3.1.34 \(\int (a+b (F^{g (e+f x)})^n)^2 (c+d x) \, dx\) [34]

Optimal. Leaf size=156 \[ \frac {a^2 (c+d x)^2}{2 d}-\frac {2 a b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}+\frac {2 a b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)} \]

[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)

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Rubi [A]
time = 0.11, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2214, 2207, 2225} \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.31, size = 117, normalized size = 0.75 \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [A]
time = 0.03, size = 174, normalized size = 1.12

method	result	size
norman	\(a^{2} c x +\frac {a^{2} d \,x^{2}}{2}+\frac {b^{2} \left (2 \ln \left (F \right ) c f g n -d \right ) {\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{4 n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}+\frac {2 a b \left (\ln \left (F \right ) c f g n -d \right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}+\frac {b^{2} d x \,{\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{2 n g f \ln \left (F \right )}+\frac {2 a b d x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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Maxima [A]
time = 0.30, size = 184, normalized size = 1.18 \begin {gather*} \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x + \frac {2 \, F^{f g n x + g n e} a b c}{f g n \log \left (F\right )} + \frac {F^{2 \, f g n x + 2 \, g n e} b^{2} c}{2 \, f g n \log \left (F\right )} + \frac {2 \, {\left (F^{g n e} f g n x \log \left (F\right ) - F^{g n e}\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 \, g n e} f g n x \log \left (F\right ) - F^{2 \, g n e}\right )} F^{2 \, f g n x} b^{2} d}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Fricas [A]
time = 0.35, size = 141, normalized size = 0.90 \begin {gather*} \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 \, g n e} - 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 + g n e}}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*g*n*e) - 8*(a*b*d - (a*b*d*f*g*n*x + a*b*c*f*g*n)*log(F))*F^(f*g*n*x + g*n*e))/(f^2*g^2*
n^2*log(F)^2)

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Sympy [A]
time = 0.11, size = 231, normalized size = 1.48 \begin {gather*} a^{2} c x + \frac {a^{2} d x^{2}}{2} + \begin {cases} \frac {\left (2 b^{2} c f^{3} g^{3} n^{3} \log {\left (F \right )}^{3} + 2 b^{2} d f^{3} g^{3} n^{3} x \log {\left (F \right )}^{3} - b^{2} d f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}\right ) \left (F^{g \left (e + f x\right )}\right )^{2 n} + \left (8 a b c f^{3} g^{3} n^{3} \log {\left (F \right )}^{3} + 8 a b d f^{3} g^{3} n^{3} x \log {\left (F \right )}^{3} - 8 a b d f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}\right ) \left (F^{g \left (e + f x\right )}\right )^{n}}{4 f^{4} g^{4} n^{4} \log {\left (F \right )}^{4}} & \text {for}\: f^{4} g^{4} n^{4} \log {\left (F \right )}^{4} \neq 0 \\x^{2} \left (a b d + \frac {b^{2} d}{2}\right ) + x \left (2 a b c + b^{2} c\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*x + a**2*d*x**2/2 + Piecewise((((2*b**2*c*f**3*g**3*n**3*log(F)**3 + 2*b**2*d*f**3*g**3*n**3*x*log(F)**
3 - b**2*d*f**2*g**2*n**2*log(F)**2)*(F**(g*(e + f*x)))**(2*n) + (8*a*b*c*f**3*g**3*n**3*log(F)**3 + 8*a*b*d*f
**3*g**3*n**3*x*log(F)**3 - 8*a*b*d*f**2*g**2*n**2*log(F)**2)*(F**(g*(e + f*x)))**n)/(4*f**4*g**4*n**4*log(F)*
*4), Ne(f**4*g**4*n**4*log(F)**4, 0)), (x**2*(a*b*d + b**2*d/2) + x*(2*a*b*c + b**2*c), True))

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Giac [C] Result contains complex when optimal does not.
time = 3.18, size = 2304, normalized size = 14.77 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*g*n*e*sgn(F) + pi*g*n*e) + ((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*g*n*e*sgn(F) + pi*g*n*e))*e^(2*f*g*n*x*log(abs(F)) + 2*g*n*e*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*g*n*e*sgn(F) - I*pi*g*n*e)/(
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*g*n*e*sgn(F) + I*pi*g*n*e)/(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*g*n*e*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*g*n*e*sgn(F) + 1/2*pi*g*n*e) + ((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*g*n*e*sgn(F) + 1/2*pi*g*n*e))*e^(f*g*n*x*log(abs(F)) + g*n*e*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*g*n*e*sgn(F) - 1/2*I*pi*g*n*e)/(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*g*n*e*sgn(F) + 1/2*I*pi*g*n*e)/(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)) + g*n*e*log(abs(F)))

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Mupad [B]
time = 3.67, size = 146, normalized size = 0.94 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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