\(\int F^{f (a+b \log (c (d+e x)^n))^2} (d g+e g x)^m \, dx\) [602]

Optimal result

Integrand size = 31, antiderivative size = 153 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\frac {e^{-\frac {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}} F^{a^2 f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^m \text {erfi}\left (\frac {1+m+2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 153, 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} \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=\frac {\sqrt {\pi } F^{a^2 f} (d+e x) (d g+e g x)^m \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \exp \left (-\frac {(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \text {erfi}\left (\frac {2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[In]

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

[Out]

(F^(a^2*f)*Sqrt[Pi]*(d + e*x)*(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*(c
*(d + e*x)^n)^((1 + m)/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)^{-m-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)^m\right ) \int F^{a^2 f+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{m+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 {1+m+2 a b f n \log (F)}{n}} (d g+e g x)^m\right ) \text {Subst}\left (\int \exp \left (a^2 f \log (F)+b^2 f x^2 \log (F)+\frac {x (1+m+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 (-\frac {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) F^{a^2 f} (d+e x) \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {1+m+2 a b f n \log (F)}{n}} (d g+e g x)^m\right ) \text {Subst}\left (\int \exp \left (\frac {\left (2 b^2 f x \log (F)+\frac {1+m+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 {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) 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)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + m + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, b^{2} f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \, {\left (b^{2} f m + b^{2} f\right )} n \log \left (F\right ) \log \left (c\right ) - 4 \, {\left (a b f m + a b f\right )} n \log \left (F\right ) - m^{2} - 2 \, m - 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e n} \]

[In]

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

[Out]

-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)*log(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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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