\(\int (d+e (F^{c (a+b x)})^n)^2 (f+g x)^2 \, dx\) [27]

Optimal result

Integrand size = 25, antiderivative size = 239 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\frac {d^2 (f+g x)^3}{3 g}+\frac {4 d e \left (F^{a c+b c x}\right )^n g^2}{b^3 c^3 n^3 \log ^3(F)}+\frac {e^2 \left (F^{a c+b c x}\right )^{2 n} g^2}{4 b^3 c^3 n^3 \log ^3(F)}-\frac {4 d e \left (F^{a c+b c x}\right )^n g (f+g x)}{b^2 c^2 n^2 \log ^2(F)}-\frac {e^2 \left (F^{a c+b c x}\right )^{2 n} g (f+g x)}{2 b^2 c^2 n^2 \log ^2(F)}+\frac {2 d e \left (F^{a c+b c x}\right )^n (f+g x)^2}{b c n \log (F)}+\frac {e^2 \left (F^{a c+b c x}\right )^{2 n} (f+g x)^2}{2 b c n \log (F)} \] Output:

1/3*d^2*(g*x+f)^3/g+4*d*e*(F^(b*c*x+a*c))^n*g^2/b^3/c^3/n^3/ln(F)^3+1/4*e^ 
2*(F^(b*c*x+a*c))^(2*n)*g^2/b^3/c^3/n^3/ln(F)^3-4*d*e*(F^(b*c*x+a*c))^n*g* 
(g*x+f)/b^2/c^2/n^2/ln(F)^2-1/2*e^2*(F^(b*c*x+a*c))^(2*n)*g*(g*x+f)/b^2/c^ 
2/n^2/ln(F)^2+2*d*e*(F^(b*c*x+a*c))^n*(g*x+f)^2/b/c/n/ln(F)+1/2*e^2*(F^(b* 
c*x+a*c))^(2*n)*(g*x+f)^2/b/c/n/ln(F)
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.72 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=d^2 f^2 x+d^2 f g x^2+\frac {1}{3} d^2 g^2 x^3+\frac {2 d e \left (F^{c (a+b x)}\right )^n \left (2 g^2-2 b c g n (f+g x) \log (F)+b^2 c^2 n^2 (f+g x)^2 \log ^2(F)\right )}{b^3 c^3 n^3 \log ^3(F)}+\frac {e^2 \left (F^{c (a+b x)}\right )^{2 n} \left (g^2-2 b c g n (f+g x) \log (F)+2 b^2 c^2 n^2 (f+g x)^2 \log ^2(F)\right )}{4 b^3 c^3 n^3 \log ^3(F)} \] Input:

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

Output:

d^2*f^2*x + d^2*f*g*x^2 + (d^2*g^2*x^3)/3 + (2*d*e*(F^(c*(a + b*x)))^n*(2* 
g^2 - 2*b*c*g*n*(f + g*x)*Log[F] + b^2*c^2*n^2*(f + g*x)^2*Log[F]^2))/(b^3 
*c^3*n^3*Log[F]^3) + (e^2*(F^(c*(a + b*x)))^(2*n)*(g^2 - 2*b*c*g*n*(f + g* 
x)*Log[F] + 2*b^2*c^2*n^2*(f + g*x)^2*Log[F]^2))/(4*b^3*c^3*n^3*Log[F]^3)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 239, 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.

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

\(\Big \downarrow \) 2614

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 d e g^2 \left (F^{a c+b c x}\right )^n}{b^3 c^3 n^3 \log ^3(F)}+\frac {e^2 g^2 \left (F^{a c+b c x}\right )^{2 n}}{4 b^3 c^3 n^3 \log ^3(F)}-\frac {4 d e g (f+g x) \left (F^{a c+b c x}\right )^n}{b^2 c^2 n^2 \log ^2(F)}-\frac {e^2 g (f+g x) \left (F^{a c+b c x}\right )^{2 n}}{2 b^2 c^2 n^2 \log ^2(F)}+\frac {2 d e (f+g x)^2 \left (F^{a c+b c x}\right )^n}{b c n \log (F)}+\frac {e^2 (f+g x)^2 \left (F^{a c+b c x}\right )^{2 n}}{2 b c n \log (F)}+\frac {d^2 (f+g x)^3}{3 g}\)

