3.723 \(\int e^x (-e^{-x}+e^x)^3 \, dx\)

Optimal. Leaf size=31 \[ 3 x+\frac{e^{-2 x}}{2}-\frac{3 e^{2 x}}{2}+\frac{e^{4 x}}{4} \]

[Out]

1/(2*E^(2*x)) - (3*E^(2*x))/2 + E^(4*x)/4 + 3*x

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Rubi [A]  time = 0.0366155, antiderivative size = 31, 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} \[ 3 x+\frac{e^{-2 x}}{2}-\frac{3 e^{2 x}}{2}+\frac{e^{4 x}}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^x*(-E^(-x) + E^x)^3,x]

[Out]

1/(2*E^(2*x)) - (3*E^(2*x))/2 + E^(4*x)/4 + 3*x

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 e^x \left (-e^{-x}+e^x\right )^3 \, dx &=\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{x^3} \, dx,x,e^x\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(-1+x)^3}{x^2} \, dx,x,e^{2 x}\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-3-\frac{1}{x^2}+\frac{3}{x}+x\right ) \, dx,x,e^{2 x}\right )\\ &=\frac{e^{-2 x}}{2}-\frac{3 e^{2 x}}{2}+\frac{e^{4 x}}{4}+3 x\\ \end{align*}

Mathematica [A]  time = 0.0126826, size = 29, normalized size = 0.94 \[ \frac{1}{2} \left (6 x+e^{-2 x}-3 e^{2 x}+\frac{e^{4 x}}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*(-E^(-x) + E^x)^3,x]

[Out]

(E^(-2*x) - 3*E^(2*x) + E^(4*x)/2 + 6*x)/2

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Maple [A]  time = 0.026, size = 25, normalized size = 0.8 \begin{align*}{\frac{ \left ({{\rm e}^{x}} \right ) ^{4}}{4}}-{\frac{3\, \left ({{\rm e}^{x}} \right ) ^{2}}{2}}+3\,\ln \left ({{\rm e}^{x}} \right ) +{\frac{1}{2\, \left ({{\rm e}^{x}} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(-1/exp(x)+exp(x))^3,x)

[Out]

1/4*exp(x)^4-3/2*exp(x)^2+3*ln(exp(x))+1/2/exp(x)^2

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Maxima [A]  time = 0.968184, size = 32, normalized size = 1.03 \begin{align*} -\frac{1}{4} \,{\left (6 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (4 \, x\right )} + 3 \, x + \frac{1}{2} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(-1/exp(x)+exp(x))^3,x, algorithm="maxima")

[Out]

-1/4*(6*e^(-2*x) - 1)*e^(4*x) + 3*x + 1/2*e^(-2*x)

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Fricas [A]  time = 0.856181, size = 74, normalized size = 2.39 \begin{align*} \frac{1}{4} \,{\left (12 \, x e^{\left (2 \, x\right )} + e^{\left (6 \, x\right )} - 6 \, e^{\left (4 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(-1/exp(x)+exp(x))^3,x, algorithm="fricas")

[Out]

1/4*(12*x*e^(2*x) + e^(6*x) - 6*e^(4*x) + 2)*e^(-2*x)

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Sympy [A]  time = 0.133643, size = 26, normalized size = 0.84 \begin{align*} 3 x + \frac{e^{4 x}}{4} - \frac{3 e^{2 x}}{2} + \frac{e^{- 2 x}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(-1/exp(x)+exp(x))**3,x)

[Out]

3*x + exp(4*x)/4 - 3*exp(2*x)/2 + exp(-2*x)/2

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Giac [A]  time = 1.20512, size = 41, normalized size = 1.32 \begin{align*} -\frac{1}{2} \,{\left (3 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )} + 3 \, x + \frac{1}{4} \, e^{\left (4 \, x\right )} - \frac{3}{2} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(-1/exp(x)+exp(x))^3,x, algorithm="giac")

[Out]

-1/2*(3*e^(2*x) - 1)*e^(-2*x) + 3*x + 1/4*e^(4*x) - 3/2*e^(2*x)