Integrand size = 21, antiderivative size = 49 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=-\frac {F^a (c+d x)^{11} \Gamma \left (\frac {11}{2},-b (c+d x)^2 \log (F)\right )}{2 d \left (-b (c+d x)^2 \log (F)\right )^{11/2}} \]
-1/2*F^a*(d*x+c)^11*(1048576/61836869254970658257624840625*GAMMA(51/2,-b*( d*x+c)^2*ln(F))-1048576/61836869254970658257624840625*(-b*(d*x+c)^2*ln(F)) ^(49/2)*exp(b*(d*x+c)^2*ln(F))-524288/1261976923570829760359690625*(-b*(d* x+c)^2*ln(F))^(47/2)*exp(b*(d*x+c)^2*ln(F))-262144/26850572841932548092759 375*(-b*(d*x+c)^2*ln(F))^(45/2)*exp(b*(d*x+c)^2*ln(F))-131072/596679396487 389957616875*(-b*(d*x+c)^2*ln(F))^(43/2)*exp(b*(d*x+c)^2*ln(F))-65536/1387 6265034590464130625*(-b*(d*x+c)^2*ln(F))^(41/2)*exp(b*(d*x+c)^2*ln(F))-327 68/338445488648547905625*(-b*(d*x+c)^2*ln(F))^(39/2)*exp(b*(d*x+c)^2*ln(F) )-16384/8678089452526869375*(-b*(d*x+c)^2*ln(F))^(37/2)*exp(b*(d*x+c)^2*ln (F))-8192/234542958176401875*(-b*(d*x+c)^2*ln(F))^(35/2)*exp(b*(d*x+c)^2*l n(F))-4096/6701227376468625*(-b*(d*x+c)^2*ln(F))^(33/2)*exp(b*(d*x+c)^2*ln (F))-2048/203067496256625*(-b*(d*x+c)^2*ln(F))^(31/2)*exp(b*(d*x+c)^2*ln(F ))-1024/6550564395375*(-b*(d*x+c)^2*ln(F))^(29/2)*exp(b*(d*x+c)^2*ln(F))-5 12/225881530875*(-b*(d*x+c)^2*ln(F))^(27/2)*exp(b*(d*x+c)^2*ln(F))-256/836 5982625*(-b*(d*x+c)^2*ln(F))^(25/2)*exp(b*(d*x+c)^2*ln(F))-128/334639305*( -b*(d*x+c)^2*ln(F))^(23/2)*exp(b*(d*x+c)^2*ln(F))-64/14549535*(-b*(d*x+c)^ 2*ln(F))^(21/2)*exp(b*(d*x+c)^2*ln(F))-32/692835*(-b*(d*x+c)^2*ln(F))^(19/ 2)*exp(b*(d*x+c)^2*ln(F))-16/36465*(-b*(d*x+c)^2*ln(F))^(17/2)*exp(b*(d*x+ c)^2*ln(F))-8/2145*(-b*(d*x+c)^2*ln(F))^(15/2)*exp(b*(d*x+c)^2*ln(F))-4/14 3*(-b*(d*x+c)^2*ln(F))^(13/2)*exp(b*(d*x+c)^2*ln(F))-2/11*(-b*(d*x+c)^2...
Time = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=-\frac {F^a (c+d x)^{11} \Gamma \left (\frac {11}{2},-b (c+d x)^2 \log (F)\right )}{2 d \left (-b (c+d x)^2 \log (F)\right )^{11/2}} \]
-1/2*(F^a*(c + d*x)^11*Gamma[11/2, -(b*(c + d*x)^2*Log[F])])/(d*(-(b*(c + d*x)^2*Log[F]))^(11/2))
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.
