3.7.7 \(\int \frac {e^x (1-\sinh (x))}{1-\cosh (x)} \, dx\) [607]

Optimal. Leaf size=15 \[ e^x-\frac {2}{1-e^x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2320, 697} \begin {gather*} e^x-\frac {2}{1-e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - Sinh[x]))/(1 - Cosh[x]),x]

[Out]

E^x - 2/(1 - E^x)

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 2320

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

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Mathematica [A]
time = 0.02, size = 20, normalized size = 1.33 \begin {gather*} \frac {2-e^x+e^{2 x}}{-1+e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - Sinh[x]))/(1 - Cosh[x]),x]

[Out]

(2 - E^x + E^(2*x))/(-1 + E^x)

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Maple [A]
time = 0.05, size = 18, normalized size = 1.20

method	result	size
risch	\({\mathrm e}^{x}+\frac {2}{-1+{\mathrm e}^{x}}\)	\(12\)
default	\(-\frac {2}{-1+\tanh \left (\frac {x}{2}\right )}+\frac {1}{\tanh \left (\frac {x}{2}\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(1-sinh(x))/(1-cosh(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 2.84, size = 11, normalized size = 0.73 \begin {gather*} \frac {2}{e^{x} - 1} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(1-sinh(x))/(1-cosh(x)),x, algorithm="maxima")

[Out]

2/(e^x - 1) + e^x

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Fricas [A]
time = 0.64, size = 22, normalized size = 1.47 \begin {gather*} -\frac {3 \, \cosh \left (x\right ) - \sinh \left (x\right ) - 1}{\cosh \left (x\right ) - \sinh \left (x\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(1-sinh(x))/(1-cosh(x)),x, algorithm="fricas")

[Out]

-(3*cosh(x) - sinh(x) - 1)/(cosh(x) - sinh(x) - 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\sinh {\left (x \right )} - 1\right ) e^{x}}{\cosh {\left (x \right )} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(1-sinh(x))/(1-cosh(x)),x)

[Out]

Integral((sinh(x) - 1)*exp(x)/(cosh(x) - 1), x)

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Giac [A]
time = 1.00, size = 11, normalized size = 0.73 \begin {gather*} \frac {2}{e^{x} - 1} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(1-sinh(x))/(1-cosh(x)),x, algorithm="giac")

[Out]

2/(e^x - 1) + e^x

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Mupad [B]
time = 0.04, size = 11, normalized size = 0.73 \begin {gather*} {\mathrm {e}}^x+\frac {2}{{\mathrm {e}}^x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(sinh(x) - 1))/(cosh(x) - 1),x)

[Out]

exp(x) + 2/(exp(x) - 1)

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