\(\int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx\) [701]

Optimal result

Integrand size = 25, antiderivative size = 25 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=e^a F^b G^c \text {Int}(F(x),x) \] Output:

exp(a)*F^b*G^c*Defer(Int)(F(x),x)
 

Mathematica [N/A]

Not integrable

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx \] Input:

Integrate[E^(a - x*(Log[F] + Log[G]))*F^(b + x)*G^(c + x)*F[x],x]
 

Output:

Integrate[E^(a - x*(Log[F] + Log[G]))*F^(b + x)*G^(c + x)*F[x], x]
 

Rubi [N/A]

Not integrable

Time = 0.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {2725, 2704, 27, 27, 7299}

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.

\(\displaystyle \int F(x) F^{b+x} G^{c+x} e^{a-x (\log (F)+\log (G))} \, dx\)

\(\Big \downarrow \) 2725

\(\displaystyle \int F(x) G^{c+x} e^{a+b \log (F)-x \log (G)}dx\)

\(\Big \downarrow \) 2704

\(\displaystyle \int F^b F(x) G^{c+x} e^{a-x \log (G)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle F^b \int e^a G^c F(x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle e^a F^b G^c \int F(x)dx\)

\(\Big \downarrow \) 7299

\(\displaystyle e^a F^b G^c \int F(x)dx\)

Input:

Int[E^(a - x*(Log[F] + Log[G]))*F^(b + x)*G^(c + x)*F[x],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2704

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)* 
z^(a*b*Log[F]), x] /; FreeQ[{F, a, b}, x]
 

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]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [N/A]

Not integrable

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

\[\int {\mathrm e}^{a -x \left (\ln \left (F \right )+\ln \left (G \right )\right )} F^{b +x} G^{c +x} F \left (x \right )d x\]

Input:

int(exp(a-x*(ln(F)+ln(G)))*F^(b+x)*G^(c+x)*F(x),x)
 

Output:

int(exp(a-x*(ln(F)+ln(G)))*F^(b+x)*G^(c+x)*F(x),x)
 

Fricas [N/A]

Not integrable

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int { F^{b + x} G^{c + x} F\left (x\right ) e^{\left (-x {\left (\log \left (F\right ) + \log \left (G\right )\right )} + a\right )} \,d x } \] Input:

integrate(exp(a-x*(log(F)+log(G)))*F^(b+x)*G^(c+x)*F(x),x, algorithm="fric 
as")
 

Output:

integral(F^(b + x)*G^(c + x)*F(x)*e^(-x*log(F) - x*log(G) + a), x)
 

Sympy [N/A]

Not integrable

Time = 16.86 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=e^{a} \int F^{b + x} G^{c + x} F{\left (x \right )} e^{- x \log {\left (F \right )}} e^{- x \log {\left (G \right )}}\, dx \] Input:

integrate(exp(a-x*(ln(F)+ln(G)))*F**(b+x)*G**(c+x)*F(x),x)
 

Output:

exp(a)*Integral(F**(b + x)*G**(c + x)*F(x)*exp(-x*log(F))*exp(-x*log(G)), 
x)
 

Maxima [N/A]

Not integrable

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int { F^{b + x} G^{c + x} F\left (x\right ) e^{\left (-x {\left (\log \left (F\right ) + \log \left (G\right )\right )} + a\right )} \,d x } \] Input:

integrate(exp(a-x*(log(F)+log(G)))*F^(b+x)*G^(c+x)*F(x),x, algorithm="maxi 
ma")
 

Output:

integrate(F^(b + x)*G^(c + x)*F(x)*e^(-x*(log(F) + log(G)) + a), x)
 

Giac [N/A]

Not integrable

Time = 0.75 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int { F^{b + x} G^{c + x} F\left (x\right ) e^{\left (-x {\left (\log \left (F\right ) + \log \left (G\right )\right )} + a\right )} \,d x } \] Input:

integrate(exp(a-x*(log(F)+log(G)))*F^(b+x)*G^(c+x)*F(x),x, algorithm="giac 
")
 

Output:

integrate(F^(b + x)*G^(c + x)*F(x)*e^(-x*(log(F) + log(G)) + a), x)
 

Mupad [N/A]

Not integrable

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int F^{b+x}\,G^{c+x}\,{\mathrm {e}}^{a-x\,\left (\ln \left (F\right )+\ln \left (G\right )\right )}\,F\left (x\right ) \,d x \] Input:

int(F^(b + x)*G^(c + x)*exp(a - x*(log(F) + log(G)))*F(x),x)
 

Output:

int(F^(b + x)*G^(c + x)*exp(a - x*(log(F) + log(G)))*F(x), x)
 

Reduce [N/A]

Not integrable

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^{a-x (\log (F)+\log (G))} F^{b+x} G^{c+x} F(x) \, dx=\int {\mathrm e}^{a -x \left (\mathrm {log}\left (F \right )+\mathrm {log}\left (G \right )\right )} F^{b +x} G^{c +x} F \left (x \right )d x \] Input:

int(exp(a-x*(log(F)+log(G)))*F^(b+x)*G^(c+x)*F(x),x)
 

Output:

int(exp(a-x*(log(F)+log(G)))*F^(b+x)*G^(c+x)*F(x),x)