Integrand size = 28, antiderivative size = 397 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} h (e g-d h) \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\frac {1}{n}+a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )}{b \sqrt {f} \sqrt {\log (F)}}\right )}{b e^3 \sqrt {f} n \sqrt {\log (F)}}+\frac {e^{-\frac {1+4 a b f n \log (F)}{4 b^2 f n^2 \log (F)}} (e g-d h)^2 \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\frac {1}{n}+2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e^3 \sqrt {f} n \sqrt {\log (F)}}+\frac {e^{-\frac {3 (3+4 a b f n \log (F))}{4 b^2 f n^2 \log (F)}} h^2 \sqrt {\pi } (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \text {erfi}\left (\frac {\frac {3}{n}+2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e^3 \sqrt {f} n \sqrt {\log (F)}} \]
h*(-d*h+e*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^3/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)+1/2*(-d*h+e*g)^2*(e*x+d)* erfi(1/2*(1/n+2*a*b*f*ln(F)+2*b^2*f*ln(F)*ln(c*(e*x+d)^n))/b/f^(1/2)/ln(F) ^(1/2))*Pi^(1/2)/b/e^3/exp(1/4*(1+4*a*b*f*n*ln(F))/b^2/f/n^2/ln(F))/n/((c* (e*x+d)^n)^(1/n))/f^(1/2)/ln(F)^(1/2)+1/2*h^2*(e*x+d)^3*erfi(1/2*(3/n+2*a* b*f*ln(F)+2*b^2*f*ln(F)*ln(c*(e*x+d)^n))/b/f^(1/2)/ln(F)^(1/2))*Pi^(1/2)/b /e^3/exp(3/4*(3+4*a*b*f*n*ln(F))/b^2/f/n^2/ln(F))/n/((c*(e*x+d)^n)^(3/n))/ f^(1/2)/ln(F)^(1/2)
Time = 0.40 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.83 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\frac {e^{-\frac {3 (3+4 a b f n \log (F))}{4 b^2 f n^2 \log (F)}} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-3/n} \left (-2 e^{\frac {5+4 a b f n \log (F)}{4 b^2 f n^2 \log (F)}} h (-e g+d h) (d+e x) \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1+b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{b \sqrt {f} n \sqrt {\log (F)}}\right )+e^{\frac {2+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} (e g-d h)^2 \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )+h^2 (d+e x)^2 \text {erfi}\left (\frac {3+2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )\right )}{2 b e^3 \sqrt {f} n \sqrt {\log (F)}} \]
(Sqrt[Pi]*(d + e*x)*(-2*E^((5 + 4*a*b*f*n*Log[F])/(4*b^2*f*n^2*Log[F]))*h* (-(e*g) + d*h)*(d + e*x)*(c*(d + e*x)^n)^n^(-1)*Erfi[(1 + b*f*n*Log[F]*(a + b*Log[c*(d + e*x)^n]))/(b*Sqrt[f]*n*Sqrt[Log[F]])] + E^((2 + 2*a*b*f*n*L og[F])/(b^2*f*n^2*Log[F]))*(e*g - d*h)^2*(c*(d + e*x)^n)^(2/n)*Erfi[(1 + 2 *b*f*n*Log[F]*(a + b*Log[c*(d + e*x)^n]))/(2*b*Sqrt[f]*n*Sqrt[Log[F]])] + h^2*(d + e*x)^2*Erfi[(3 + 2*b*f*n*Log[F]*(a + b*Log[c*(d + e*x)^n]))/(2*b* Sqrt[f]*n*Sqrt[Log[F]])]))/(2*b*e^3*E^((3*(3 + 4*a*b*f*n*Log[F]))/(4*b^2*f *n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(3/n)*Sqrt[Log[F]])
Time = 0.85 (sec) , antiderivative size = 392, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2713, 2009}
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.
