\(\int \frac {(d+e (F^{c (a+b x)})^n)^2}{(f+g x)^2} \, dx\) [31]

Optimal result

Integrand size = 25, antiderivative size = 202 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^2} \, dx=-\frac {d^2}{g (f+g x)}-\frac {2 d e \left (F^{a c+b c x}\right )^n}{g (f+g x)}-\frac {e^2 \left (F^{a c+b c x}\right )^{2 n}}{g (f+g x)}+\frac {2 b c d e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n n \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log (F)}{g^2}+\frac {2 b c e^2 F^{2 c \left (a-\frac {b f}{g}\right ) n-2 c n (a+b x)} \left (F^{a c+b c x}\right )^{2 n} n \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right ) \log (F)}{g^2} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^2} \, dx=\frac {-\frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 g}{f+g x}+2 b c d e F^{-\frac {b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^n n \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log (F)+2 b c e^2 F^{-\frac {2 b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^{2 n} n \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right ) \log (F)}{g^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 202, 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.

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

\(\Big \downarrow \) 2614

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^2} \, dx=-\frac {2 \, F^{b c n x + a c n} d e g + F^{2 \, b c n x + 2 \, a c n} e^{2} g + d^{2} g - \frac {2 \, {\left (b c e^{2} g n x + b c e^{2} f n\right )} {\rm Ei}\left (\frac {2 \, {\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) \log \left (F\right )}{F^{\frac {2 \, {\left (b c f - a c g\right )} n}{g}}} - \frac {2 \, {\left (b c d e g n x + b c d e f n\right )} {\rm Ei}\left (\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) \log \left (F\right )}{F^{\frac {{\left (b c f - a c g\right )} n}{g}}}}{g^{3} x + f g^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**3*n + 2*log(f)*b*c*f**2*g* 
n*x + log(f)*b*c*f*g**2*n*x**2 - f**2*g - 2*f*g**2*x - g**3*x**2),x)*log(f 
)*b*c*e**2*f**3*n + f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**3*n + 2 
*log(f)*b*c*f**2*g*n*x + log(f)*b*c*f*g**2*n*x**2 - f**2*g - 2*f*g**2*x - 
g**3*x**2),x)*log(f)*b*c*e**2*f**2*g*n*x - f**(2*a*c*n)*int(f**(2*b*c*n*x) 
/(log(f)*b*c*f**3*n + 2*log(f)*b*c*f**2*g*n*x + log(f)*b*c*f*g**2*n*x**2 - 
 f**2*g - 2*f*g**2*x - g**3*x**2),x)*e**2*f**2*g - f**(2*a*c*n)*int(f**(2* 
b*c*n*x)/(log(f)*b*c*f**3*n + 2*log(f)*b*c*f**2*g*n*x + log(f)*b*c*f*g**2* 
n*x**2 - f**2*g - 2*f*g**2*x - g**3*x**2),x)*e**2*f*g**2*x + 4*f**(a*c*n)* 
int(f**(b*c*n*x)/(2*log(f)*b*c*f**3*n + 4*log(f)*b*c*f**2*g*n*x + 2*log(f) 
*b*c*f*g**2*n*x**2 - f**2*g - 2*f*g**2*x - g**3*x**2),x)*log(f)*b*c*d*e*f* 
*3*n + 4*f**(a*c*n)*int(f**(b*c*n*x)/(2*log(f)*b*c*f**3*n + 4*log(f)*b*c*f 
**2*g*n*x + 2*log(f)*b*c*f*g**2*n*x**2 - f**2*g - 2*f*g**2*x - g**3*x**2), 
x)*log(f)*b*c*d*e*f**2*g*n*x - 2*f**(a*c*n)*int(f**(b*c*n*x)/(2*log(f)*b*c 
*f**3*n + 4*log(f)*b*c*f**2*g*n*x + 2*log(f)*b*c*f*g**2*n*x**2 - f**2*g - 
2*f*g**2*x - g**3*x**2),x)*d*e*f**2*g - 2*f**(a*c*n)*int(f**(b*c*n*x)/(2*l 
og(f)*b*c*f**3*n + 4*log(f)*b*c*f**2*g*n*x + 2*log(f)*b*c*f*g**2*n*x**2 - 
f**2*g - 2*f*g**2*x - g**3*x**2),x)*d*e*f*g**2*x + d**2*x)/(f*(f + g*x))