\(\int F^{f (a+b \log (c (d+e x)^n))^2} (d g+e g x)^2 \, dx\) [603]

Optimal result

Integrand size = 31, antiderivative size = 133 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\frac {e^{-\frac {3 (3+4 a b f n \log (F))}{4 b^2 f n^2 \log (F)}} g^2 \sqrt {\pi } (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \text {erfi}\left (\frac {\frac {3}{n}+2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

1/2*g^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/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)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2314, 2308, 2266, 2235} \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\frac {\sqrt {\pi } g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \exp \left (-\frac {3 (4 a b f n \log (F)+3)}{4 b^2 f n^2 \log (F)}\right ) \text {erfi}\left (\frac {2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {3}{n}}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[In]

Int[F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(d*g + e*g*x)^2,x]

[Out]

(g^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[Lo
g[F]])])/(2*b*e*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
]])

Rule 2235

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]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(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*Lo
g[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]

Rule 2314

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \left ((d+e x)^{-2-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)} (d g+e g x)^2\right ) \int F^{a^2 f+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{2+2 a b f n \log (F)} \, dx \\ & = \frac {\left ((d+e x) \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {3+2 a b f n \log (F)}{n}} (d g+e g x)^2\right ) \text {Subst}\left (\int \exp \left (a^2 f \log (F)+b^2 f x^2 \log (F)+\frac {x (3+2 a b f n \log (F))}{n}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n} \\ & = \frac {\left (\exp \left (a^2 f \log (F)-\frac {(3+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) (d+e x) \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {3+2 a b f n \log (F)}{n}} (d g+e g x)^2\right ) \text {Subst}\left (\int \exp \left (\frac {\left (2 b^2 f x \log (F)+\frac {3+2 a b f n \log (F)}{n}\right )^2}{4 b^2 f \log (F)}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n} \\ & = \frac {\exp \left (-\frac {3 (3+4 a b f n \log (F))}{4 b^2 f n^2 \log (F)}\right ) g^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 \sqrt {f} n \sqrt {\log (F)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\frac {e^{-\frac {3 \left (3+4 b f n \log (F) \left (a+b \left (-n \log (d+e x)+\log \left (c (d+e x)^n\right )\right )\right )\right )}{4 b^2 f n^2 \log (F)}} g^2 \sqrt {\pi } \text {erfi}\left (\frac {3+2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[In]

Integrate[F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(d*g + e*g*x)^2,x]

[Out]

(g^2*Sqrt[Pi]*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*E^((3
*(3 + 4*b*f*n*Log[F]*(a + b*(-(n*Log[d + e*x]) + Log[c*(d + e*x)^n]))))/(4*b^2*f*n^2*Log[F]))*Sqrt[f]*n*Sqrt[L
og[F]])

Maple [F]

\[\int F^{f {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}} \left (e g x +d g \right )^{2}d x\]

[In]

int(F^(f*(a+b*ln(c*(e*x+d)^n))^2)*(e*g*x+d*g)^2,x)

[Out]

int(F^(f*(a+b*ln(c*(e*x+d)^n))^2)*(e*g*x+d*g)^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.01 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} g^{2} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 3\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {3 \, {\left (4 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) + 3\right )}}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e n} \]

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(e*g*x+d*g)^2,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*g^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)))/(b*e*n)

Sympy [F(-1)]

Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\text {Timed out} \]

[In]

integrate(F**(f*(a+b*ln(c*(e*x+d)**n))**2)*(e*g*x+d*g)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\int { {\left (e g x + d g\right )}^{2} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(e*g*x+d*g)^2,x, algorithm="maxima")

[Out]

integrate((e*g*x + d*g)^2*F^((b*log((e*x + d)^n*c) + a)^2*f), x)

Giac [F]

\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\int { {\left (e g x + d g\right )}^{2} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(e*g*x+d*g)^2,x, algorithm="giac")

[Out]

integrate((e*g*x + d*g)^2*F^((b*log((e*x + d)^n*c) + a)^2*f), x)

Mupad [F(-1)]

Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^2 \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,{\left (d\,g+e\,g\,x\right )}^2 \,d x \]

[In]

int(F^(f*(a + b*log(c*(d + e*x)^n))^2)*(d*g + e*g*x)^2,x)

[Out]

int(exp(f*log(F)*(a + b*log(c*(d + e*x)^n))^2)*(d*g + e*g*x)^2, x)