\(\int \frac {F^{f (a+b \log ^2(c (d+e x)^n))}}{(d g+e g x)^2} \, dx\) [592]

Optimal result

Integrand size = 31, antiderivative size = 121 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^2} \, dx=-\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2308, 2266, 2235} \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^2} \, dx=-\frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n \sqrt {\log (F)} (d+e x)} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c (d+e x)^n\right )^{\frac {1}{n}} \text {Subst}\left (\int e^{-\frac {x}{n}+a f \log (F)+b f x^2 \log (F)} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g n (d g+e g x)} \\ & = \frac {\left (e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \left (c (d+e x)^n\right )^{\frac {1}{n}}\right ) \text {Subst}\left (\int e^{\frac {\left (-\frac {1}{n}+2 b f x \log (F)\right )^2}{4 b f \log (F)}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g n (d g+e g x)} \\ & = -\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^2} \, dx=\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {-1+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}} \]

[In]

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

[Out]

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

Maple [F]

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

[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 = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b f n \log \left (F\right ) \log \left (c\right ) - 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{2 \, b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \left (F\right )^{2} + 4 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{4 \, b f n^{2} \log \left (F\right )}\right )}}{2 \, e g^{2} n} \]

[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*f*n^2*log(F))*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c) - 1)*sqrt(-
b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^(1/4*(4*a*b*f^2*n^2*log(F)^2 + 4*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F))
)/(e*g^2*n)

Sympy [A] (verification not implemented)

Time = 62.64 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^2} \, dx=\begin {cases} - \frac {2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n^{2} \log {\left (F \right )}}{d e g^{2} + e^{2} g^{2} x} - \frac {2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{d e g^{2} + e^{2} g^{2} x} - \frac {F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}}}{d e g^{2} + e^{2} g^{2} x} & \text {for}\: e \neq 0 \\\frac {F^{f \left (a + b \log {\left (c d^{n} \right )}^{2}\right )} x}{d^{2} g^{2}} & \text {otherwise} \end {cases} \]

[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*f + b*f*log(c*(d + e*x)**n)**2)*b*f*n**2*log(F)/(d*e*g**2 + e**2*g**2*x) - 2*F**(a*f + b*f
*log(c*(d + e*x)**n)**2)*b*f*n*log(F)*log(c*(d + e*x)**n)/(d*e*g**2 + e**2*g**2*x) - F**(a*f + b*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), True))

Maxima [F]

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

[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)^2 + a)*f)/(e*g*x + d*g)^2, x)

Giac [F]

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

[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)^2 + a)*f)/(e*g*x + d*g)^2, x)

Mupad [F(-1)]

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

[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)