\(\int (a+b (F^{g (e+f x)})^n)^2 (c+d x) \, dx\) [34]
Optimal result
Integrand size = 23, antiderivative size = 156 \[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {a^2 (c+d x)^2}{2 d}-\frac {2 a b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}+\frac {2 a b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}
\]
[Out]
1/2*a^2*(d*x+c)^2/d-2*a*b*d*(F^(f*g*x+e*g))^n/f^2/g^2/n^2/ln(F)^2-1/4*b^2*d*(F^(f*g*x+e*g))^(2*n)/f^2/g^2/n^2/
ln(F)^2+2*a*b*(F^(f*g*x+e*g))^n*(d*x+c)/f/g/n/ln(F)+1/2*b^2*(F^(f*g*x+e*g))^(2*n)*(d*x+c)/f/g/n/ln(F)
Rubi [A] (verified)
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2214, 2207, 2225}
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {a^2 (c+d x)^2}{2 d}+\frac {2 a b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac {2 a b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}+\frac {b^2 (c+d x) \left (F^{e g+f g x}\right )^{2 n}}{2 f g n \log (F)}-\frac {b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}
\]
[In]
Int[(a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x),x]
[Out]
(a^2*(c + d*x)^2)/(2*d) - (2*a*b*d*(F^(e*g + f*g*x))^n)/(f^2*g^2*n^2*Log[F]^2) - (b^2*d*(F^(e*g + f*g*x))^(2*n
))/(4*f^2*g^2*n^2*Log[F]^2) + (2*a*b*(F^(e*g + f*g*x))^n*(c + d*x))/(f*g*n*Log[F]) + (b^2*(F^(e*g + f*g*x))^(2
*n)*(c + d*x))/(2*f*g*n*Log[F])
Rule 2207
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] && !TrueQ[$UseGamma]
Rule 2214
Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[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]
Rule 2225
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]
Rubi steps \begin{align*}
\text {integral}& = \int \left (a^2 (c+d x)+2 a b \left (F^{e g+f g x}\right )^n (c+d x)+b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)\right ) \, dx \\ & = \frac {a^2 (c+d x)^2}{2 d}+(2 a b) \int \left (F^{e g+f g x}\right )^n (c+d x) \, dx+b^2 \int \left (F^{e g+f g x}\right )^{2 n} (c+d x) \, dx \\ & = \frac {a^2 (c+d x)^2}{2 d}+\frac {2 a b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}-\frac {(2 a b d) \int \left (F^{e g+f g x}\right )^n \, dx}{f g n \log (F)}-\frac {\left (b^2 d\right ) \int \left (F^{e g+f g x}\right )^{2 n} \, dx}{2 f g n \log (F)} \\ & = \frac {a^2 (c+d x)^2}{2 d}-\frac {2 a b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}+\frac {2 a b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {-b d \left (F^{g (e+f x)}\right )^n \left (8 a+b \left (F^{g (e+f x)}\right )^n\right )+2 b f \left (F^{g (e+f x)}\right )^n \left (4 a+b \left (F^{g (e+f x)}\right )^n\right ) g n (c+d x) \log (F)+2 a^2 f^2 g^2 n^2 x (2 c+d x) \log ^2(F)}{4 f^2 g^2 n^2 \log ^2(F)}
\]
[In]
Integrate[(a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x),x]
[Out]
(-(b*d*(F^(g*(e + f*x)))^n*(8*a + b*(F^(g*(e + f*x)))^n)) + 2*b*f*(F^(g*(e + f*x)))^n*(4*a + b*(F^(g*(e + f*x)
))^n)*g*n*(c + d*x)*Log[F] + 2*a^2*f^2*g^2*n^2*x*(2*c + d*x)*Log[F]^2)/(4*f^2*g^2*n^2*Log[F]^2)
Maple [A] (verified)
Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.12
| | |
| method | result | size |
| | |
| norman |
\(a^{2} c x +\frac {a^{2} d \,x^{2}}{2}+\frac {b^{2} \left (2 \ln \left (F \right ) c f g n -d \right ) {\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{4 n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}+\frac {2 a b \left (\ln \left (F \right ) c f g n -d \right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}+\frac {b^{2} d x \,{\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{2 n g f \ln \left (F \right )}+\frac {2 a b d x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}\) |
