Integrand size = 44, antiderivative size = 80 \[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\frac {\left (F^{e (c+d x)}\right )^{-n} \left (a+b \left (F^{e (c+d x)}\right )^n\right )^{1+p} \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}}}{b d e n (1+p) \log (F)} \]
(a+b*(F^(e*(d*x+c)))^n)^(p+1)*(G^(h*(g*x+f)))^(d*e*n*ln(F)/g/h/ln(G))/b/d/ e/((F^(e*(d*x+c)))^n)/n/(p+1)/ln(F)
\[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx \]
Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {2677, 2676, 17}
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.
\(\displaystyle \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx\) |
\(\Big \downarrow \) 2677 |
\(\displaystyle \left (F^{e (c+d x)}\right )^{-n} \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \int \left (F^{e (c+d x)}\right )^n \left (b \left (F^{e (c+d x)}\right )^n+a\right )^pdx\) |
\(\Big \downarrow \) 2676 |
\(\displaystyle \frac {\left (F^{e (c+d x)}\right )^{-n} \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \int \left (b \left (F^{e (c+d x)}\right )^n+a\right )^pd\left (F^{e (c+d x)}\right )^n}{d e n \log (F)}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {\left (F^{e (c+d x)}\right )^{-n} \left (a+b \left (F^{e (c+d x)}\right )^n\right )^{p+1} \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}}}{b d e n (p+1) \log (F)}\) |
((a + b*(F^(e*(c + d*x)))^n)^(1 + p)*(G^(h*(f + g*x)))^((d*e*n*Log[F])/(g* h*Log[G])))/(b*d*e*(F^(e*(c + d*x)))^n*n*(1 + p)*Log[F])
3.1.17.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*(( c_.) + (d_.)*(x_))))^(n_.))^(p_.), x_Symbol] :> Simp[1/(d*e*n*Log[F]) Sub st[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d , e, n, p}, x]
Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^(( h_.)*((f_.) + (g_.)*(x_))))^(m_.), x_Symbol] :> Simp[(G^(h*(f + g*x)))^m/(F ^(e*(c + d*x)))^n Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Lo g[F], g*h*m*Log[G]]
\[\int {\left (a +b \left (F^{e \left (d x +c \right )}\right )^{n}\right )}^{p} \left (G^{h \left (g x +f \right )}\right )^{\frac {d e n \ln \left (F \right )}{g h \ln \left (G \right )}}d x\]
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.10 \[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\frac {{\left (F^{d e n x + c e n} F^{\frac {{\left (d e f - c e g\right )} n}{g}} b + F^{\frac {{\left (d e f - c e g\right )} n}{g}} a\right )} {\left (F^{d e n x + c e n} b + a\right )}^{p}}{{\left (b d e n p + b d e n\right )} \log \left (F\right )} \]
integrate((a+b*(F^(e*(d*x+c)))^n)^p*(G^(h*(g*x+f)))^(d*e*n*log(F)/g/h/log( G)),x, algorithm="fricas")
(F^(d*e*n*x + c*e*n)*F^((d*e*f - c*e*g)*n/g)*b + F^((d*e*f - c*e*g)*n/g)*a )*(F^(d*e*n*x + c*e*n)*b + a)^p/((b*d*e*n*p + b*d*e*n)*log(F))
\[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\int \left (a + b \left (F^{c e + d e x}\right )^{n}\right )^{p} \left (G^{f h + g h x}\right )^{\frac {d e n \log {\left (F \right )}}{g h \log {\left (G \right )}}}\, dx \]
Integral((a + b*(F**(c*e + d*e*x))**n)**p*(G**(f*h + g*h*x))**(d*e*n*log(F )/(g*h*log(G))), x)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\frac {{\left (F^{d e n x} F^{c e n + \frac {d e f n}{g}} b + F^{\frac {d e f n}{g}} a\right )} {\left (F^{d e n x} F^{c e n} b + a\right )}^{p}}{F^{c e n} b d e n {\left (p + 1\right )} \log \left (F\right )} \]
integrate((a+b*(F^(e*(d*x+c)))^n)^p*(G^(h*(g*x+f)))^(d*e*n*log(F)/g/h/log( G)),x, algorithm="maxima")
(F^(d*e*n*x)*F^(c*e*n + d*e*f*n/g)*b + F^(d*e*f*n/g)*a)*(F^(d*e*n*x)*F^(c* e*n)*b + a)^p/(F^(c*e*n)*b*d*e*n*(p + 1)*log(F))
Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.79 \[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\frac {F^{\frac {d e f n}{g}} b e^{\left (2 \, d e n x \log \left (F\right ) + c e n \log \left (F\right ) + p \log \left (b e^{\left (d e n x \log \left (F\right ) + c e n \log \left (F\right )\right )} + a\right )\right )} + F^{\frac {d e f n}{g}} a e^{\left (d e n x \log \left (F\right ) + p \log \left (b e^{\left (d e n x \log \left (F\right ) + c e n \log \left (F\right )\right )} + a\right )\right )}}{b d e n p e^{\left (d e n x \log \left (F\right ) + c e n \log \left (F\right )\right )} \log \left (F\right ) + b d e n e^{\left (d e n x \log \left (F\right ) + c e n \log \left (F\right )\right )} \log \left (F\right )} \]
integrate((a+b*(F^(e*(d*x+c)))^n)^p*(G^(h*(g*x+f)))^(d*e*n*log(F)/g/h/log( G)),x, algorithm="giac")
(F^(d*e*f*n/g)*b*e^(2*d*e*n*x*log(F) + c*e*n*log(F) + p*log(b*e^(d*e*n*x*l og(F) + c*e*n*log(F)) + a)) + F^(d*e*f*n/g)*a*e^(d*e*n*x*log(F) + p*log(b* e^(d*e*n*x*log(F) + c*e*n*log(F)) + a)))/(b*d*e*n*p*e^(d*e*n*x*log(F) + c* e*n*log(F))*log(F) + b*d*e*n*e^(d*e*n*x*log(F) + c*e*n*log(F))*log(F))
Timed out. \[ \int \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \left (G^{h (f+g x)}\right )^{\frac {d e n \log (F)}{g h \log (G)}} \, dx=\int {\left (G^{h\,\left (f+g\,x\right )}\right )}^{\frac {d\,e\,n\,\ln \left (F\right )}{g\,h\,\ln \left (G\right )}}\,{\left (a+b\,{\left (F^{e\,\left (c+d\,x\right )}\right )}^n\right )}^p \,d x \]