Input:

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

Output:

(d^2*(f + g*x)^3)/(3*g) + (4*d*e*(F^(a*c + b*c*x))^n*g^2)/(b^3*c^3*n^3*Log 
[F]^3) + (e^2*(F^(a*c + b*c*x))^(2*n)*g^2)/(4*b^3*c^3*n^3*Log[F]^3) - (4*d 
*e*(F^(a*c + b*c*x))^n*g*(f + g*x))/(b^2*c^2*n^2*Log[F]^2) - (e^2*(F^(a*c 
+ b*c*x))^(2*n)*g*(f + g*x))/(2*b^2*c^2*n^2*Log[F]^2) + (2*d*e*(F^(a*c + b 
*c*x))^n*(f + g*x)^2)/(b*c*n*Log[F]) + (e^2*(F^(a*c + b*c*x))^(2*n)*(f + g 
*x)^2)/(2*b*c*n*Log[F])
 

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 [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.77

method	result	size
parallelrisch	\(\frac {4 d^{2} g^{2} x^{3} \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+12 d^{2} g f \,x^{2} \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+12 d^{2} f^{2} x \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+6 x^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n} e^{2} g^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+24 x^{2} \left (F^{c \left (b x +a \right )}\right )^{n} d e \,g^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+12 \ln \left (F \right )^{2} x \left (F^{c \left (b x +a \right )}\right )^{2 n} b^{2} c^{2} e^{2} f g \,n^{2}+48 \ln \left (F \right )^{2} x \left (F^{c \left (b x +a \right )}\right )^{n} b^{2} c^{2} d e f g \,n^{2}+6 \ln \left (F \right )^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n} b^{2} c^{2} e^{2} f^{2} n^{2}+24 \ln \left (F \right )^{2} \left (F^{c \left (b x +a \right )}\right )^{n} b^{2} c^{2} d e \,f^{2} n^{2}-6 \ln \left (F \right ) x \left (F^{c \left (b x +a \right )}\right )^{2 n} b c \,e^{2} g^{2} n -48 \ln \left (F \right ) x \left (F^{c \left (b x +a \right )}\right )^{n} b c d e \,g^{2} n -6 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{2 n} b c \,e^{2} f g n -48 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{n} b c d e f g n +3 \left (F^{c \left (b x +a \right )}\right )^{2 n} e^{2} g^{2}+48 \left (F^{c \left (b x +a \right )}\right )^{n} d e \,g^{2}}{12 \ln \left (F \right )^{3} b^{3} c^{3} n^{3}}\)	\(424\)
orering	\(\frac {\left (4 \ln \left (F \right )^{4} b^{4} c^{4} g^{4} n^{4} x^{5}+20 \ln \left (F \right )^{4} b^{4} c^{4} f \,g^{3} n^{4} x^{4}+40 \ln \left (F \right )^{4} b^{4} c^{4} f^{2} g^{2} n^{4} x^{3}+36 \ln \left (F \right )^{4} b^{4} c^{4} f^{3} g \,n^{4} x^{2}+12 \ln \left (F \right )^{4} b^{4} c^{4} f^{4} n^{4} x +12 \ln \left (F \right )^{3} b^{3} c^{3} g^{4} n^{3} x^{4}+48 \ln \left (F \right )^{3} b^{3} c^{3} f \,g^{3} n^{3} x^{3}+90 \ln \left (F \right )^{3} b^{3} c^{3} f^{2} g^{2} n^{3} x^{2}+72 \ln \left (F \right )^{3} b^{3} c^{3} f^{3} g \,n^{3} x +18 \ln \left (F \right )^{3} b^{3} c^{3} f^{4} n^{3}-30 \ln \left (F \right )^{2} b^{2} c^{2} g^{4} n^{2} x^{3}-132 \ln \left (F \right )^{2} b^{2} c^{2} f \,g^{3} n^{2} x^{2}-120 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} g^{2} n^{2} x -30 \ln \left (F \right )^{2} b^{2} c^{2} f^{3} g \,n^{2}+81 \ln \left (F \right ) b c \,g^{4} n \,x^{2}+36 \ln \left (F \right ) b c f \,g^{3} n x +9 \ln \left (F \right ) b c \,f^{2} g^{2} n +168 g^{4} x +42 f \,g^{3}\right ) {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2}}{12 b^{4} c^{4} n^{4} \ln \left (F \right )^{4} \left (g x +f \right )^{2}}-\frac {\left (6 \ln \left (F \right )^{3} b^{3} c^{3} g^{3} n^{3} x^{4}+24 \ln \left (F \right )^{3} b^{3} c^{3} f \,g^{2} n^{3} x^{3}+36 \ln \left (F \right )^{3} b^{3} c^{3} f^{2} g \,n^{3} x^{2}+18 \ln \left (F \right )^{3} b^{3} c^{3} f^{3} n^{3} x -13 \ln \left (F \right )^{2} b^{2} c^{2} g^{3} n^{2} x^{3}-39 \ln \left (F \right )^{2} b^{2} c^{2} f \,g^{2} n^{2} x^{2}-12 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} g \,n^{2} x +6 \ln \left (F \right )^{2} b^{2} c^{2} f^{3} n^{2}+9 g^{3} \ln \left (F \right ) b c n \,x^{2}-45 g^{2} \ln \left (F \right ) b c n x f -18 g \ln \left (F \right ) b c \,f^{2} n +105 g^{3} x +21 g^{2} f \right ) \left (2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right )^{2} e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c n +2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2} \left (g x +f \right ) g \right )}{12 b^{4} c^{4} n^{4} \ln \left (F \right )^{4} \left (g x +f \right )^{3}}+\frac {x \left (2 g^{2} x^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+6 \ln \left (F \right )^{2} b^{2} c^{2} f g \,n^{2} x +6 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} n^{2}-9 \ln \left (F \right ) b c \,g^{2} n x -18 \ln \left (F \right ) b c f g n +21 g^{2}\right ) \left (2 e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2} \left (g x +f \right )^{2}+8 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right ) e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c n g +2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right )^{2} e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2} g^{2}\right )}{12 b^{4} c^{4} n^{4} \ln \left (F \right )^{4} \left (g x +f \right )^{2}}\)	\(948\)

