\(\int \frac {F^{f (a+b \log ^2(c (d+e x)^n))}}{d g+e g x} \, dx\) [591]

Optimal result

Integrand size = 31, antiderivative size = 67 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{d g+e g x} \, dx=\frac {F^{a f} \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2308, 2235} \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{d g+e g x} \, dx=\frac {\sqrt {\pi } F^{a f} \text {erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{d g+e g x} \, dx=\frac {F^{a f} \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.22 (sec) , antiderivative size = 382, normalized size of antiderivative = 5.70

method	result	size
risch	\(\frac {\sqrt {\pi }\, F^{f \left (-b \,\pi ^{2}+b \,\pi ^{2} \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )+b \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi ^{2} \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )-b \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi ^{2} \operatorname {csgn}\left (i c \right )+i b \ln \left (c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi -i b \ln \left (c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \ln \left (c \right ) \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+i b \ln \left (c \right ) \pi \,\operatorname {csgn}\left (i c \right )+b \ln \left (c \right )^{2}+a \right )} F^{-\frac {f b {\left (i \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \pi +i \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )-i \pi \,\operatorname {csgn}\left (i c \right )-2 \ln \left (c \right )\right )}^{2}}{4}} \operatorname {erf}\left (\sqrt {-\ln \left (F \right ) b f}\, \ln \left (\left (e x +d \right )^{n}\right )-\frac {f b \left (2 \ln \left (c \right )-i \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \left (-\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )\right )\right ) \ln \left (F \right )}{2 \sqrt {-\ln \left (F \right ) b f}}\right )}{2 g e n \sqrt {-\ln \left (F \right ) b f}}\)	\(382\)

[In]

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

[Out]

1/2/g/e/n*Pi^(1/2)*F^(f*(-b*Pi^2+b*Pi^2*csgn(I*c*(e*x+d)^n)*csgn(I*c)+b*csgn(I*(e*x+d)^n)*Pi^2*csgn(I*c*(e*x+d
)^n)-b*csgn(I*(e*x+d)^n)*Pi^2*csgn(I*c)+I*b*ln(c)*csgn(I*(e*x+d)^n)*Pi-I*b*ln(c)*csgn(I*(e*x+d)^n)*Pi*csgn(I*c
*(e*x+d)^n)*csgn(I*c)-I*b*ln(c)*Pi*csgn(I*c*(e*x+d)^n)+I*b*ln(c)*Pi*csgn(I*c)+b*ln(c)^2+a))*F^(-1/4*f*b*(I*csg
n(I*(e*x+d)^n)*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)-I*csgn(I*(e*x+d)^n)*Pi+I*Pi*csgn(I*c*(e*x+d)^n)-I*Pi*csgn(I*c)
-2*ln(c))^2)/(-ln(F)*b*f)^(1/2)*erf((-ln(F)*b*f)^(1/2)*ln((e*x+d)^n)-1/2*f*b*(2*ln(c)-I*Pi*csgn(I*c*(e*x+d)^n)
*(-csgn(I*c*(e*x+d)^n)+csgn(I*c))*(-csgn(I*c*(e*x+d)^n)+csgn(I*(e*x+d)^n)))*ln(F)/(-ln(F)*b*f)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{d g+e g x} \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} F^{a f} \operatorname {erf}\left (\frac {\sqrt {-b f n^{2} \log \left (F\right )} {\left (n \log \left (e x + d\right ) + \log \left (c\right )\right )}}{n}\right )}{2 \, e g n} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{d g+e g x} \, dx=\frac {F^{a\,f}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,f\,\ln \left (F\right )\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {b\,f\,\ln \left (F\right )}}\right )}{2\,e\,g\,n\,\sqrt {b\,f\,\ln \left (F\right )}} \]

[In]

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

[Out]

(F^(a*f)*pi^(1/2)*erfi((b*f*log(F)*log(c*(d + e*x)^n))/(b*f*log(F))^(1/2)))/(2*e*g*n*(b*f*log(F))^(1/2))