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

Optimal. Leaf size=77 \[ \frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac{b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)} \]

[Out]

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

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Rubi [A]  time = 0.07627, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2183, 2176, 2194} \[ \frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac{b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2183

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 2176

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] &&  !$UseGamma === True

Rule 2194

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

Mathematica [A]  time = 0.161687, size = 73, normalized size = 0.95 \[ \frac{1}{2} a x (2 c+d x)+\frac{b (c+d x) \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}-\frac{b d \left (F^{g (e+f x)}\right )^n}{f^2 g^2 n^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 105, normalized size = 1.4 \begin{align*} acx+{\frac{b{{\rm e}^{n\ln \left ({{\rm e}^{g \left ( fx+e \right ) \ln \left ( F \right ) }} \right ) }}c}{ngf\ln \left ( F \right ) }}-{\frac{b{{\rm e}^{n\ln \left ({{\rm e}^{g \left ( fx+e \right ) \ln \left ( F \right ) }} \right ) }}d}{ \left ( \ln \left ( F \right ) \right ) ^{2}{f}^{2}{g}^{2}{n}^{2}}}+{\frac{bdx{{\rm e}^{n\ln \left ({{\rm e}^{g \left ( fx+e \right ) \ln \left ( F \right ) }} \right ) }}}{ngf\ln \left ( F \right ) }}+{\frac{ad{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80532, size = 200, normalized size = 2.6 \begin{align*} \frac{{\left (a d f^{2} g^{2} n^{2} x^{2} + 2 \, a c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 2 \,{\left (b d -{\left (b d f g n x + b c f g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n}}{2 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.461776, size = 94, normalized size = 1.22 \begin{align*} a c x + \frac{a d x^{2}}{2} + \begin{cases} \frac{\left (b c f g n \log{\left (F \right )} + b d f g n x \log{\left (F \right )} - b d\right ) \left (F^{g \left (e + f x\right )}\right )^{n}}{f^{2} g^{2} n^{2} \log{\left (F \right )}^{2}} & \text{for}\: f^{2} g^{2} n^{2} \log{\left (F \right )}^{2} \neq 0 \\b c x + \frac{b d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.54084, size = 1508, normalized size = 19.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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