\(\int \frac {(d+e (F^{c (a+b x)})^n)^2}{f+g x} \, dx\) [30]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{f+g x} \, dx=\frac {2 d e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right )}{g}+\frac {e^2 F^{2 c \left (a-\frac {b f}{g}\right ) n-2 c n (a+b x)} \left (F^{a c+b c x}\right )^{2 n} \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right )}{g}+\frac {d^2 \log (f+g x)}{g} \] Output:

2*d*e*F^(c*(a-b*f/g)*n-c*n*(b*x+a))*(F^(b*c*x+a*c))^n*Ei(b*c*n*(g*x+f)*ln( 
F)/g)/g+e^2*F^(2*c*(a-b*f/g)*n-2*c*n*(b*x+a))*(F^(b*c*x+a*c))^(2*n)*Ei(2*b 
*c*n*(g*x+f)*ln(F)/g)/g+d^2*ln(g*x+f)/g
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{f+g x} \, dx=\frac {2 d e F^{-\frac {b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^n \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right )+e^2 F^{-\frac {2 b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^{2 n} \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right )+d^2 \log (f+g x)}{g} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2614, 2009}

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.

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

\(\Big \downarrow \) 2614

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(2*d*e*F^(c*(a - (b*f)/g)*n - c*n*(a + b*x))*(F^(a*c + b*c*x))^n*ExpIntegr 
alEi[(b*c*n*(f + g*x)*Log[F])/g])/g + (e^2*F^(2*c*(a - (b*f)/g)*n - 2*c*n* 
(a + b*x))*(F^(a*c + b*c*x))^(2*n)*ExpIntegralEi[(2*b*c*n*(f + g*x)*Log[F] 
)/g])/g + (d^2*Log[f + g*x])/g
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2614

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + 
 (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*(F 
^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n}, x] && 
 IGtQ[p, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{f+g x} \, dx=\frac {d^{2} \log \left (g x + f\right ) + \frac {e^{2} {\rm Ei}\left (\frac {2 \, {\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right )}{F^{\frac {2 \, {\left (b c f - a c g\right )} n}{g}}} + \frac {2 \, d e {\rm Ei}\left (\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right )}{F^{\frac {{\left (b c f - a c g\right )} n}{g}}}}{g} \] Input:

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

Output:

(d^2*log(g*x + f) + e^2*Ei(2*(b*c*g*n*x + b*c*f*n)*log(F)/g)/F^(2*(b*c*f - 
 a*c*g)*n/g) + 2*d*e*Ei((b*c*g*n*x + b*c*f*n)*log(F)/g)/F^((b*c*f - a*c*g) 
*n/g))/g
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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