Integrand size = 21, antiderivative size = 49 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\frac {F^a (c+d x)^{11} \Gamma \left (-\frac {11}{2},-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{11/2}}{2 d} \]
1/2*F^a*(d*x+c)^11*(64/10395*Pi^(1/2)*erfc((-b*ln(F)/(d*x+c)^2)^(1/2))-64/ 10395/(-b*ln(F)/(d*x+c)^2)^(1/2)*exp(b*ln(F)/(d*x+c)^2)+32/10395/(-b*ln(F) /(d*x+c)^2)^(3/2)*exp(b*ln(F)/(d*x+c)^2)-16/3465/(-b*ln(F)/(d*x+c)^2)^(5/2 )*exp(b*ln(F)/(d*x+c)^2)+8/693/(-b*ln(F)/(d*x+c)^2)^(7/2)*exp(b*ln(F)/(d*x +c)^2)-4/99/(-b*ln(F)/(d*x+c)^2)^(9/2)*exp(b*ln(F)/(d*x+c)^2)+2/11/(-b*ln( F)/(d*x+c)^2)^(11/2)*exp(b*ln(F)/(d*x+c)^2))*(-b*ln(F)/(d*x+c)^2)^(11/2)/d
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\frac {F^a (c+d x)^{11} \Gamma \left (-\frac {11}{2},-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{11/2}}{2 d} \]
(F^a*(c + d*x)^11*Gamma[-11/2, -((b*Log[F])/(c + d*x)^2)]*(-((b*Log[F])/(c + d*x)^2))^(11/2))/(2*d)
Time = 0.22 (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.
| \(\displaystyle \int (c+d x)^{10} F^{a+\frac {b}{(c+d x)^2}} \, dx\) |
| \(\Big \downarrow \) 2648 |
| \(\displaystyle \frac {F^a (c+d x)^{11} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{11/2} \Gamma \left (-\frac {11}{2},-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d}\) |
(F^a*(c + d*x)^11*Gamma[-11/2, -((b*Log[F])/(c + d*x)^2)]*(-((b*Log[F])/(c + d*x)^2))^(11/2))/(2*d)
3.4.27.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(1172\) vs. \(2(218)=436\).
Time = 4.19 (sec) , antiderivative size = 1173, normalized size of antiderivative = 23.94
F^a*d^9*F^(b/(d*x+c)^2)*c*x^10+5*F^a*d^8*F^(b/(d*x+c)^2)*c^2*x^9+15*F^a*d^ 7*F^(b/(d*x+c)^2)*c^3*x^8+30*F^a*d^6*F^(b/(d*x+c)^2)*c^4*x^7+42*F^a*d^5*F^ (b/(d*x+c)^2)*c^5*x^6+42*F^a*d^4*F^(b/(d*x+c)^2)*c^6*x^5+30*F^a*d^3*F^(b/( d*x+c)^2)*c^7*x^4+15*F^a*d^2*F^(b/(d*x+c)^2)*c^8*x^3+5*F^a*d*F^(b/(d*x+c)^ 2)*c^9*x^2+32/10395*F^a*b^5*ln(F)^5*F^(b/(d*x+c)^2)*x+F^a*F^(b/(d*x+c)^2)* c^10*x+1/11*F^a/d*F^(b/(d*x+c)^2)*c^11+1/11*F^a*d^10*F^(b/(d*x+c)^2)*x^11+ 4/693*F^a*d^6*b^2*ln(F)^2*F^(b/(d*x+c)^2)*x^7+8/3465*F^a*d^4*b^3*ln(F)^3*F ^(b/(d*x+c)^2)*x^5+16/10395*F^a*d^2*b^4*ln(F)^4*F^(b/(d*x+c)^2)*x^3+2/99*F ^a*d^8*b*ln(F)*F^(b/(d*x+c)^2)*x^9+4/693*F^a/d*b^2*ln(F)^2*F^(b/(d*x+c)^2) *c^7+8/3465*F^a/d*b^3*ln(F)^3*F^(b/(d*x+c)^2)*c^5+16/10395*F^a/d*b^4*ln(F) ^4*F^(b/(d*x+c)^2)*c^3+32/10395*F^a/d*b^5*ln(F)^5*F^(b/(d*x+c)^2)*c+2/99*F ^a/d*b*ln(F)*F^(b/(d*x+c)^2)*c^9+2/11*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c^8*x+4/ 99*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c^6*x+8/693*F^a*b^3*ln(F)^3*F^(b/(d*x+c )^2)*c^4*x+16/3465*F^a*b^4*ln(F)^4*F^(b/(d*x+c)^2)*c^2*x+16/3465*F^a*d*b^4 *ln(F)^4*F^(b/(d*x+c)^2)*c*x^2+2/11*F^a*d^7*b*ln(F)*F^(b/(d*x+c)^2)*c*x^8+ 8/11*F^a*d^6*b*ln(F)*F^(b/(d*x+c)^2)*c^2*x^7+56/33*F^a*d^5*b*ln(F)*F^(b/(d *x+c)^2)*c^3*x^6+28/11*F^a*d^4*b*ln(F)*F^(b/(d*x+c)^2)*c^4*x^5+28/11*F^a*d ^3*b*ln(F)*F^(b/(d*x+c)^2)*c^5*x^4+56/33*F^a*d^2*b*ln(F)*F^(b/(d*x+c)^2)*c ^6*x^3+8/11*F^a*d*b*ln(F)*F^(b/(d*x+c)^2)*c^7*x^2-32/10395*F^a/d*b^6*ln(F) ^6*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))+4/99*F^a*d^5...
