3.35 \(\int (a+b (F^{g (e+f x)})^n)^2 \, dx\)

Optimal. Leaf size=67 \[ a^2 x+\frac{2 a b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac{b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.0366563, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2282, 266, 43} \[ a^2 x+\frac{2 a b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac{b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^2,x]

[Out]

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0371011, size = 52, normalized size = 0.78 \[ a^2 x+\frac{b \left (F^{g (e+f x)}\right )^n \left (4 a+b \left (F^{g (e+f x)}\right )^n\right )}{2 f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 90, normalized size = 1.3 \begin{align*}{\frac{ \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{2}{b}^{2}}{2\,ngf\ln \left ( F \right ) }}+2\,{\frac{ab \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n}}{ngf\ln \left ( F \right ) }}+{\frac{{a}^{2}\ln \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.29884, size = 101, normalized size = 1.51 \begin{align*} a^{2} x + \frac{2 \,{\left (F^{f g x}\right )}^{n}{\left (F^{e g}\right )}^{n} a b}{f g n \log \left (F\right )} + \frac{{\left (F^{f g x}\right )}^{2 \, n}{\left (F^{e g}\right )}^{2 \, n} b^{2}}{2 \, f g n \log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.51511, size = 136, normalized size = 2.03 \begin{align*} \frac{2 \, a^{2} f g n x \log \left (F\right ) + 4 \, F^{f g n x + e g n} a b + F^{2 \, f g n x + 2 \, e g n} b^{2}}{2 \, f g n \log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.210824, size = 94, normalized size = 1.4 \begin{align*} a^{2} x + \begin{cases} \frac{4 a b f g n \left (F^{g \left (e + f x\right )}\right )^{n} \log{\left (F \right )} + b^{2} f g n \left (F^{g \left (e + f x\right )}\right )^{2 n} \log{\left (F \right )}}{2 f^{2} g^{2} n^{2} \log{\left (F \right )}^{2}} & \text{for}\: 2 f^{2} g^{2} n^{2} \log{\left (F \right )}^{2} \neq 0 \\x \left (2 a b + b^{2}\right ) & \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)**2,x)

[Out]

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

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Giac [C]  time = 1.42809, size = 914, normalized size = 13.64 \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)^2,x, algorithm="giac")

[Out]

a^2*x + (2*b^2*f*g*n*cos(-pi*f*g*n*x*sgn(F) + pi*f*g*n*x - pi*g*n*e*sgn(F) + pi*g*n*e)*log(abs(F))/(4*f^2*g^2*
n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2) - (pi*f*g*n*sgn(F) - pi*f*g*n)*b^2*sin(-pi*f*g*n*x*sgn(F)
+ pi*f*g*n*x - pi*g*n*e*sgn(F) + pi*g*n*e)/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2))*e^(
2*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) - 1/2*I*(-I*b^2*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)/(I*pi*f*g*n*sgn(F) - I*pi*f*g*n + 2*f*g*n*log(abs(F))) + I*b^2*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)/(-I*pi*f*g*n*sgn(F) + I*pi*f*g*n + 2*f*g*n*log(abs(F))))*e^(2
*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) + 4*(2*a*b*f*g*n*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)*log(abs(F))/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2) -
(pi*f*g*n*sgn(F) - pi*f*g*n)*a*b*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)/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2))*e^(f*g*n*x*log(abs(F)) + g*n*e*log(abs(F)
)) - 1/2*I*(-4*I*a*b*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)/(
I*pi*f*g*n*sgn(F) - I*pi*f*g*n + 2*f*g*n*log(abs(F))) + 4*I*a*b*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)/(-I*pi*f*g*n*sgn(F) + I*pi*f*g*n + 2*f*g*n*log(abs(F))))*e^(f*g*n*x
*log(abs(F)) + g*n*e*log(abs(F)))