Input:

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

Output:

1/12*(4*d^2*g^2*x^3*ln(F)^3*b^3*c^3*n^3+12*d^2*g*f*x^2*ln(F)^3*b^3*c^3*n^3 
+12*d^2*f^2*x*ln(F)^3*b^3*c^3*n^3+6*x^2*((F^(c*(b*x+a)))^n)^2*e^2*g^2*ln(F 
)^2*b^2*c^2*n^2+24*x^2*(F^(c*(b*x+a)))^n*d*e*g^2*ln(F)^2*b^2*c^2*n^2+12*ln 
(F)^2*x*((F^(c*(b*x+a)))^n)^2*b^2*c^2*e^2*f*g*n^2+48*ln(F)^2*x*(F^(c*(b*x+ 
a)))^n*b^2*c^2*d*e*f*g*n^2+6*ln(F)^2*((F^(c*(b*x+a)))^n)^2*b^2*c^2*e^2*f^2 
*n^2+24*ln(F)^2*(F^(c*(b*x+a)))^n*b^2*c^2*d*e*f^2*n^2-6*ln(F)*x*((F^(c*(b* 
x+a)))^n)^2*b*c*e^2*g^2*n-48*ln(F)*x*(F^(c*(b*x+a)))^n*b*c*d*e*g^2*n-6*ln( 
F)*((F^(c*(b*x+a)))^n)^2*b*c*e^2*f*g*n-48*ln(F)*(F^(c*(b*x+a)))^n*b*c*d*e* 
f*g*n+3*((F^(c*(b*x+a)))^n)^2*e^2*g^2+48*(F^(c*(b*x+a)))^n*d*e*g^2)/ln(F)^ 
3/b^3/c^3/n^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.20 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\frac {4 \, {\left (b^{3} c^{3} d^{2} g^{2} n^{3} x^{3} + 3 \, b^{3} c^{3} d^{2} f g n^{3} x^{2} + 3 \, b^{3} c^{3} d^{2} f^{2} n^{3} x\right )} \log \left (F\right )^{3} + 3 \, {\left (e^{2} g^{2} + 2 \, {\left (b^{2} c^{2} e^{2} g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} e^{2} f g n^{2} x + b^{2} c^{2} e^{2} f^{2} n^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b c e^{2} g^{2} n x + b c e^{2} f g n\right )} \log \left (F\right )\right )} F^{2 \, b c n x + 2 \, a c n} + 24 \, {\left (2 \, d e g^{2} + {\left (b^{2} c^{2} d e g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} d e f g n^{2} x + b^{2} c^{2} d e f^{2} n^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b c d e g^{2} n x + b c d e f g n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{12 \, b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \] Input:

