Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {F^a \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{9/2}}{2 d (c+d x)^9} \]
-1/2*F^a*(-32/945*Pi^(1/2)*erfc((-b*(d*x+c)^2*ln(F))^(1/2))+32/945/(-b*(d* x+c)^2*ln(F))^(1/2)*exp(b*(d*x+c)^2*ln(F))-16/945/(-b*(d*x+c)^2*ln(F))^(3/ 2)*exp(b*(d*x+c)^2*ln(F))+8/315/(-b*(d*x+c)^2*ln(F))^(5/2)*exp(b*(d*x+c)^2 *ln(F))-4/63/(-b*(d*x+c)^2*ln(F))^(7/2)*exp(b*(d*x+c)^2*ln(F))+2/9/(-b*(d* x+c)^2*ln(F))^(9/2)*exp(b*(d*x+c)^2*ln(F)))*(-b*(d*x+c)^2*ln(F))^(9/2)/d/( d*x+c)^9
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {F^a \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{9/2}}{2 d (c+d x)^9} \]
-1/2*(F^a*Gamma[-9/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^( 9/2))/(d*(c + d*x)^9)
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.
\(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle -\frac {F^a \left (-b \log (F) (c+d x)^2\right )^{9/2} \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right )}{2 d (c+d x)^9}\) |
-1/2*(F^a*Gamma[-9/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^( 9/2))/(d*(c + d*x)^9)
3.3.78.3.1 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 = 0.84 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.98
method | result | size |
risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{9 d \left (d x +c \right )^{9}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{63 d \left (d x +c \right )^{7}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{315 d \left (d x +c \right )^{5}}-\frac {8 b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{945 d \left (d x +c \right )^{3}}-\frac {16 b^{4} \ln \left (F \right )^{4} F^{b \left (d x +c \right )^{2}} F^{a}}{945 d \left (d x +c \right )}+\frac {16 b^{5} \ln \left (F \right )^{5} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{945 d \sqrt {-b \ln \left (F \right )}}\) | \(195\) |
-1/9/d/(d*x+c)^9*F^(b*(d*x+c)^2)*F^a-2/63/d*b*ln(F)/(d*x+c)^7*F^(b*(d*x+c) ^2)*F^a-4/315/d*b^2*ln(F)^2/(d*x+c)^5*F^(b*(d*x+c)^2)*F^a-8/945/d*b^3*ln(F )^3/(d*x+c)^3*F^(b*(d*x+c)^2)*F^a-16/945/d*b^4*ln(F)^4/(d*x+c)*F^(b*(d*x+c )^2)*F^a+16/945/d*b^5*ln(F)^5*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. 598 vs. \(2 (185) = 370\).
Time = 0.27 (sec) , antiderivative size = 598, normalized size of antiderivative = 12.20 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {16 \, \sqrt {\pi } {\left (b^{4} d^{9} x^{9} + 9 \, b^{4} c d^{8} x^{8} + 36 \, b^{4} c^{2} d^{7} x^{7} + 84 \, b^{4} c^{3} d^{6} x^{6} + 126 \, b^{4} c^{4} d^{5} x^{5} + 126 \, b^{4} c^{5} d^{4} x^{4} + 84 \, b^{4} c^{6} d^{3} x^{3} + 36 \, b^{4} c^{7} d^{2} x^{2} + 9 \, b^{4} c^{8} d x + b^{4} c^{9}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right )^{4} + {\left (16 \, {\left (b^{4} d^{9} x^{8} + 8 \, b^{4} c d^{8} x^{7} + 28 \, b^{4} c^{2} d^{7} x^{6} + 56 \, b^{4} c^{3} d^{6} x^{5} + 70 \, b^{4} c^{4} d^{5} x^{4} + 56 \, b^{4} c^{5} d^{4} x^{3} + 28 \, b^{4} c^{6} d^{3} x^{2} + 8 \, b^{4} c^{7} d^{2} x + b^{4} c^{8} d\right )} \log \left (F\right )^{4} + 8 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \left (F\right )^{3} + 12 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 30 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 105 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{945 \, {\left (d^{11} x^{9} + 9 \, c d^{10} x^{8} + 36 \, c^{2} d^{9} x^{7} + 84 \, c^{3} d^{8} x^{6} + 126 \, c^{4} d^{7} x^{5} + 126 \, c^{5} d^{6} x^{4} + 84 \, c^{6} d^{5} x^{3} + 36 \, c^{7} d^{4} x^{2} + 9 \, c^{8} d^{3} x + c^{9} d^{2}\right )}} \]
-1/945*(16*sqrt(pi)*(b^4*d^9*x^9 + 9*b^4*c*d^8*x^8 + 36*b^4*c^2*d^7*x^7 + 84*b^4*c^3*d^6*x^6 + 126*b^4*c^4*d^5*x^5 + 126*b^4*c^5*d^4*x^4 + 84*b^4*c^ 6*d^3*x^3 + 36*b^4*c^7*d^2*x^2 + 9*b^4*c^8*d*x + b^4*c^9)*sqrt(-b*d^2*log( F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d)*log(F)^4 + (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)*log(F)^4 + 8*(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 + 12*(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 + 30*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + 105*d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^11*x^9 + 9*c*d^10*x^8 + 36*c^2*d^9*x^7 + 84*c^3*d^8*x^6 + 126*c^4*d^7*x^5 + 126*c^5*d^6*x^4 + 84* c^6*d^5*x^3 + 36*c^7*d^4*x^2 + 9*c^8*d^3*x + c^9*d^2)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{10}}\, dx \]
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
Time = 0.81 (sec) , antiderivative size = 234, normalized size of antiderivative = 4.78 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\frac {16\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{9/2}}{945\,d\,{\left (c+d\,x\right )}^9}-\frac {16\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{9/2}}{945\,d\,{\left (c+d\,x\right )}^9}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{315\,d\,{\left (c+d\,x\right )}^5}-\frac {8\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3}{945\,d\,{\left (c+d\,x\right )}^3}-\frac {16\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^4\,{\ln \left (F\right )}^4}{945\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{63\,d\,{\left (c+d\,x\right )}^7}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{9\,d\,{\left (c+d\,x\right )}^9} \]
(16*F^a*pi^(1/2)*erfc((-b*log(F)*(c + d*x)^2)^(1/2))*(-b*log(F)*(c + d*x)^ 2)^(9/2))/(945*d*(c + d*x)^9) - (16*F^a*pi^(1/2)*(-b*log(F)*(c + d*x)^2)^( 9/2))/(945*d*(c + d*x)^9) - (4*F^a*F^(b*(c + d*x)^2)*b^2*log(F)^2)/(315*d* (c + d*x)^5) - (8*F^a*F^(b*(c + d*x)^2)*b^3*log(F)^3)/(945*d*(c + d*x)^3) - (16*F^a*F^(b*(c + d*x)^2)*b^4*log(F)^4)/(945*d*(c + d*x)) - (2*F^a*F^(b* (c + d*x)^2)*b*log(F))/(63*d*(c + d*x)^7) - (F^a*F^(b*(c + d*x)^2))/(9*d*( c + d*x)^9)