\(174\) |
| parallelrisch |
\(\frac {2 a^{2} d \,x^{2} n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+4 a^{2} c x \,n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+2 x \left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2} d n g f \ln \left (F \right )+8 x \left (F^{g \left (f x +e \right )}\right )^{n} a b d n g f \ln \left (F \right )+2 \ln \left (F \right ) \left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2} c f g n +8 \ln \left (F \right ) \left (F^{g \left (f x +e \right )}\right )^{n} a b c f g n -\left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2} d -8 \left (F^{g \left (f x +e \right )}\right )^{n} a b d}{4 n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}\) |
\(186\) |
| | |
|
|
|
|
[In]
int((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c),x,method=_RETURNVERBOSE)
[Out]
a^2*c*x+1/2*a^2*d*x^2+1/4*b^2*(2*ln(F)*c*f*g*n-d)/n^2/g^2/f^2/ln(F)^2*exp(n*ln(exp(g*(f*x+e)*ln(F))))^2+2*a*b*
(ln(F)*c*f*g*n-d)/n^2/g^2/f^2/ln(F)^2*exp(n*ln(exp(g*(f*x+e)*ln(F))))+1/2/n/g/f/ln(F)*b^2*d*x*exp(n*ln(exp(g*(
f*x+e)*ln(F))))^2+2/n/g/f/ln(F)*a*b*d*x*exp(n*ln(exp(g*(f*x+e)*ln(F))))
Fricas [A] (verification not implemented)
none
Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {2 \, {\left (a^{2} d f^{2} g^{2} n^{2} x^{2} + 2 \, a^{2} c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - {\left (b^{2} d - 2 \, {\left (b^{2} d f g n x + b^{2} c f g n\right )} \log \left (F\right )\right )} F^{2 \, f g n x + 2 \, e g n} - 8 \, {\left (a b d - {\left (a b d f g n x + a b c f g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n}}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}}
\]
[In]
integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c),x, algorithm="fricas")
[Out]
1/4*(2*(a^2*d*f^2*g^2*n^2*x^2 + 2*a^2*c*f^2*g^2*n^2*x)*log(F)^2 - (b^2*d - 2*(b^2*d*f*g*n*x + b^2*c*f*g*n)*log
(F))*F^(2*f*g*n*x + 2*e*g*n) - 8*(a*b*d - (a*b*d*f*g*n*x + a*b*c*f*g*n)*log(F))*F^(f*g*n*x + e*g*n))/(f^2*g^2*
n^2*log(F)^2)
Sympy [A] (verification not implemented)
Time = 0.72 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.65
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\begin {cases} \left (a + b\right )^{2} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \wedge f = 0 \wedge g = 0 \wedge n = 0 \\\left (a + b \left (F^{e g}\right )^{n}\right )^{2} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: f = 0 \\\left (a + b\right )^{2} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \vee g = 0 \vee n = 0 \\a^{2} c x + \frac {a^{2} d x^{2}}{2} + \frac {2 a b c \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} + \frac {2 a b d x \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} - \frac {2 a b d \left (F^{e g + f g x}\right )^{n}}{f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} + \frac {b^{2} c \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} + \frac {b^{2} d x \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} - \frac {b^{2} d \left (F^{e g + f g x}\right )^{2 n}}{4 f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases}
\]
[In]
integrate((a+b*(F**(g*(f*x+e)))**n)**2*(d*x+c),x)
[Out]
Piecewise(((a + b)**2*(c*x + d*x**2/2), Eq(F, 1) & Eq(f, 0) & Eq(g, 0) & Eq(n, 0)), ((a + b*(F**(e*g))**n)**2*
