3.497 \(\int (-e^{-x}+e^x)^2 \, dx\)

Optimal. Leaf size=22 \[ -2 x-\frac{e^{-2 x}}{2}+\frac{e^{2 x}}{2} \]

[Out]

-1/(2*E^(2*x)) + E^(2*x)/2 - 2*x

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Rubi [A]  time = 0.0177986, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2282, 266, 43} \[ -2 x-\frac{e^{-2 x}}{2}+\frac{e^{2 x}}{2} \]

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + E^x)^2,x]

[Out]

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

Mathematica [A]  time = 0.0058902, size = 22, normalized size = 1. \[ -2 x-\frac{e^{-2 x}}{2}+\frac{e^{2 x}}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-x) + E^x)^2,x]

[Out]

-1/(2*E^(2*x)) + E^(2*x)/2 - 2*x

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Maple [A]  time = 0.006, size = 19, normalized size = 0.9 \begin{align*}{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}}{2}}-2\,\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((-1/exp(x)+exp(x))^2,x)

[Out]

1/2*exp(x)^2-2*ln(exp(x))-1/2/exp(x)^2

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Maxima [A]  time = 0.926443, size = 22, normalized size = 1. \begin{align*} -2 \, x + \frac{1}{2} \, e^{\left (2 \, x\right )} - \frac{1}{2} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))^2,x, algorithm="maxima")

[Out]

-2*x + 1/2*e^(2*x) - 1/2*e^(-2*x)

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Fricas [A]  time = 1.86786, size = 58, normalized size = 2.64 \begin{align*} -\frac{1}{2} \,{\left (4 \, x e^{\left (2 \, x\right )} - e^{\left (4 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.101106, size = 17, normalized size = 0.77 \begin{align*} - 2 x + \frac{e^{2 x}}{2} - \frac{e^{- 2 x}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))**2,x)

[Out]

-2*x + exp(2*x)/2 - exp(-2*x)/2

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Giac [A]  time = 1.05804, size = 32, normalized size = 1.45 \begin{align*} \frac{1}{2} \,{\left (2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )} - 2 \, x + \frac{1}{2} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))^2,x, algorithm="giac")

[Out]

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