\(\displaystyle \int (c+d x)^{10} F^{a+b (c+d x)^2} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle -\frac {F^a (c+d x)^{11} \Gamma \left (\frac {11}{2},-b (c+d x)^2 \log (F)\right )}{2 d \left (-b \log (F) (c+d x)^2\right )^{11/2}}\) |
-1/2*(F^a*(c + d*x)^11*Gamma[11/2, -(b*(c + d*x)^2*Log[F])])/(d*(-(b*(c + d*x)^2*Log[F]))^(11/2))
3.3.68.3.1 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. \(1373\) vs. \(2(606)=1212\).
Time = 1.59 (sec) , antiderivative size = 1374, normalized size of antiderivative = 28.04
63/8*F^(b*c^2)*F^a*d^4/ln(F)^3/b^3*x^5*F^(b*d^2*x^2)*F^(2*b*c*d*x)-9/4*F^( b*c^2)*F^a*d^6/ln(F)^2/b^2*x^7*F^(b*d^2*x^2)*F^(2*b*c*d*x)-315/16*F^(b*c^2 )*F^a*d^2/ln(F)^4/b^4*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)+945/32*F^(b*c^2)*F^a /d*c/ln(F)^5/b^5*F^(b*d^2*x^2)*F^(2*b*c*d*x)-315/16*F^(b*c^2)*F^a/d*c^3/ln (F)^4/b^4*F^(b*d^2*x^2)*F^(2*b*c*d*x)+1/2*F^(b*c^2)*F^a/d*c^9/ln(F)/b*F^(b *d^2*x^2)*F^(2*b*c*d*x)-9/4*F^(b*c^2)*F^a/d*c^7/ln(F)^2/b^2*F^(b*d^2*x^2)* F^(2*b*c*d*x)+63/8*F^(b*c^2)*F^a/d*c^5/ln(F)^3/b^3*F^(b*d^2*x^2)*F^(2*b*c* d*x)+1/2*F^(b*c^2)*F^a*d^8/ln(F)/b*x^9*F^(b*d^2*x^2)*F^(2*b*c*d*x)+9/2*F^( b*c^2)*F^a*c^8/ln(F)/b*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)-63/4*F^(b*c^2)*F^a*c^ 6/ln(F)^2/b^2*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)+315/8*F^(b*c^2)*F^a*c^4/ln(F)^ 3/b^3*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)-945/16*F^(b*c^2)*F^a*c^2/ln(F)^4/b^4*x *F^(b*d^2*x^2)*F^(2*b*c*d*x)-315/4*F^(b*c^2)*F^a*d^3*c^3/ln(F)^2/b^2*x^4*F ^(b*d^2*x^2)*F^(2*b*c*d*x)+315/4*F^(b*c^2)*F^a*d*c^3/ln(F)^3/b^3*x^2*F^(b* d^2*x^2)*F^(2*b*c*d*x)+42*F^(b*c^2)*F^a*d^2*c^6/ln(F)/b*x^3*F^(b*d^2*x^2)* F^(2*b*c*d*x)+18*F^(b*c^2)*F^a*d*c^7/ln(F)/b*x^2*F^(b*d^2*x^2)*F^(2*b*c*d* x)-189/4*F^(b*c^2)*F^a*d*c^5/ln(F)^2/b^2*x^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)-3 15/4*F^(b*c^2)*F^a*d^2*c^4/ln(F)^2/b^2*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)+18* F^(b*c^2)*F^a*d^6*c^2/ln(F)/b*x^7*F^(b*d^2*x^2)*F^(2*b*c*d*x)+315/4*F^(b*c ^2)*F^a*d^2*c^2/ln(F)^3/b^3*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)+945/64*F^(b*c^ 2)*F^a/d/ln(F)^5/b^5*Pi^(1/2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*erf(-d*(-b*ln...