(c*x + d*x**2/2), Eq(f, 0)), ((a + b)**2*(c*x + d*x**2/2), Eq(F, 1) | Eq(g, 0) | Eq(n, 0)), (a**2*c*x + a**2*d
*x**2/2 + 2*a*b*c*(F**(e*g + f*g*x))**n/(f*g*n*log(F)) + 2*a*b*d*x*(F**(e*g + f*g*x))**n/(f*g*n*log(F)) - 2*a*
b*d*(F**(e*g + f*g*x))**n/(f**2*g**2*n**2*log(F)**2) + b**2*c*(F**(e*g + f*g*x))**(2*n)/(2*f*g*n*log(F)) + b**
2*d*x*(F**(e*g + f*g*x))**(2*n)/(2*f*g*n*log(F)) - b**2*d*(F**(e*g + f*g*x))**(2*n)/(4*f**2*g**2*n**2*log(F)**
2), True))
Maxima [A] (verification not implemented)
none
Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {1}{2} \, a^{2} d x^{2} + a^{2} c x + \frac {2 \, F^{f g n x + e g n} a b c}{f g n \log \left (F\right )} + \frac {F^{2 \, f g n x + 2 \, e g n} b^{2} c}{2 \, f g n \log \left (F\right )} + \frac {2 \, {\left (F^{e g n} f g n x \log \left (F\right ) - F^{e g n}\right )} F^{f g n x} a b d}{f^{2} g^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (2 \, F^{2 \, e g n} f g n x \log \left (F\right ) - F^{2 \, e g n}\right )} F^{2 \, f g n x} b^{2} d}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}}
\]
[In]
integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c),x, algorithm="maxima")
[Out]
1/2*a^2*d*x^2 + a^2*c*x + 2*F^(f*g*n*x + e*g*n)*a*b*c/(f*g*n*log(F)) + 1/2*F^(2*f*g*n*x + 2*e*g*n)*b^2*c/(f*g*
n*log(F)) + 2*(F^(e*g*n)*f*g*n*x*log(F) - F^(e*g*n))*F^(f*g*n*x)*a*b*d/(f^2*g^2*n^2*log(F)^2) + 1/4*(2*F^(2*e*
g*n)*f*g*n*x*log(F) - F^(2*e*g*n))*F^(2*f*g*n*x)*b^2*d/(f^2*g^2*n^2*log(F)^2)
Giac [C] (verification not implemented)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 2284, normalized size of antiderivative = 14.64
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\text {Too large to display}
\]
[In]
integrate((a+b*(F^(g*(f*x+e)))^n)^2*(d*x+c),x, algorithm="giac")
[Out]
1/2*a^2*d*x^2 + a^2*c*x + 1/2*((2*(pi*b^2*d*f*g*n*x*sgn(F) - pi*b^2*d*f*g*n*x + pi*b^2*c*f*g*n*sgn(F) - pi*b^2
*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^
2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F))
)^2) + (pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)*(2*b^2*d*f*g*n*x*log(abs(F))
+ 2*b^2*c*f*g*n*log(abs(F)) - b^2*d)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))
^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*cos(-pi*f*g*n*x*sgn(F) + pi*f*g
*n*x - pi*e*g*n*sgn(F) + pi*e*g*n) + ((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^
2)*(pi*b^2*d*f*g*n*x*sgn(F) - pi*b^2*d*f*g*n*x + pi*b^2*c*f*g*n*sgn(F) - pi*b^2*c*f*g*n)/((pi^2*f^2*g^2*n^2*sg
n(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n
^2*log(abs(F)))^2) - 2*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))*(2*b^2*d*f*g*n*x*log(a
bs(F)) + 2*b^2*c*f*g*n*log(abs(F)) - b^2*d)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(a
bs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*sin(-pi*f*g*n*x*sgn(F) +
pi*f*g*n*x - pi*e*g*n*sgn(F) + pi*e*g*n))*e^(2*f*g*n*x*log(abs(F)) + 2*e*g*n*log(abs(F))) - 1/4*I*((pi*b^2*d*f
*g*n*x*sgn(F) - pi*b^2*d*f*g*n*x - 2*I*b^2*d*f*g*n*x*log(abs(F)) + pi*b^2*c*f*g*n*sgn(F) - pi*b^2*c*f*g*n - 2*
I*b^2*c*f*g*n*log(abs(F)) + I*b^2*d)*e^(I*pi*f*g*n*x*sgn(F) - I*pi*f*g*n*x + I*pi*e*g*n*sgn(F) - I*pi*e*g*n)/(
pi^2*f^2*g^2*n^2*sgn(F) + 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*g^2*n^2 - 2*I*pi*f^2*g^2*n^2*log(ab
s(F)) + 2*f^2*g^2*n^2*log(abs(F))^2) + (pi*b^2*d*f*g*n*x*sgn(F) - pi*b^2*d*f*g*n*x + 2*I*b^2*d*f*g*n*x*log(abs
