\(\int \frac {F^{f (a+b \log ^2(c (d+e x)^n))}}{(d g+e g x)^3} \, dx\) [516]

Optimal result

Integrand size = 31, antiderivative size = 118 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=-\frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {1-b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n (d+e x)^2 \sqrt {\log (F)}} \] Output:

-1/2*F^(a*f)*Pi^(1/2)*(c*(e*x+d)^n)^(2/n)*erfi((1-b*f*n*ln(F)*ln(c*(e*x+d) 
^n))/b^(1/2)/f^(1/2)/n/ln(F)^(1/2))/b^(1/2)/e/exp(1/b/f/n^2/ln(F))/f^(1/2) 
/g^3/n/(e*x+d)^2/ln(F)^(1/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {-1+b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n (d+e x)^2 \sqrt {\log (F)}} \] Input:

Integrate[F^(f*(a + b*Log[c*(d + e*x)^n]^2))/(d*g + e*g*x)^3,x]
 

Output:

(F^(a*f)*Sqrt[Pi]*(c*(d + e*x)^n)^(2/n)*Erfi[(-1 + b*f*n*Log[F]*Log[c*(d + 
 e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/(2*Sqrt[b]*e*E^(1/(b*f*n^2*Lo 
g[F]))*Sqrt[f]*g^3*n*(d + e*x)^2*Sqrt[Log[F]])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2706, 2725, 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.

\(\displaystyle \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx\)

\(\Big \downarrow \) 2706

\(\displaystyle \frac {\left (c (d+e x)^n\right )^{2/n} \int F^{b f \log ^2\left (c (d+e x)^n\right )+a f} \left (c (d+e x)^n\right )^{-2/n}d\log \left (c (d+e x)^n\right )}{e g^3 n (d+e x)^2}\)

\(\Big \downarrow \) 2725

\(\displaystyle \frac {\left (c (d+e x)^n\right )^{2/n} \int \exp \left (b f \log (F) \log ^2\left (c (d+e x)^n\right )-\frac {2 \log \left (c (d+e x)^n\right )}{n}+a f \log (F)\right )d\log \left (c (d+e x)^n\right )}{e g^3 n (d+e x)^2}\)

\(\Big \downarrow \) 2664

\(\displaystyle \frac {F^{a f} e^{-\frac {1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{2/n} \int \exp \left (\frac {\left (1-b f n \log (F) \log \left (c (d+e x)^n\right )\right )^2}{b f n^2 \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e g^3 n (d+e x)^2}\)

\(\Big \downarrow \) 2633

\(\displaystyle -\frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {1-b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n \sqrt {\log (F)} (d+e x)^2}\)

Input:

Int[F^(f*(a + b*Log[c*(d + e*x)^n]^2))/(d*g + e*g*x)^3,x]
 

Output:

-1/2*(F^(a*f)*Sqrt[Pi]*(c*(d + e*x)^n)^(2/n)*Erfi[(1 - b*f*n*Log[F]*Log[c* 
(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/(Sqrt[b]*e*E^(1/(b*f*n^2* 
Log[F]))*Sqrt[f]*g^3*n*(d + e*x)^2*Sqrt[Log[F]])
 

Defintions of rubi rules used

rule 2633

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 2664

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]
 

rule 2706

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]
 

rule 2725

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, 
 Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, 
 x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
 

Maple [F(-1)]

Timed out.

\[\int \frac {F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )^{2}\right )}}{\left (e g x +d g \right )^{3}}d x\]

Input:

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x)
 

Output:

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + b f n \log \left (F\right ) \log \left (c\right ) - 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {a b f^{2} n^{2} \log \left (F\right )^{2} + 2 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{b f n^{2} \log \left (F\right )}\right )}}{2 \, e g^{3} n} \] Input:

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x, algorithm="frica 
s")
 

Output:

-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*erf((b*f*n^2*log(e*x + d)*log(F) + b*f 
*n*log(F)*log(c) - 1)*sqrt(-b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^((a*b*f^2* 
n^2*log(F)^2 + 2*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F)))/(e*g^3*n)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\text {Timed out} \] Input:

integrate(F**(f*(a+b*ln(c*(e*x+d)**n)**2))/(e*g*x+d*g)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\int { \frac {F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f}}{{\left (e g x + d g\right )}^{3}} \,d x } \] Input:

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x, algorithm="maxim 
a")
 

Output:

integrate(F^((b*log((e*x + d)^n*c)^2 + a)*f)/(e*g*x + d*g)^3, x)
 

Giac [F]

\[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\int { \frac {F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f}}{{\left (e g x + d g\right )}^{3}} \,d x } \] Input:

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x, algorithm="giac" 
)
 

Output:

integrate(F^((b*log((e*x + d)^n*c)^2 + a)*f)/(e*g*x + d*g)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\int \frac {{\mathrm {e}}^{f\,\ln \left (F\right )\,\left (b\,{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2+a\right )}}{{\left (d\,g+e\,g\,x\right )}^3} \,d x \] Input:

int(F^(f*(a + b*log(c*(d + e*x)^n)^2))/(d*g + e*g*x)^3,x)
 

Output:

int(exp(f*log(F)*(a + b*log(c*(d + e*x)^n)^2))/(d*g + e*g*x)^3, x)
 

Reduce [F]

\[ \int \frac {F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )}}{(d g+e g x)^3} \, dx=\frac {f^{a f} \left (\int \frac {f^{\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b f}}{e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}}d x \right )}{g^{3}} \] Input:

int(F^(f*(a+b*log(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x)
 

Output:

(f**(a*f)*int(f**(log((d + e*x)**n*c)**2*b*f)/(d**3 + 3*d**2*e*x + 3*d*e** 
2*x**2 + e**3*x**3),x))/g**3