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

Output:

1/12*(4*(b^3*c^3*d^2*g^2*n^3*x^3 + 3*b^3*c^3*d^2*f*g*n^3*x^2 + 3*b^3*c^3*d 
^2*f^2*n^3*x)*log(F)^3 + 3*(e^2*g^2 + 2*(b^2*c^2*e^2*g^2*n^2*x^2 + 2*b^2*c 
^2*e^2*f*g*n^2*x + b^2*c^2*e^2*f^2*n^2)*log(F)^2 - 2*(b*c*e^2*g^2*n*x + b* 
c*e^2*f*g*n)*log(F))*F^(2*b*c*n*x + 2*a*c*n) + 24*(2*d*e*g^2 + (b^2*c^2*d* 
e*g^2*n^2*x^2 + 2*b^2*c^2*d*e*f*g*n^2*x + b^2*c^2*d*e*f^2*n^2)*log(F)^2 - 
2*(b*c*d*e*g^2*n*x + b*c*d*e*f*g*n)*log(F))*F^(b*c*n*x + a*c*n))/(b^3*c^3* 
n^3*log(F)^3)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (231) = 462\).

Time = 1.71 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.21 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\begin {cases} \left (d + e\right )^{2} \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge n = 0 \\\left (d + e \left (F^{a c}\right )^{n}\right )^{2} \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: b = 0 \\\left (d + e\right )^{2} \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: F = 1 \vee c = 0 \vee n = 0 \\d^{2} f^{2} x + d^{2} f g x^{2} + \frac {d^{2} g^{2} x^{3}}{3} + \frac {2 d e f^{2} \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {4 d e f g x \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {2 d e g^{2} x^{2} \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {e^{2} f^{2} \left (F^{a c + b c x}\right )^{2 n}}{2 b c n \log {\left (F \right )}} + \frac {e^{2} f g x \left (F^{a c + b c x}\right )^{2 n}}{b c n \log {\left (F \right )}} + \frac {e^{2} g^{2} x^{2} \left (F^{a c + b c x}\right )^{2 n}}{2 b c n \log {\left (F \right )}} - \frac {4 d e f g \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} - \frac {4 d e g^{2} x \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} - \frac {e^{2} f g \left (F^{a c + b c x}\right )^{2 n}}{2 b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} - \frac {e^{2} g^{2} x \left (F^{a c + b c x}\right )^{2 n}}{2 b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} + \frac {4 d e g^{2} \left (F^{a c + b c x}\right )^{n}}{b^{3} c^{3} n^{3} \log {\left (F \right )}^{3}} + \frac {e^{2} g^{2} \left (F^{a c + b c x}\right )^{2 n}}{4 b^{3} c^{3} n^{3} \log {\left (F \right )}^{3}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((d + e)**2*(f**2*x + f*g*x**2 + g**2*x**3/3), Eq(F, 1) & Eq(b, 
0) & Eq(c, 0) & Eq(n, 0)), ((d + e*(F**(a*c))**n)**2*(f**2*x + f*g*x**2 + 
g**2*x**3/3), Eq(b, 0)), ((d + e)**2*(f**2*x + f*g*x**2 + g**2*x**3/3), Eq 
(F, 1) | Eq(c, 0) | Eq(n, 0)), (d**2*f**2*x + d**2*f*g*x**2 + d**2*g**2*x* 
*3/3 + 2*d*e*f**2*(F**(a*c + b*c*x))**n/(b*c*n*log(F)) + 4*d*e*f*g*x*(F**( 
