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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, 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 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])

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 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]]]

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*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.91 \[ \frac {1}{2} \left (4 x-e^{-2 x}+e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x) + E^x)^2,x]

[Out]

(-E^(-2*x) + E^(2*x) + 4*x)/2

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fricas [A]  time = 0.40, size = 19, normalized size = 0.86 \[ \frac {1}{2} \, {\left (4 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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giac [A]  time = 0.20, size = 24, normalized size = 1.09 \[ -\frac {1}{2} \, {\left (2 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )} + 2 \, x + \frac {1}{2} \, e^{\left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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maple [A]  time = 0.03, size = 17, normalized size = 0.77 \[ 2 x -\frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)+exp(x))^2,x)

[Out]

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

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maxima [A]  time = 0.74, size = 16, normalized size = 0.73 \[ 2 \, x + \frac {1}{2} \, e^{\left (2 \, x\right )} - \frac {1}{2} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [B]  time = 3.58, size = 8, normalized size = 0.36 \[ 2\,x+\mathrm {sinh}\left (2\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x) + exp(x))^2,x)

[Out]

2*x + sinh(2*x)

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sympy [A]  time = 0.11, size = 17, normalized size = 0.77 \[ 2 x + \frac {e^{2 x}}{2} - \frac {e^{- 2 x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + exp(2*x)/2 - exp(-2*x)/2

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