\(\int F^{f (a+b \log ^2(c (d+e x)^n))} (d g+e g x)^m \, dx\) [510]

Optimal result

Integrand size = 31, antiderivative size = 137 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\frac {e^{-\frac {(1+m)^2}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^{1+m} \text {erfi}\left (\frac {1+m+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 137, 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.

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

\(\Big \downarrow \) 2706

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

\(\Big \downarrow \) 2725

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

\(\Big \downarrow \) 2664

\(\displaystyle \frac {F^{a f} (d g+e g x)^{m+1} e^{-\frac {(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \int \exp \left (\frac {\left (m+2 b f n \log (F) \log \left (c (d+e x)^n\right )+1\right )^2}{4 b f n^2 \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e g n}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {\sqrt {\pi } F^{a f} (d g+e g x)^{m+1} e^{-\frac {(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}}\)

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b f n \log \left (F\right ) \log \left (c\right ) + m + 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{2 \, b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \left (F\right )^{2} + 4 \, b f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \, {\left (b f m + b f\right )} n \log \left (F\right ) \log \left (c\right ) - m^{2} - 2 \, m - 1}{4 \, b f n^{2} \log \left (F\right )}\right )}}{2 \, e n} \] Input:

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

Output:

-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) + m + 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*m*n^2*log(F)*log(g) - 4*(b*f*m + b* 
f)*n*log(F)*log(c) - m^2 - 2*m - 1)/(b*f*n^2*log(F)))/(e*n)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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