Time = 0.09 (sec) , antiderivative size = 456, normalized size of antiderivative = 9.31 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=\frac {945 \, \sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) + 2 \, {\left (16 \, {\left (b^{5} d^{10} x^{9} + 9 \, b^{5} c d^{9} x^{8} + 36 \, b^{5} c^{2} d^{8} x^{7} + 84 \, b^{5} c^{3} d^{7} x^{6} + 126 \, b^{5} c^{4} d^{6} x^{5} + 126 \, b^{5} c^{5} d^{5} x^{4} + 84 \, b^{5} c^{6} d^{4} x^{3} + 36 \, b^{5} c^{7} d^{3} x^{2} + 9 \, b^{5} c^{8} d^{2} x + b^{5} c^{9} d\right )} \log \left (F\right )^{5} - 72 \, {\left (b^{4} d^{8} x^{7} + 7 \, b^{4} c d^{7} x^{6} + 21 \, b^{4} c^{2} d^{6} x^{5} + 35 \, b^{4} c^{3} d^{5} x^{4} + 35 \, b^{4} c^{4} d^{4} x^{3} + 21 \, b^{4} c^{5} d^{3} x^{2} + 7 \, b^{4} c^{6} d^{2} x + b^{4} c^{7} d\right )} \log \left (F\right )^{4} + 252 \, {\left (b^{3} d^{6} x^{5} + 5 \, b^{3} c d^{5} x^{4} + 10 \, b^{3} c^{2} d^{4} x^{3} + 10 \, b^{3} c^{3} d^{3} x^{2} + 5 \, b^{3} c^{4} d^{2} x + b^{3} c^{5} d\right )} \log \left (F\right )^{3} - 630 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} + 945 \, {\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{64 \, b^{6} d^{2} \log \left (F\right )^{6}} \]
1/64*(945*sqrt(pi)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d) + 2*(16*(b^5*d^10*x^9 + 9*b^5*c*d^9*x^8 + 36*b^5*c^2*d^8*x^7 + 84*b^ 5*c^3*d^7*x^6 + 126*b^5*c^4*d^6*x^5 + 126*b^5*c^5*d^5*x^4 + 84*b^5*c^6*d^4 *x^3 + 36*b^5*c^7*d^3*x^2 + 9*b^5*c^8*d^2*x + b^5*c^9*d)*log(F)^5 - 72*(b^ 4*d^8*x^7 + 7*b^4*c*d^7*x^6 + 21*b^4*c^2*d^6*x^5 + 35*b^4*c^3*d^5*x^4 + 35 *b^4*c^4*d^4*x^3 + 21*b^4*c^5*d^3*x^2 + 7*b^4*c^6*d^2*x + b^4*c^7*d)*log(F )^4 + 252*(b^3*d^6*x^5 + 5*b^3*c*d^5*x^4 + 10*b^3*c^2*d^4*x^3 + 10*b^3*c^3 *d^3*x^2 + 5*b^3*c^4*d^2*x + b^3*c^5*d)*log(F)^3 - 630*(b^2*d^4*x^3 + 3*b^ 2*c*d^3*x^2 + 3*b^2*c^2*d^2*x + b^2*c^3*d)*log(F)^2 + 945*(b*d^2*x + b*c*d )*log(F))*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(b^6*d^2*log(F)^6)
\[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=\int F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{10}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 4471 vs. \(2 (586) = 1172\).
