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

Optimal result

Integrand size = 21, antiderivative size = 77 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\frac {a (c+d x)^2}{2 d}-\frac {b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}+\frac {b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2214, 2207, 2225} \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\frac {a (c+d x)^2}{2 d}+\frac {b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac {b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)} \]

[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 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 (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] (verified)

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\frac {1}{2} a x (2 c+d x)-\frac {b d \left (F^{g (e+f x)}\right )^n}{f^2 g^2 n^2 \log ^2(F)}+\frac {b \left (F^{g (e+f x)}\right )^n (c+d x)}{f g n \log (F)} \]

[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])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09

method	result	size
norman	\(c a x +\frac {b \left (\ln \left (F \right ) c f g n -d \right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}+\frac {b d x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}+\frac {a d \,x^{2}}{2}\)	\(84\)
parallelrisch	\(\frac {a d \,x^{2} n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+2 c a x \,n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+2 \left (F^{g \left (f x +e \right )}\right )^{n} \ln \left (F \right ) b d f g n x +2 \left (F^{g \left (f x +e \right )}\right )^{n} \ln \left (F \right ) b c f g n -2 \left (F^{g \left (f x +e \right )}\right )^{n} b d}{2 n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\frac {{\left (a d f^{2} g^{2} n^{2} x^{2} + 2 \, a c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 2 \, {\left (b d - {\left (b d f g n x + b c f g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n}}{2 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (68) = 136\).

Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.87 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\begin {cases} \left (a + b\right ) \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \wedge f = 0 \wedge g = 0 \wedge n = 0 \\\left (a + b \left (F^{e g}\right )^{n}\right ) \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: f = 0 \\\left (a + b\right ) \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \vee g = 0 \vee n = 0 \\a c x + \frac {a d x^{2}}{2} + \frac {b c \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} + \frac {b d x \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} - \frac {b d \left (F^{e g + f g x}\right )^{n}}{f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\frac {1}{2} \, a d x^{2} + a c x + \frac {F^{f g n x + e g n} b c}{f g n \log \left (F\right )} + \frac {{\left (F^{e g n} f g n x \log \left (F\right ) - F^{e g n}\right )} F^{f g n x} b d}{f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 1101, normalized size of antiderivative = 14.30 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=\text {Too large to display} \]

[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*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*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*
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))) - 1/2*I*((pi*b*d*f*g*n*x*sgn(F) - pi
*b*d*f*g*n*x - 2*I*b*d*f*g*n*x*log(abs(F)) + pi*b*c*f*g*n*sgn(F) - pi*b*c*f*g*n - 2*I*b*c*f*g*n*log(abs(F)) +
2*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*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*b*d*f*g*n*x*sgn(F) - pi*b*d*f*g*n*x + 2*I*b*d*f*g*n*x*log(abs(F)) + pi*b*c*f*g*
n*sgn(F) - pi*b*c*f*g*n + 2*I*b*c*f*g*n*log(abs(F)) - 2*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*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) - p
i^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 = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x) \, dx=a\,c\,x-\left (\frac {b\,\left (d-c\,f\,g\,n\,\ln \left (F\right )\right )}{f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {b\,d\,x}{f\,g\,n\,\ln \left (F\right )}\right )\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n+\frac {a\,d\,x^2}{2} \]

[In]

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

[Out]

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