{3} x + c^{11} d^{2}\right )}} \]
-1/10395*(32*sqrt(pi)*(b^5*d^11*x^11 + 11*b^5*c*d^10*x^10 + 55*b^5*c^2*d^9 *x^9 + 165*b^5*c^3*d^8*x^8 + 330*b^5*c^4*d^7*x^7 + 462*b^5*c^5*d^6*x^6 + 4 62*b^5*c^6*d^5*x^5 + 330*b^5*c^7*d^4*x^4 + 165*b^5*c^8*d^3*x^3 + 55*b^5*c^ 9*d^2*x^2 + 11*b^5*c^10*d*x + b^5*c^11)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(- b*d^2*log(F))*(d*x + c)/d)*log(F)^5 + (32*(b^5*d^11*x^10 + 10*b^5*c*d^10*x ^9 + 45*b^5*c^2*d^9*x^8 + 120*b^5*c^3*d^8*x^7 + 210*b^5*c^4*d^7*x^6 + 252* b^5*c^5*d^6*x^5 + 210*b^5*c^6*d^5*x^4 + 120*b^5*c^7*d^4*x^3 + 45*b^5*c^8*d ^3*x^2 + 10*b^5*c^9*d^2*x + b^5*c^10*d)*log(F)^5 + 16*(b^4*d^9*x^8 + 8*b^4 *c*d^8*x^7 + 28*b^4*c^2*d^7*x^6 + 56*b^4*c^3*d^6*x^5 + 70*b^4*c^4*d^5*x^4 + 56*b^4*c^5*d^4*x^3 + 28*b^4*c^6*d^3*x^2 + 8*b^4*c^7*d^2*x + b^4*c^8*d)*l og(F)^4 + 24*(b^3*d^7*x^6 + 6*b^3*c*d^6*x^5 + 15*b^3*c^2*d^5*x^4 + 20*b^3* c^3*d^4*x^3 + 15*b^3*c^4*d^3*x^2 + 6*b^3*c^5*d^2*x + b^3*c^6*d)*log(F)^3 + 60*(b^2*d^5*x^4 + 4*b^2*c*d^4*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c^3*d^2*x + b^2*c^4*d)*log(F)^2 + 210*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + 94 5*d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^13*x^11 + 11*c*d^12*x^10 + 55*c^2*d^11*x^9 + 165*c^3*d^10*x^8 + 330*c^4*d^9*x^7 + 462*c^5*d^8*x^6 + 4 62*c^6*d^7*x^5 + 330*c^7*d^6*x^4 + 165*c^8*d^5*x^3 + 55*c^9*d^4*x^2 + 11*c ^10*d^3*x + c^11*d^2)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{12}}\, dx \]
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{12}} \,d x } \]
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{12}} \,d x } \]
Time = 0.92 (sec) , antiderivative size = 267, normalized size of antiderivative = 5.45 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\frac {32\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{11/2}}{10395\,d\,{\left (c+d\,x\right )}^{11}}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{11\,d\,{\left (c+d\,x\right )}^{11}}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{693\,d\,{\left (c+d\,x\right )}^7}-\frac {8\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3}{3465\,d\,{\left (c+d\,x\right )}^5}-\frac {16\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^4\,{\ln \left (F\right )}^4}{10395\,d\,{\left (c+d\,x\right )}^3}-\frac {32\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^5\,{\ln \left (F\right )}^5}{10395\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{99\,d\,{\left (c+d\,x\right )}^9}-\frac {32\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{11/2}}{10395\,d\,{\left (c+d\,x\right )}^{11}} \]
(32*F^a*pi^(1/2)*(-b*log(F)*(c + d*x)^2)^(11/2))/(10395*d*(c + d*x)^11) - (F^a*F^(b*(c + d*x)^2))/(11*d*(c + d*x)^11) - (4*F^a*F^(b*(c + d*x)^2)*b^2 *log(F)^2)/(693*d*(c + d*x)^7) - (8*F^a*F^(b*(c + d*x)^2)*b^3*log(F)^3)/(3 465*d*(c + d*x)^5) - (16*F^a*F^(b*(c + d*x)^2)*b^4*log(F)^4)/(10395*d*(c + d*x)^3) - (32*F^a*F^(b*(c + d*x)^2)*b^5*log(F)^5)/(10395*d*(c + d*x)) - ( 2*F^a*F^(b*(c + d*x)^2)*b*log(F))/(99*d*(c + d*x)^9) - (32*F^a*pi^(1/2)*er fc((-b*log(F)*(c + d*x)^2)^(1/2))*(-b*log(F)*(c + d*x)^2)^(11/2))/(10395*d *(c + d*x)^11)