Leaf count of result is larger than twice the leaf count of optimal. 561 vs. \(2 (212) = 424\).
Time = 0.29 (sec) , antiderivative size = 561, normalized size of antiderivative = 11.45 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\frac {32 \, \sqrt {\pi } F^{a} b^{5} d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) \log \left (F\right )^{5} + {\left (945 \, d^{11} x^{11} + 10395 \, c d^{10} x^{10} + 51975 \, c^{2} d^{9} x^{9} + 155925 \, c^{3} d^{8} x^{8} + 311850 \, c^{4} d^{7} x^{7} + 436590 \, c^{5} d^{6} x^{6} + 436590 \, c^{6} d^{5} x^{5} + 311850 \, c^{7} d^{4} x^{4} + 155925 \, c^{8} d^{3} x^{3} + 51975 \, c^{9} d^{2} x^{2} + 10395 \, c^{10} d x + 945 \, c^{11} + 32 \, {\left (b^{5} d x + b^{5} c\right )} \log \left (F\right )^{5} + 16 \, {\left (b^{4} d^{3} x^{3} + 3 \, b^{4} c d^{2} x^{2} + 3 \, b^{4} c^{2} d x + b^{4} c^{3}\right )} \log \left (F\right )^{4} + 24 \, {\left (b^{3} d^{5} x^{5} + 5 \, b^{3} c d^{4} x^{4} + 10 \, b^{3} c^{2} d^{3} x^{3} + 10 \, b^{3} c^{3} d^{2} x^{2} + 5 \, b^{3} c^{4} d x + b^{3} c^{5}\right )} \log \left (F\right )^{3} + 60 \, {\left (b^{2} d^{7} x^{7} + 7 \, b^{2} c d^{6} x^{6} + 21 \, b^{2} c^{2} d^{5} x^{5} + 35 \, b^{2} c^{3} d^{4} x^{4} + 35 \, b^{2} c^{4} d^{3} x^{3} + 21 \, b^{2} c^{5} d^{2} x^{2} + 7 \, b^{2} c^{6} d x + b^{2} c^{7}\right )} \log \left (F\right )^{2} + 210 \, {\left (b d^{9} x^{9} + 9 \, b c d^{8} x^{8} + 36 \, b c^{2} d^{7} x^{7} + 84 \, b c^{3} d^{6} x^{6} + 126 \, b c^{4} d^{5} x^{5} + 126 \, b c^{5} d^{4} x^{4} + 84 \, b c^{6} d^{3} x^{3} + 36 \, b c^{7} d^{2} x^{2} + 9 \, b c^{8} d x + b c^{9}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{10395 \, d} \]
1/10395*(32*sqrt(pi)*F^a*b^5*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^ 2)/(d*x + c))*log(F)^5 + (945*d^11*x^11 + 10395*c*d^10*x^10 + 51975*c^2*d^ 9*x^9 + 155925*c^3*d^8*x^8 + 311850*c^4*d^7*x^7 + 436590*c^5*d^6*x^6 + 436 590*c^6*d^5*x^5 + 311850*c^7*d^4*x^4 + 155925*c^8*d^3*x^3 + 51975*c^9*d^2* x^2 + 10395*c^10*d*x + 945*c^11 + 32*(b^5*d*x + b^5*c)*log(F)^5 + 16*(b^4* d^3*x^3 + 3*b^4*c*d^2*x^2 + 3*b^4*c^2*d*x + b^4*c^3)*log(F)^4 + 24*(b^3*d^ 5*x^5 + 5*b^3*c*d^4*x^4 + 10*b^3*c^2*d^3*x^3 + 10*b^3*c^3*d^2*x^2 + 5*b^3* c^4*d*x + b^3*c^5)*log(F)^3 + 60*(b^2*d^7*x^7 + 7*b^2*c*d^6*x^6 + 21*b^2*c ^2*d^5*x^5 + 35*b^2*c^3*d^4*x^4 + 35*b^2*c^4*d^3*x^3 + 21*b^2*c^5*d^2*x^2 + 7*b^2*c^6*d*x + b^2*c^7)*log(F)^2 + 210*(b*d^9*x^9 + 9*b*c*d^8*x^8 + 36* b*c^2*d^7*x^7 + 84*b*c^3*d^6*x^6 + 126*b*c^4*d^5*x^5 + 126*b*c^5*d^4*x^4 + 84*b*c^6*d^3*x^3 + 36*b*c^7*d^2*x^2 + 9*b*c^8*d*x + b*c^9)*log(F))*F^((a* d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/d
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{10}\, dx \]
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\int { {\left (d x + c\right )}^{10} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
