Integrand size = 31, antiderivative size = 70 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\frac {\sqrt {\pi } \text {erfi}\left (a \sqrt {f} \sqrt {\log (F)}+b \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \] Output:
1/2*Pi^(1/2)*erfi(a*f^(1/2)*ln(F)^(1/2)+b*f^(1/2)*ln(F)^(1/2)*ln(c*(e*x+d) ^n))/b/e/f^(1/2)/g/n/ln(F)^(1/2)
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {f} \sqrt {\log (F)} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \] Input:
Integrate[F^(f*(a + b*Log[c*(d + e*x)^n])^2)/(d*g + e*g*x),x]
Output:
(Sqrt[Pi]*Erfi[Sqrt[f]*Sqrt[Log[F]]*(a + b*Log[c*(d + e*x)^n])])/(2*b*e*Sq rt[f]*g*n*Sqrt[Log[F]])
Time = 0.55 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2712, 2706, 2664, 2633}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx\) |
\(\Big \downarrow \) 2712 |
\(\displaystyle \frac {(d+e x)^{-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)} \int F^{f a^2+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{2 a b f n \log (F)-1}dx}{g}\) |
\(\Big \downarrow \) 2706 |
\(\displaystyle \frac {\int F^{f a^2+2 b f \log \left (c (d+e x)^n\right ) a+b^2 f \log ^2\left (c (d+e x)^n\right )}d\log \left (c (d+e x)^n\right )}{e g n}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle \frac {\int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}d\log \left (c (d+e x)^n\right )}{e g n}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {\sqrt {\pi } \text {erfi}\left (\sqrt {f} \sqrt {\log (F)} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}}\) |
Input:
Int[F^(f*(a + b*Log[c*(d + e*x)^n])^2)/(d*g + e*g*x),x]
Output:
(Sqrt[Pi]*Erfi[Sqrt[f]*Sqrt[Log[F]]*(a + b*Log[c*(d + e*x)^n])])/(2*b*e*Sq rt[f]*g*n*Sqrt[Log[F]])
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]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(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*Log[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]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.94 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.79
method | result | size |
risch | \(-\frac {\sqrt {\pi }\, \operatorname {erf}\left (-b \sqrt {-f \ln \left (F \right )}\, \ln \left (\left (e x +d \right )^{n}\right )+\frac {f \left (a +b \left (\ln \left (c \right )-\frac {i \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \left (-\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )+\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )\right )}{2}\right )\right ) \ln \left (F \right )}{\sqrt {-f \ln \left (F \right )}}\right )}{2 g e n b \sqrt {-f \ln \left (F \right )}}\) | \(125\) |
Input:
int(F^(f*(a+b*ln(c*(e*x+d)^n))^2)/(e*g*x+d*g),x,method=_RETURNVERBOSE)
Output:
-1/2/g/e/n*Pi^(1/2)/b/(-f*ln(F))^(1/2)*erf(-b*(-f*ln(F))^(1/2)*ln((e*x+d)^ n)+f*(a+b*(ln(c)-1/2*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)/(-f*ln(F))^(1/2))
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {\sqrt {-b^{2} f n^{2} \log \left (F\right )} {\left (b n \log \left (e x + d\right ) + b \log \left (c\right ) + a\right )}}{b n}\right )}{2 \, b e g n} \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)/(e*g*x+d*g),x, algorithm="fricas" )
Output:
-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*erf(sqrt(-b^2*f*n^2*log(F))*(b*n*log (e*x + d) + b*log(c) + a)/(b*n))/(b*e*g*n)
\[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\frac {\int \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}}}{d + e x}\, dx}{g} \] Input:
integrate(F**(f*(a+b*ln(c*(e*x+d)**n))**2)/(e*g*x+d*g),x)
Output:
Integral(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*x), x)/g
\[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\int { \frac {F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}}{e g x + d g} \,d x } \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)/(e*g*x+d*g),x, algorithm="maxima" )
Output:
integrate(F^((b*log((e*x + d)^n*c) + a)^2*f)/(e*g*x + d*g), x)
\[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\int { \frac {F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}}{e g x + d g} \,d x } \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)/(e*g*x+d*g),x, algorithm="giac")
Output:
integrate(F^((b*log((e*x + d)^n*c) + a)^2*f)/(e*g*x + d*g), x)
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=-\frac {\sqrt {\pi }\,\mathrm {erf}\left (\frac {1{}\mathrm {i}\,f\,\ln \left (F\right )\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,b^2+1{}\mathrm {i}\,a\,f\,\ln \left (F\right )\,b}{\sqrt {b^2\,f\,\ln \left (F\right )}}\right )\,1{}\mathrm {i}}{2\,e\,g\,n\,\sqrt {b^2\,f\,\ln \left (F\right )}} \] Input:
int(F^(f*(a + b*log(c*(d + e*x)^n))^2)/(d*g + e*g*x),x)
Output:
-(pi^(1/2)*erf((b^2*f*log(F)*log(c*(d + e*x)^n)*1i + a*b*f*log(F)*1i)/(b^2 *f*log(F))^(1/2))*1i)/(2*e*g*n*(b^2*f*log(F))^(1/2))
\[ \int \frac {F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2}}{d g+e g x} \, dx=\frac {f^{a^{2} f} \left (\int \frac {f^{\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b^{2} f +2 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) a b f}}{e x +d}d x \right )}{g} \] Input:
int(F^(f*(a+b*log(c*(e*x+d)^n))^2)/(e*g*x+d*g),x)
Output:
(f**(a**2*f)*int(f**(log((d + e*x)**n*c)**2*b**2*f + 2*log((d + e*x)**n*c) *a*b*f)/(d + e*x),x))/g