Time = 1.50 (sec) , antiderivative size = 4471, normalized size of antiderivative = 91.24 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=\text {Too large to display} \]
-5*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/( b*d^2))) - 1)*log(F)^2/((b*log(F))^(3/2)*d^2*sqrt(-(b*d^2*x + b*c*d)^2*log (F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*log(F)/((b*log(F))^(3/2) *d))*F^a*c^9/sqrt(b*log(F)) + 45/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*(er f(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^3/((b*log(F))^(5/ 2)*d^3*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*F^((b*d^2*x + b*c*d) ^2/(b*d^2))*b^2*c*log(F)^2/((b*log(F))^(5/2)*d^2) - (b*d^2*x + b*c*d)^3*ga mma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F))^(5/2)*d ^5*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F^a*c^8*d/sqrt(b*log(F)) - 60*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*lo g(F)/(b*d^2))) - 1)*log(F)^4/((b*log(F))^(7/2)*d^4*sqrt(-(b*d^2*x + b*c*d) ^2*log(F)/(b*d^2))) - 3*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^3*c^2*log(F)^3/( (b*log(F))^(7/2)*d^3) - 3*(b*d^2*x + b*c*d)^3*b*c*gamma(3/2, -(b*d^2*x + b *c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*log(F))^(7/2)*d^6*(-(b*d^2*x + b*c*d) ^2*log(F)/(b*d^2))^(3/2)) + b^2*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^ 2))*log(F)^2/((b*log(F))^(7/2)*d^3))*F^a*c^7*d^2/sqrt(b*log(F)) + 105*(sqr t(pi)*(b*d^2*x + b*c*d)*b^4*c^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d ^2))) - 1)*log(F)^5/((b*log(F))^(9/2)*d^5*sqrt(-(b*d^2*x + b*c*d)^2*log(F) /(b*d^2))) - 4*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^4*c^3*log(F)^4/((b*log(F) )^(9/2)*d^4) - 6*(b*d^2*x + b*c*d)^3*b^2*c^2*gamma(3/2, -(b*d^2*x + b*c...
Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.55 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=\frac {{\left (16 \, b^{4} d^{8} {\left (x + \frac {c}{d}\right )}^{9} \log \left (F\right )^{4} - 72 \, b^{3} d^{6} {\left (x + \frac {c}{d}\right )}^{7} \log \left (F\right )^{3} + 252 \, b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{5} \log \left (F\right )^{2} - 630 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{3} \log \left (F\right ) + 945 \, x + \frac {945 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{32 \, b^{5} \log \left (F\right )^{5}} + \frac {945 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{64 \, \sqrt {-b \log \left (F\right )} b^{5} d \log \left (F\right )^{5}} \]
1/32*(16*b^4*d^8*(x + c/d)^9*log(F)^4 - 72*b^3*d^6*(x + c/d)^7*log(F)^3 + 252*b^2*d^4*(x + c/d)^5*log(F)^2 - 630*b*d^2*(x + c/d)^3*log(F) + 945*x + 945*c/d)*e^(b*d^2*x^2*log(F) + 2*b*c*d*x*log(F) + b*c^2*log(F) + a*log(F)) /(b^5*log(F)^5) + 945/64*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*d*(x + c/d))/(s qrt(-b*log(F))*b^5*d*log(F)^5)