a*c + b*c*x))**n/(b*c*n*log(F)) + 2*d*e*g**2*x**2*(F**(a*c + b*c*x))**n/(b 
*c*n*log(F)) + e**2*f**2*(F**(a*c + b*c*x))**(2*n)/(2*b*c*n*log(F)) + e**2 
*f*g*x*(F**(a*c + b*c*x))**(2*n)/(b*c*n*log(F)) + e**2*g**2*x**2*(F**(a*c 
+ b*c*x))**(2*n)/(2*b*c*n*log(F)) - 4*d*e*f*g*(F**(a*c + b*c*x))**n/(b**2* 
c**2*n**2*log(F)**2) - 4*d*e*g**2*x*(F**(a*c + b*c*x))**n/(b**2*c**2*n**2* 
log(F)**2) - e**2*f*g*(F**(a*c + b*c*x))**(2*n)/(2*b**2*c**2*n**2*log(F)** 
2) - e**2*g**2*x*(F**(a*c + b*c*x))**(2*n)/(2*b**2*c**2*n**2*log(F)**2) + 
4*d*e*g**2*(F**(a*c + b*c*x))**n/(b**3*c**3*n**3*log(F)**3) + e**2*g**2*(F 
**(a*c + b*c*x))**(2*n)/(4*b**3*c**3*n**3*log(F)**3), True))
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.45 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\frac {1}{3} \, d^{2} g^{2} x^{3} + d^{2} f g x^{2} + d^{2} f^{2} x + \frac {2 \, F^{b c n x + a c n} d e f^{2}}{b c n \log \left (F\right )} + \frac {F^{2 \, b c n x + 2 \, a c n} e^{2} f^{2}}{2 \, b c n \log \left (F\right )} + \frac {4 \, {\left (F^{a c n} b c n x \log \left (F\right ) - F^{a c n}\right )} F^{b c n x} d e f g}{b^{2} c^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (2 \, F^{2 \, a c n} b c n x \log \left (F\right ) - F^{2 \, a c n}\right )} F^{2 \, b c n x} e^{2} f g}{2 \, b^{2} c^{2} n^{2} \log \left (F\right )^{2}} + \frac {2 \, {\left (F^{a c n} b^{2} c^{2} n^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c n} b c n x \log \left (F\right ) + 2 \, F^{a c n}\right )} F^{b c n x} d e g^{2}}{b^{3} c^{3} n^{3} \log \left (F\right )^{3}} + \frac {{\left (2 \, F^{2 \, a c n} b^{2} c^{2} n^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{2 \, a c n} b c n x \log \left (F\right ) + F^{2 \, a c n}\right )} F^{2 \, b c n x} e^{2} g^{2}}{4 \, b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \] Input:

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

Output:

1/3*d^2*g^2*x^3 + d^2*f*g*x^2 + d^2*f^2*x + 2*F^(b*c*n*x + a*c*n)*d*e*f^2/ 
(b*c*n*log(F)) + 1/2*F^(2*b*c*n*x + 2*a*c*n)*e^2*f^2/(b*c*n*log(F)) + 4*(F 
^(a*c*n)*b*c*n*x*log(F) - F^(a*c*n))*F^(b*c*n*x)*d*e*f*g/(b^2*c^2*n^2*log( 
F)^2) + 1/2*(2*F^(2*a*c*n)*b*c*n*x*log(F) - F^(2*a*c*n))*F^(2*b*c*n*x)*e^2 
*f*g/(b^2*c^2*n^2*log(F)^2) + 2*(F^(a*c*n)*b^2*c^2*n^2*x^2*log(F)^2 - 2*F^ 
(a*c*n)*b*c*n*x*log(F) + 2*F^(a*c*n))*F^(b*c*n*x)*d*e*g^2/(b^3*c^3*n^3*log 
(F)^3) + 1/4*(2*F^(2*a*c*n)*b^2*c^2*n^2*x^2*log(F)^2 - 2*F^(2*a*c*n)*b*c*n 
*x*log(F) + F^(2*a*c*n))*F^(2*b*c*n*x)*e^2*g^2/(b^3*c^3*n^3*log(F)^3)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 5675, normalized size of antiderivative = 23.74 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\text {Too large to display} \] Input:

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

Output:

1/3*d^2*g^2*x^3 + d^2*f*g*x^2 + d^2*f^2*x - 1/2*(((2*pi*b^2*c^2*e^2*g^2*n^ 
2*x^2*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*e^2*g^2*n^2*x^2*log(abs(F)) + 4*pi 
*b^2*c^2*e^2*f*g*n^2*x*log(abs(F))*sgn(F) - 4*pi*b^2*c^2*e^2*f*g*n^2*x*log 
(abs(F)) + 2*pi*b^2*c^2*e^2*f^2*n^2*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*e^2* 
f^2*n^2*log(abs(F)) - pi*b*c*e^2*g^2*n*x*sgn(F) + pi*b*c*e^2*g^2*n*x - pi* 
b*c*e^2*f*g*n*sgn(F) + pi*b*c*e^2*f*g*n)*(pi^3*b^3*c^3*n^3*sgn(F) - 3*pi*b 
^3*c^3*n^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3*n^3 + 3*pi*b^3*c^3*n^3*log( 
abs(F))^2)/((pi^3*b^3*c^3*n^3*sgn(F) - 3*pi*b^3*c^3*n^3*log(abs(F))^2*sgn( 
F) - pi^3*b^3*c^3*n^3 + 3*pi*b^3*c^3*n^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^ 
3*n^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3* 
log(abs(F))^3)^2) - (pi^2*b^2*c^2*e^2*g^2*n^2*x^2*sgn(F) - pi^2*b^2*c^2*e^ 
2*g^2*n^2*x^2 + 2*b^2*c^2*e^2*g^2*n^2*x^2*log(abs(F))^2 + 2*pi^2*b^2*c^2*e 
^2*f*g*n^2*x*sgn(F) - 2*pi^2*b^2*c^2*e^2*f*g*n^2*x + 4*b^2*c^2*e^2*f*g*n^2 
*x*log(abs(F))^2 + pi^2*b^2*c^2*e^2*f^2*n^2*sgn(F) - pi^2*b^2*c^2*e^2*f^2* 
n^2 + 2*b^2*c^2*e^2*f^2*n^2*log(abs(F))^2 - 2*b*c*e^2*g^2*n*x*log(abs(F)) 
- 2*b*c*e^2*f*g*n*log(abs(F)) + e^2*g^2)*(3*pi^2*b^3*c^3*n^3*log(abs(F))*s 
gn(F) - 3*pi^2*b^3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3*log(abs(F))^3)/((pi 
^3*b^3*c^3*n^3*sgn(F) - 3*pi*b^3*c^3*n^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c 
^3*n^3 + 3*pi*b^3*c^3*n^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*n^3*log(abs(F 
))*sgn(F) - 3*pi^2*b^3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3*log(abs(F))^...
 

Mupad [B] (verification not implemented)

Time = 23.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.12 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx={\left (F^{b\,c\,x}\,F^{a\,c}\right )}^{2\,n}\,\left (\frac {e^2\,\left (2\,b^2\,c^2\,f^2\,n^2\,{\ln \left (F\right )}^2-2\,b\,c\,f\,g\,n\,\ln \left (F\right )+g^2\right )}{4\,b^3\,c^3\,n^3\,{\ln \left (F\right )}^3}+\frac {e^2\,g^2\,x^2}{2\,b\,c\,n\,\ln \left (F\right )}-\frac {e^2\,g\,x\,\left (g-2\,b\,c\,f\,n\,\ln \left (F\right )\right )}{2\,b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}\right )+{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\,\left (\frac {2\,d\,e\,\left (b^2\,c^2\,f^2\,n^2\,{\ln \left (F\right )}^2-2\,b\,c\,f\,g\,n\,\ln \left (F\right )+2\,g^2\right )}{b^3\,c^3\,n^3\,{\ln \left (F\right )}^3}+\frac {2\,d\,e\,g^2\,x^2}{b\,c\,n\,\ln \left (F\right )}-\frac {4\,d\,e\,g\,x\,\left (g-b\,c\,f\,n\,\ln \left (F\right )\right )}{b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}\right )+d^2\,f^2\,x+\frac {d^2\,g^2\,x^3}{3}+d^2\,f\,g\,x^2 \] Input:

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

Output:

(F^(b*c*x)*F^(a*c))^(2*n)*((e^2*(g^2 + 2*b^2*c^2*f^2*n^2*log(F)^2 - 2*b*c* 
f*g*n*log(F)))/(4*b^3*c^3*n^3*log(F)^3) + (e^2*g^2*x^2)/(2*b*c*n*log(F)) - 
 (e^2*g*x*(g - 2*b*c*f*n*log(F)))/(2*b^2*c^2*n^2*log(F)^2)) + (F^(b*c*x)*F 
^(a*c))^n*((2*d*e*(2*g^2 + b^2*c^2*f^2*n^2*log(F)^2 - 2*b*c*f*g*n*log(F))) 
/(b^3*c^3*n^3*log(F)^3) + (2*d*e*g^2*x^2)/(b*c*n*log(F)) - (4*d*e*g*x*(g - 
 b*c*f*n*log(F)))/(b^2*c^2*n^2*log(F)^2)) + d^2*f^2*x + (d^2*g^2*x^3)/3 + 
d^2*f*g*x^2
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.82 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^2 \, dx=\frac {6 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e^{2} f^{2} n^{2}+12 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e^{2} f g \,n^{2} x +6 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e^{2} g^{2} n^{2} x^{2}-6 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right ) b c \,e^{2} f g n -6 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right ) b c \,e^{2} g^{2} n x +3 f^{2 b c n x +2 a c n} e^{2} g^{2}+24 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d e \,f^{2} n^{2}+48 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d e f g \,n^{2} x +24 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d e \,g^{2} n^{2} x^{2}-48 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c d e f g n -48 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c d e \,g^{2} n x +48 f^{b c n x +a c n} d e \,g^{2}+12 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} d^{2} f^{2} n^{3} x +12 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} d^{2} f g \,n^{3} x^{2}+4 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} d^{2} g^{2} n^{3} x^{3}}{12 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} n^{3}} \] Input:

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

Output:

(6*f**(2*a*c*n + 2*b*c*n*x)*log(f)**2*b**2*c**2*e**2*f**2*n**2 + 12*f**(2* 
a*c*n + 2*b*c*n*x)*log(f)**2*b**2*c**2*e**2*f*g*n**2*x + 6*f**(2*a*c*n + 2 
*b*c*n*x)*log(f)**2*b**2*c**2*e**2*g**2*n**2*x**2 - 6*f**(2*a*c*n + 2*b*c* 
n*x)*log(f)*b*c*e**2*f*g*n - 6*f**(2*a*c*n + 2*b*c*n*x)*log(f)*b*c*e**2*g* 
*2*n*x + 3*f**(2*a*c*n + 2*b*c*n*x)*e**2*g**2 + 24*f**(a*c*n + b*c*n*x)*lo 
g(f)**2*b**2*c**2*d*e*f**2*n**2 + 48*f**(a*c*n + b*c*n*x)*log(f)**2*b**2*c 
**2*d*e*f*g*n**2*x + 24*f**(a*c*n + b*c*n*x)*log(f)**2*b**2*c**2*d*e*g**2* 
n**2*x**2 - 48*f**(a*c*n + b*c*n*x)*log(f)*b*c*d*e*f*g*n - 48*f**(a*c*n + 
b*c*n*x)*log(f)*b*c*d*e*g**2*n*x + 48*f**(a*c*n + b*c*n*x)*d*e*g**2 + 12*l 
og(f)**3*b**3*c**3*d**2*f**2*n**3*x + 12*log(f)**3*b**3*c**3*d**2*f*g*n**3 
*x**2 + 4*log(f)**3*b**3*c**3*d**2*g**2*n**3*x**3)/(12*log(f)**3*b**3*c**3 
*n**3)