| \(\displaystyle \int (g+h x)^2 F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 2713 |
| \(\displaystyle \frac {\int \left ((e g-d h)^2 F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}+h^2 (d+e x)^2 F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}+2 h (e g-d h) (d+e x) F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}\right )d(d+e x)}{e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sqrt {\pi } h^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \exp \left (-\frac {3 (4 a b f n \log (F)+3)}{4 b^2 f n^2 \log (F)}\right ) \text {erfi}\left (\frac {2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {3}{n}}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}+\frac {\sqrt {\pi } h (d+e x)^2 (e g-d h) \left (c (d+e x)^n\right )^{-2/n} e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \text {erfi}\left (\frac {a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{b \sqrt {f} \sqrt {\log (F)}}\right )}{b \sqrt {f} n \sqrt {\log (F)}}+\frac {\sqrt {\pi } (d+e x) (e g-d h)^2 \left (c (d+e x)^n\right )^{-1/n} e^{-\frac {4 a b f n \log (F)+1}{4 b^2 f n^2 \log (F)}} \text {erfi}\left (\frac {2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}}{e^3}\) |
((h*(e*g - d*h)*Sqrt[Pi]*(d + e*x)^2*Erfi[(n^(-1) + a*b*f*Log[F] + b^2*f*L og[F]*Log[c*(d + e*x)^n])/(b*Sqrt[f]*Sqrt[Log[F]])])/(b*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]]) + ((e*g - d*h)^2*Sqrt[Pi]*(d + e*x)*Erfi[(n^(-1) + 2*a*b*f*Log[F] + 2*b^2* f*Log[F]*Log[c*(d + e*x)^n])/(2*b*Sqrt[f]*Sqrt[Log[F]])])/(2*b*E^((1 + 4*a *b*f*n*Log[F])/(4*b^2*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt [Log[F]]) + (h^2*Sqrt[Pi]*(d + e*x)^3*Erfi[(3/n + 2*a*b*f*Log[F] + 2*b^2*f *Log[F]*Log[c*(d + e*x)^n])/(2*b*Sqrt[f]*Sqrt[Log[F]])])/(2*b*E^((3*(3 + 4 *a*b*f*n*Log[F]))/(4*b^2*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(3/n)*Sq rt[Log[F]]))/e^3
3.7.11.3.1 Defintions of rubi rules used
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[1/e^(m + 1) Subst[Int[ExpandI ntegrand[F^(f*(a + b*Log[c*x^n])^2), (e*g - d*h + h*x)^m, x], x], x, d + e* x], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, n}, x] && IGtQ[m, 0]
\[\int F^{f {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}} \left (h x +g \right )^{2}d x\]
Time = 0.31 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.03 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} h^{2} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 3\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {3 \, {\left (4 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) + 3\right )}}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )} + \sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} {\left (e^{2} g^{2} - 2 \, d e g h + d^{2} h^{2}\right )} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {4 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) + 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )} + 2 \, \sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} {\left (e g h - d h^{2}\right )} \operatorname {erf}\left (\frac {{\left (b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + b^{2} f n \log \left (F\right ) \log \left (c\right ) + a b f n \log \left (F\right ) + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 1}{b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e^{3} n} \]
-1/2*(sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*h^2*erf(1/2*(2*b^2*f*n^2*log(e*x + d)*log(F) + 2*b^2*f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + 3)*sqrt(-b^2*f*n^ 2*log(F))/(b^2*f*n^2*log(F)))*e^(-3/4*(4*b^2*f*n*log(F)*log(c) + 4*a*b*f*n *log(F) + 3)/(b^2*f*n^2*log(F))) + sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*(e^2*g ^2 - 2*d*e*g*h + d^2*h^2)*erf(1/2*(2*b^2*f*n^2*log(e*x + d)*log(F) + 2*b^2 *f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + 1)*sqrt(-b^2*f*n^2*log(F))/(b^2*f* n^2*log(F)))*e^(-1/4*(4*b^2*f*n*log(F)*log(c) + 4*a*b*f*n*log(F) + 1)/(b^2 *f*n^2*log(F))) + 2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*(e*g*h - d*h^2)*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^3*n)
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\text {Timed out} \]
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\text {Timed out} \]
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\int { {\left (h x + g\right )}^{2} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^2 \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,{\left (g+h\,x\right )}^2 \,d x \]