1/10395*(945*F^a*d^10*x^11 + 10395*F^a*c*d^9*x^10 + 105*(495*F^a*c^2*d^8 + 2*F^a*b*d^8*log(F))*x^9 + 945*(165*F^a*c^3*d^7 + 2*F^a*b*c*d^7*log(F))*x^ 8 + 30*(10395*F^a*c^4*d^6 + 252*F^a*b*c^2*d^6*log(F) + 2*F^a*b^2*d^6*log(F )^2)*x^7 + 210*(2079*F^a*c^5*d^5 + 84*F^a*b*c^3*d^5*log(F) + 2*F^a*b^2*c*d ^5*log(F)^2)*x^6 + 6*(72765*F^a*c^6*d^4 + 4410*F^a*b*c^4*d^4*log(F) + 210* F^a*b^2*c^2*d^4*log(F)^2 + 4*F^a*b^3*d^4*log(F)^3)*x^5 + 30*(10395*F^a*c^7 *d^3 + 882*F^a*b*c^5*d^3*log(F) + 70*F^a*b^2*c^3*d^3*log(F)^2 + 4*F^a*b^3* c*d^3*log(F)^3)*x^4 + (155925*F^a*c^8*d^2 + 17640*F^a*b*c^6*d^2*log(F) + 2 100*F^a*b^2*c^4*d^2*log(F)^2 + 240*F^a*b^3*c^2*d^2*log(F)^3 + 16*F^a*b^4*d ^2*log(F)^4)*x^3 + 3*(17325*F^a*c^9*d + 2520*F^a*b*c^7*d*log(F) + 420*F^a* b^2*c^5*d*log(F)^2 + 80*F^a*b^3*c^3*d*log(F)^3 + 16*F^a*b^4*c*d*log(F)^4)* x^2 + (10395*F^a*c^10 + 1890*F^a*b*c^8*log(F) + 420*F^a*b^2*c^6*log(F)^2 + 120*F^a*b^3*c^4*log(F)^3 + 48*F^a*b^4*c^2*log(F)^4 + 32*F^a*b^5*log(F)^5) *x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(2/10395*(32*F^a*b^6*d*x*lo g(F)^6 - 945*F^a*b*c^11*log(F) - 210*F^a*b^2*c^9*log(F)^2 - 60*F^a*b^3*c^7 *log(F)^3 - 24*F^a*b^4*c^5*log(F)^4 - 16*F^a*b^5*c^3*log(F)^5)*F^(b/(d^2*x ^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\int { {\left (d x + c\right )}^{10} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
Time = 0.95 (sec) , antiderivative size = 265, normalized size of antiderivative = 5.41 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^{10} \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^{11}}{11\,d}-\frac {32\,F^a\,\sqrt {\pi }\,{\left (c+d\,x\right )}^{11}\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{11/2}}{10395\,d}+\frac {4\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^7}{693\,d}+\frac {8\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \left (F\right )}^3\,{\left (c+d\,x\right )}^5}{3465\,d}+\frac {16\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^4\,{\ln \left (F\right )}^4\,{\left (c+d\,x\right )}^3}{10395\,d}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^9}{99\,d}+\frac {32\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^5\,{\ln \left (F\right )}^5\,\left (c+d\,x\right )}{10395\,d}+\frac {32\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}}\right )\,{\left (c+d\,x\right )}^{11}\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{11/2}}{10395\,d} \]
(F^a*F^(b/(c + d*x)^2)*(c + d*x)^11)/(11*d) - (32*F^a*pi^(1/2)*(c + d*x)^1 1*(-(b*log(F))/(c + d*x)^2)^(11/2))/(10395*d) + (4*F^a*F^(b/(c + d*x)^2)*b ^2*log(F)^2*(c + d*x)^7)/(693*d) + (8*F^a*F^(b/(c + d*x)^2)*b^3*log(F)^3*( c + d*x)^5)/(3465*d) + (16*F^a*F^(b/(c + d*x)^2)*b^4*log(F)^4*(c + d*x)^3) /(10395*d) + (2*F^a*F^(b/(c + d*x)^2)*b*log(F)*(c + d*x)^9)/(99*d) + (32*F ^a*F^(b/(c + d*x)^2)*b^5*log(F)^5*(c + d*x))/(10395*d) + (32*F^a*pi^(1/2)* erfc((-(b*log(F))/(c + d*x)^2)^(1/2))*(c + d*x)^11*(-(b*log(F))/(c + d*x)^ 2)^(11/2))/(10395*d)