\(\int \frac {F^{f (a+b \log (c (d+e x)^n))^2}}{(d g+e g x)^2} \, dx\) [607]

Optimal result

Integrand size = 31, antiderivative size = 128 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{(d g+e g x)^2} \, dx=-\frac {e^{\frac {a}{b n}-\frac {1}{4 b^2 f n^2 \log (F)}} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {\frac {1}{n}-2 a b f \log (F)-2 b^2 f \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}} \]

[Out]

1/2*exp(a/b/n-1/4/b^2/f/n^2/ln(F))*(c*(e*x+d)^n)^(1/n)*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/g^2/n/(e*x+d)/f^(1/2)/ln(F)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 128, 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} \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{(d g+e g x)^2} \, dx=-\frac {\sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} e^{\frac {a}{b n}-\frac {1}{4 b^2 f n^2 \log (F)}} \text {erfi}\left (\frac {-2 a b f \log (F)-2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} g^2 n \sqrt {\log (F)} (d+e x)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{(d g+e g x)^2} \, dx=\frac {e^{\frac {-1+4 a b f n \log (F)}{4 b^2 f n^2 \log (F)}} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {-1+2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}} \]

[In]

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

[Out]

(E^((-1 + 4*a*b*f*n*Log[F])/(4*b^2*f*n^2*Log[F]))*Sqrt[Pi]*(c*(d + e*x)^n)^n^(-1)*Erfi[(-1 + 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*Sqrt[f]*g^2*n*(d + e*x)*Sqrt[Log[F]])

Maple [F]

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

[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 = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.05 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{(d g+e g x)^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) - 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) - 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e g^{2} n} \]

[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))*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)))/(b*e*g^2*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (112) = 224\).

Time = 151.13 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.31 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{(d g+e g x)^2} \, dx=\begin {cases} - \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b f n \log {\left (F \right )}}{d e g^{2} + e^{2} g^{2} x} - \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} f n^{2} \log {\left (F \right )}}{d e g^{2} + e^{2} g^{2} x} - \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} f n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{d e g^{2} + e^{2} g^{2} x} - \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}}}{d e g^{2} + e^{2} g^{2} x} & \text {for}\: e \neq 0 \\\frac {F^{f \left (a + b \log {\left (c d^{n} \right )}\right )^{2}} x}{d^{2} g^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-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)*a*b*f*n*log(F)/(d*e*g*
*2 + e**2*g**2*x) - 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*f*n**2*lo
g(F)/(d*e*g**2 + e**2*g**2*x) - 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*f*n*log(F)*log(c*(d + e*x)**n)/(d*e*g**2 + e**2*g**2*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)/(d*e*g**2 + e**2*g**2*x), Ne(e, 0)), (F**(f*(a + b*log(c*d**n))**2)*x/(d**2*g**2), Tr
ue))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)