Integrand size = 31, antiderivative size = 153 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\frac {e^{-\frac {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}} F^{a^2 f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^m \text {erfi}\left (\frac {1+m+2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
1/2*F^(a^2*f)*(e*x+d)*(e*g*x+d*g)^m*erfi(1/2*(1+m+2*a*b*f*n*ln(F)+2*b^2*f* n*ln(F)*ln(c*(e*x+d)^n))/b/n/f^(1/2)/ln(F)^(1/2))*Pi^(1/2)/b/e/exp(1/4*(1+ m+2*a*b*f*n*ln(F))^2/b^2/f/n^2/ln(F))/n/((c*(e*x+d)^n)^((1+m)/n))/f^(1/2)/ ln(F)^(1/2)
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx \]
Time = 0.59 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2712, 2706, 2725, 2664, 2633}
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.
| \(\displaystyle \int (d g+e g x)^m F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 2712 |
| \(\displaystyle (d g+e g x)^m \left (c (d+e x)^n\right )^{2 a b f \log (F)} (d+e x)^{-2 a b f n \log (F)-m} \int F^{f a^2+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{m+2 a b f n \log (F)}dx\) |
| \(\Big \downarrow \) 2706 |
| \(\displaystyle \frac {(d+e x) (d g+e g x)^m \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 a b f n \log (F)+m+1}{n}} \int F^{f a^2+2 b f \log \left (c (d+e x)^n\right ) a+b^2 f \log ^2\left (c (d+e x)^n\right )} \left (c (d+e x)^n\right )^{\frac {m+1}{n}}d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2725 |
| \(\displaystyle \frac {(d+e x) (d g+e g x)^m \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 a b f n \log (F)+m+1}{n}} \int \exp \left (f \log (F) a^2+b^2 f \log (F) \log ^2\left (c (d+e x)^n\right )+\frac {(m+2 a b f n \log (F)+1) \log \left (c (d+e x)^n\right )}{n}\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2664 |
| \(\displaystyle \frac {F^{a^2 f} (d+e x) (d g+e g x)^m \exp \left (-\frac {(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 a b f n \log (F)+m+1}{n}} \int \exp \left (\frac {\left (2 f n \log (F) \log \left (c (d+e x)^n\right ) b^2+2 a f n \log (F) b+m+1\right )^2}{4 b^2 f n^2 \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle \frac {\sqrt {\pi } F^{a^2 f} (d+e x) (d g+e g x)^m \exp \left (-\frac {(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 a b f n \log (F)+m+1}{n}} \text {erfi}\left (\frac {2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}}\) |
(F^(a^2*f)*Sqrt[Pi]*(d + e*x)*(c*(d + e*x)^n)^(2*a*b*f*Log[F] - (1 + m + 2 *a*b*f*n*Log[F])/n)*(d*g + e*g*x)^m*Erfi[(1 + m + 2*a*b*f*n*Log[F] + 2*b^2 *f*n*Log[F]*Log[c*(d + e*x)^n])/(2*b*Sqrt[f]*n*Sqrt[Log[F]])])/(2*b*e*E^(( 1 + m + 2*a*b*f*n*Log[F])^2/(4*b^2*f*n^2*Log[F]))*Sqrt[f]*n*Sqrt[Log[F]])
3.7.2.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)) Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Log[F] *x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(g + h*x)^m*((c*(d + e*x)^n)^(2 *a*b*f*Log[F])/(d + e*x)^(m + 2*a*b*f*n*Log[F]))*Int[(d + e*x)^(m + 2*a*b*f *n*Log[F])*F^(a^2*f + b^2*f*Log[c*(d + e*x)^n]^2), x], x] /; FreeQ[{F, a, b , c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]
Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
Timed out.
\[\int F^{f {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}} \left (e g x +d g \right )^{m}d x\]
Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + m + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, b^{2} f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \, {\left (b^{2} f m + b^{2} f\right )} n \log \left (F\right ) \log \left (c\right ) - 4 \, {\left (a b f m + a b f\right )} n \log \left (F\right ) - m^{2} - 2 \, m - 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e n} \]
-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*erf(1/2*(2*b^2*f*n^2*log(e*x + d)*lo g(F) + 2*b^2*f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + m + 1)*sqrt(-b^2*f*n^2 *log(F))/(b^2*f*n^2*log(F)))*e^(1/4*(4*b^2*f*m*n^2*log(F)*log(g) - 4*(b^2* f*m + b^2*f)*n*log(F)*log(c) - 4*(a*b*f*m + a*b*f)*n*log(F) - m^2 - 2*m - 1)/(b^2*f*n^2*log(F)))/(b*e*n)
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\int F^{f \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}} \left (g \left (d + e x\right )\right )^{m}\, dx \]
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\int { {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\int { {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,{\left (d\,g+e\,g\,x\right )}^m \,d x \]