Integrand size = 29, antiderivative size = 122 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} g \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\frac {1}{n}+a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
1/2*g*(e*x+d)^2*erfi((1/n+a*b*f*ln(F)+b^2*f*ln(F)*ln(c*(e*x+d)^n))/b/f^(1/ 2)/ln(F)^(1/2))*Pi^(1/2)/b/e/exp((1+2*a*b*f*n*ln(F))/b^2/f/n^2/ln(F))/n/(( c*(e*x+d)^n)^(2/n))/f^(1/2)/ln(F)^(1/2)
Time = 0.72 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} g \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {1+b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
(g*Sqrt[Pi]*(d + e*x)^2*Erfi[(1 + b*f*n*Log[F]*(a + b*Log[c*(d + e*x)^n])) /(b*Sqrt[f]*n*Sqrt[Log[F]])])/(2*b*e*E^((1 + 2*a*b*f*n*Log[F])/(b^2*f*n^2* Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(2/n)*Sqrt[Log[F]])
Time = 0.51 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int (d g+e g x) F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 2712 |
| \(\displaystyle g (d+e x)^{-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)} \int F^{f a^2+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{2 a b f n \log (F)+1}dx\) |
| \(\Big \downarrow \) 2706 |
| \(\displaystyle \frac {g (d+e x)^2 \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 (a b f n \log (F)+1)}{n}} \int F^{f a^2+2 b f \log \left (c (d+e x)^n\right ) a+b^2 f \log ^2\left (c (d+e x)^n\right )} \left (c (d+e x)^n\right )^{2/n}d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2725 |
| \(\displaystyle \frac {g (d+e x)^2 \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 (a b f n \log (F)+1)}{n}} \int \exp \left (f \log (F) a^2+b^2 f \log (F) \log ^2\left (c (d+e x)^n\right )+2 \left (a b f \log (F)+\frac {1}{n}\right ) \log \left (c (d+e x)^n\right )\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2664 |
| \(\displaystyle \frac {g (d+e x)^2 e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 (a b f n \log (F)+1)}{n}} \int \exp \left (\frac {\left (f \log (F) \log \left (c (d+e x)^n\right ) b^2+a f \log (F) b+\frac {1}{n}\right )^2}{b^2 f \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle \frac {\sqrt {\pi } g (d+e x)^2 e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2 (a b f n \log (F)+1)}{n}} \text {erfi}\left (\frac {a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}}\) |
(g*Sqrt[Pi]*(d + e*x)^2*(c*(d + e*x)^n)^(2*a*b*f*Log[F] - (2*(1 + a*b*f*n* Log[F]))/n)*Erfi[(n^(-1) + a*b*f*Log[F] + b^2*f*Log[F]*Log[c*(d + e*x)^n]) /(b*Sqrt[f]*Sqrt[Log[F]])])/(2*b*e*E^((1 + 2*a*b*f*n*Log[F])/(b^2*f*n^2*Lo g[F]))*Sqrt[f]*n*Sqrt[Log[F]])
3.7.4.3.1 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]
\[\int F^{f {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}} \left (e g x +d g \right )d x\]
Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} g \operatorname {erf}\left (\frac {{\left (b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + b^{2} f n \log \left (F\right ) \log \left (c\right ) + a b f n \log \left (F\right ) + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 1}{b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e n} \]
-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*g*erf((b^2*f*n^2*log(e*x + d)*log(F) + b^2*f*n*log(F)*log(c) + a*b*f*n*log(F) + 1)*sqrt(-b^2*f*n^2*log(F))/(b^ 2*f*n^2*log(F)))*e^(-(2*b^2*f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + 1)/(b^2 *f*n^2*log(F)))/(b*e*n)
Leaf count of result is larger than twice the leaf count of optimal. 724 vs. \(2 (116) = 232\).
Time = 65.99 (sec) , antiderivative size = 724, normalized size of antiderivative = 5.93 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\begin {cases} \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b d^{2} f g n \log {\left (F \right )}}{e} - F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b d f g n x \log {\left (F \right )} - \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b e f g n x^{2} \log {\left (F \right )}}{2} - \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d^{2} f g n^{2} \log {\left (F \right )}}{2 e} - \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d^{2} f g n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e} + \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d f g n^{2} x \log {\left (F \right )}}{2} - F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d f g n x \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )} + \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} e f g n^{2} x^{2} \log {\left (F \right )}}{4} - \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} e f g n x^{2} \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} + \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d^{2} g}{2 e} + F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d g x + \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} e g x^{2}}{2} & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}\right )^{2}} d g x & \text {otherwise} \end {cases} \]
Piecewise((F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e* x)**n)**2)*a*b*d**2*f*g*n*log(F)/e - F**(a**2*f + 2*a*b*f*log(c*(d + e*x)* *n) + b**2*f*log(c*(d + e*x)**n)**2)*a*b*d*f*g*n*x*log(F) - F**(a**2*f + 2 *a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*a*b*e*f*g*n*x* *2*log(F)/2 - F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*d**2*f*g*n**2*log(F)/(2*e) - F**(a**2*f + 2*a*b*f*log(c *(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*d**2*f*g*n*log(F)*log (c*(d + e*x)**n)/(2*e) + F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f *log(c*(d + e*x)**n)**2)*b**2*d*f*g*n**2*x*log(F)/2 - F**(a**2*f + 2*a*b*f *log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*d*f*g*n*x*log(F )*log(c*(d + e*x)**n) + F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f* log(c*(d + e*x)**n)**2)*b**2*e*f*g*n**2*x**2*log(F)/4 - F**(a**2*f + 2*a*b *f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*e*f*g*n*x**2* log(F)*log(c*(d + e*x)**n)/2 + F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*d**2*g/(2*e) + F**(a**2*f + 2*a*b*f*log(c*( d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*d*g*x + F**(a**2*f + 2*a*b*f *log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*e*g*x**2/2, Ne(e, 0) ), (F**(f*(a + b*log(c*d**n))**2)*d*g*x, True))
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\int { {\left (e g x + d g\right )} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\int { {\left (e g x + d g\right )} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,\left (d\,g+e\,g\,x\right ) \,d x \]