Time = 0.77 (sec) , antiderivative size = 730, normalized size of antiderivative = 14.90 \[ \int F^{a+b (c+d x)^2} (c+d x)^{10} \, dx=\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {b^4\,c^9\,{\ln \left (F\right )}^4}{2}-\frac {9\,b^3\,c^7\,{\ln \left (F\right )}^3}{4}+\frac {63\,b^2\,c^5\,{\ln \left (F\right )}^2}{8}-\frac {315\,b\,c^3\,\ln \left (F\right )}{16}+\frac {945\,c}{32}\right )}{b^5\,d\,{\ln \left (F\right )}^5}-\frac {945\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )}{64\,b^5\,{\ln \left (F\right )}^5\,\sqrt {b\,d^2\,\ln \left (F\right )}}+\frac {63\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^4\,\left (8\,b^2\,c^5\,d^3\,{\ln \left (F\right )}^2-10\,b\,c^3\,d^3\,\ln \left (F\right )+5\,c\,d^3\right )}{8\,b^3\,{\ln \left (F\right )}^3}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^8\,x^9}{2\,b\,\ln \left (F\right )}-\frac {9\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^2\,\left (-32\,d\,b^3\,c^7\,{\ln \left (F\right )}^3+84\,d\,b^2\,c^5\,{\ln \left (F\right )}^2-140\,d\,b\,c^3\,\ln \left (F\right )+105\,d\,c\right )}{16\,b^4\,{\ln \left (F\right )}^4}+\frac {63\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^5\,\left (8\,b^2\,c^4\,d^4\,{\ln \left (F\right )}^2-6\,b\,c^2\,d^4\,\ln \left (F\right )+d^4\right )}{8\,b^3\,{\ln \left (F\right )}^3}-\frac {21\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^6\,\left (3\,c\,d^5-8\,b\,c^3\,d^5\,\ln \left (F\right )\right )}{4\,b^2\,{\ln \left (F\right )}^2}-\frac {21\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^3\,\left (-32\,b^3\,c^6\,d^2\,{\ln \left (F\right )}^3+60\,b^2\,c^4\,d^2\,{\ln \left (F\right )}^2-60\,b\,c^2\,d^2\,\ln \left (F\right )+15\,d^2\right )}{16\,b^4\,{\ln \left (F\right )}^4}+\frac {9\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (16\,b^4\,c^8\,{\ln \left (F\right )}^4-56\,b^3\,c^6\,{\ln \left (F\right )}^3+140\,b^2\,c^4\,{\ln \left (F\right )}^2-210\,b\,c^2\,\ln \left (F\right )+105\right )}{32\,b^5\,{\ln \left (F\right )}^5}+\frac {9\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,c\,d^7\,x^8}{2\,b\,\ln \left (F\right )}+\frac {9\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^6\,x^7\,\left (8\,b\,c^2\,\ln \left (F\right )-1\right )}{4\,b^2\,{\ln \left (F\right )}^2} \]
(F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*((945*c)/32 - (315*b*c^3*log(F) )/16 + (63*b^2*c^5*log(F)^2)/8 - (9*b^3*c^7*log(F)^3)/4 + (b^4*c^9*log(F)^ 4)/2))/(b^5*d*log(F)^5) - (945*F^a*pi^(1/2)*erfi((b*c*d*log(F) + b*d^2*x*l og(F))/(b*d^2*log(F))^(1/2)))/(64*b^5*log(F)^5*(b*d^2*log(F))^(1/2)) + (63 *F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x^4*(5*c*d^3 + 8*b^2*c^5*d^3*lo g(F)^2 - 10*b*c^3*d^3*log(F)))/(8*b^3*log(F)^3) + (F^(b*d^2*x^2)*F^a*F^(b* c^2)*F^(2*b*c*d*x)*d^8*x^9)/(2*b*log(F)) - (9*F^(b*d^2*x^2)*F^a*F^(b*c^2)* F^(2*b*c*d*x)*x^2*(105*c*d + 84*b^2*c^5*d*log(F)^2 - 32*b^3*c^7*d*log(F)^3 - 140*b*c^3*d*log(F)))/(16*b^4*log(F)^4) + (63*F^(b*d^2*x^2)*F^a*F^(b*c^2 )*F^(2*b*c*d*x)*x^5*(d^4 + 8*b^2*c^4*d^4*log(F)^2 - 6*b*c^2*d^4*log(F)))/( 8*b^3*log(F)^3) - (21*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x^6*(3*c*d ^5 - 8*b*c^3*d^5*log(F)))/(4*b^2*log(F)^2) - (21*F^(b*d^2*x^2)*F^a*F^(b*c^ 2)*F^(2*b*c*d*x)*x^3*(15*d^2 + 60*b^2*c^4*d^2*log(F)^2 - 32*b^3*c^6*d^2*lo g(F)^3 - 60*b*c^2*d^2*log(F)))/(16*b^4*log(F)^4) + (9*F^(b*d^2*x^2)*F^a*F^ (b*c^2)*F^(2*b*c*d*x)*x*(140*b^2*c^4*log(F)^2 - 210*b*c^2*log(F) - 56*b^3* c^6*log(F)^3 + 16*b^4*c^8*log(F)^4 + 105))/(32*b^5*log(F)^5) + (9*F^(b*d^2 *x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*c*d^7*x^8)/(2*b*log(F)) + (9*F^(b*d^2*x^ 2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*d^6*x^7*(8*b*c^2*log(F) - 1))/(4*b^2*log(F) ^2)