(F)) + pi*b^2*c*f*g*n*sgn(F) - pi*b^2*c*f*g*n + 2*I*b^2*c*f*g*n*log(abs(F)) - I*b^2*d)*e^(-I*pi*f*g*n*x*sgn(F)
+ I*pi*f*g*n*x - I*pi*e*g*n*sgn(F) + I*pi*e*g*n)/(pi^2*f^2*g^2*n^2*sgn(F) - 2*I*pi*f^2*g^2*n^2*log(abs(F))*sg
n(F) - pi^2*f^2*g^2*n^2 + 2*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2*log(abs(F))^2))*e^(2*f*g*n*x*log(abs(
F)) + 2*e*g*n*log(abs(F))) + 2*(2*((pi*a*b*d*f*g*n*x*sgn(F) - pi*a*b*d*f*g*n*x + pi*a*b*c*f*g*n*sgn(F) - pi*a*
b*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f
^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)
))^2) + (pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)*(a*b*d*f*g*n*x*log(abs(F))
+ a*b*c*f*g*n*log(abs(F)) - a*b*d)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)
^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*cos(-1/2*pi*f*g*n*x*sgn(F) + 1/2*p
i*f*g*n*x - 1/2*pi*e*g*n*sgn(F) + 1/2*pi*e*g*n) + ((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2
*log(abs(F))^2)*(pi*a*b*d*f*g*n*x*sgn(F) - pi*a*b*d*f*g*n*x + pi*a*b*c*f*g*n*sgn(F) - pi*a*b*c*f*g*n)/((pi^2*f
^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) -
pi*f^2*g^2*n^2*log(abs(F)))^2) - 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))*(a*b*d*f*
g*n*x*log(abs(F)) + a*b*c*f*g*n*log(abs(F)) - a*b*d)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*
n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*sin(-1/2*pi*f*g*
n*x*sgn(F) + 1/2*pi*f*g*n*x - 1/2*pi*e*g*n*sgn(F) + 1/2*pi*e*g*n))*e^(f*g*n*x*log(abs(F)) + e*g*n*log(abs(F)))
- I*((pi*a*b*d*f*g*n*x*sgn(F) - pi*a*b*d*f*g*n*x - 2*I*a*b*d*f*g*n*x*log(abs(F)) + pi*a*b*c*f*g*n*sgn(F) - pi
*a*b*c*f*g*n - 2*I*a*b*c*f*g*n*log(abs(F)) + 2*I*a*b*d)*e^(1/2*I*pi*f*g*n*x*sgn(F) - 1/2*I*pi*f*g*n*x + 1/2*I*
pi*e*g*n*sgn(F) - 1/2*I*pi*e*g*n)/(pi^2*f^2*g^2*n^2*sgn(F) + 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*
g^2*n^2 - 2*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2*log(abs(F))^2) + (pi*a*b*d*f*g*n*x*sgn(F) - pi*a*b*d*
f*g*n*x + 2*I*a*b*d*f*g*n*x*log(abs(F)) + pi*a*b*c*f*g*n*sgn(F) - pi*a*b*c*f*g*n + 2*I*a*b*c*f*g*n*log(abs(F))
- 2*I*a*b*d)*e^(-1/2*I*pi*f*g*n*x*sgn(F) + 1/2*I*pi*f*g*n*x - 1/2*I*pi*e*g*n*sgn(F) + 1/2*I*pi*e*g*n)/(pi^2*f
^2*g^2*n^2*sgn(F) - 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*g^2*n^2 + 2*I*pi*f^2*g^2*n^2*log(abs(F))
+ 2*f^2*g^2*n^2*log(abs(F))^2))*e^(f*g*n*x*log(abs(F)) + e*g*n*log(abs(F)))
Mupad [B] (verification not implemented)
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94
\[
\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 (c+d x) \, dx=\frac {a^2\,d\,x^2}{2}-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}\,\left (\frac {b^2\,\left (d-2\,c\,f\,g\,n\,\ln \left (F\right )\right )}{4\,f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {b^2\,d\,x}{2\,f\,g\,n\,\ln \left (F\right )}\right )-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\,\left (\frac {2\,a\,b\,\left (d-c\,f\,g\,n\,\ln \left (F\right )\right )}{f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {2\,a\,b\,d\,x}{f\,g\,n\,\ln \left (F\right )}\right )+a^2\,c\,x
\]
[In]
int((a + b*(F^(g*(e + f*x)))^n)^2*(c + d*x),x)
[Out]
(a^2*d*x^2)/2 - (F^(f*g*x)*F^(e*g))^(2*n)*((b^2*(d - 2*c*f*g*n*log(F)))/(4*f^2*g^2*n^2*log(F)^2) - (b^2*d*x)/(
2*f*g*n*log(F))) - (F^(f*g*x)*F^(e*g))^n*((2*a*b*(d - c*f*g*n*log(F)))/(f^2*g^2*n^2*log(F)^2) - (2*a*b*d*x)/(f
*g*n*log(F))) + a^2*c*x