### 3.222 $$\int \log (\cosh ^2(x)) \sinh (x) \, dx$$

Optimal. Leaf size=13 $\cosh (x) \log \left (\cosh ^2(x)\right )-2 \cosh (x)$

[Out]

-2*Cosh[x] + Cosh[x]*Log[Cosh[x]^2]

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Rubi [A]  time = 0.0191519, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.375, Rules used = {2638, 2554, 12} $\cosh (x) \log \left (\cosh ^2(x)\right )-2 \cosh (x)$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[Cosh[x]^2]*Sinh[x],x]

[Out]

-2*Cosh[x] + Cosh[x]*Log[Cosh[x]^2]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int \log \left (\cosh ^2(x)\right ) \sinh (x) \, dx &=\cosh (x) \log \left (\cosh ^2(x)\right )-\int 2 \sinh (x) \, dx\\ &=\cosh (x) \log \left (\cosh ^2(x)\right )-2 \int \sinh (x) \, dx\\ &=-2 \cosh (x)+\cosh (x) \log \left (\cosh ^2(x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0078101, size = 13, normalized size = 1. $\cosh (x) \log \left (\cosh ^2(x)\right )-2 \cosh (x)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[Cosh[x]^2]*Sinh[x],x]

[Out]

-2*Cosh[x] + Cosh[x]*Log[Cosh[x]^2]

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Maple [A]  time = 0.013, size = 14, normalized size = 1.1 \begin{align*} -2\,\cosh \left ( x \right ) +\cosh \left ( x \right ) \ln \left ( \left ( \cosh \left ( x \right ) \right ) ^{2} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(ln(cosh(x)^2)*sinh(x),x)

[Out]

-2*cosh(x)+cosh(x)*ln(cosh(x)^2)

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Maxima [A]  time = 1.02456, size = 16, normalized size = 1.23 \begin{align*} 2 \, \cosh \left (x\right ) \log \left (\cosh \left (x\right )\right ) - 2 \, \cosh \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cosh(x)^2)*sinh(x),x, algorithm="maxima")

[Out]

2*cosh(x)*log(cosh(x)) - 2*cosh(x)

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Fricas [B]  time = 1.79697, size = 228, normalized size = 17.54 \begin{align*} -\frac{2 \, \cosh \left (x\right )^{2} -{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \cosh \left (x\right )^{2} + \frac{1}{2} \, \sinh \left (x\right )^{2} + \frac{1}{2}\right ) + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} + 2}{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cosh(x)^2)*sinh(x),x, algorithm="fricas")

[Out]

-1/2*(2*cosh(x)^2 - (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*log(1/2*cosh(x)^2 + 1/2*sinh(x)^2 + 1/2) +
4*cosh(x)*sinh(x) + 2*sinh(x)^2 + 2)/(cosh(x) + sinh(x))

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Sympy [A]  time = 1.00068, size = 14, normalized size = 1.08 \begin{align*} 2 \log{\left (\cosh{\left (x \right )} \right )} \cosh{\left (x \right )} - 2 \cosh{\left (x \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(cosh(x)**2)*sinh(x),x)

[Out]

2*log(cosh(x))*cosh(x) - 2*cosh(x)

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Giac [B]  time = 1.25137, size = 42, normalized size = 3.23 \begin{align*}{\left (e^{\left (-x\right )} + e^{x}\right )} \log \left (\frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x}\right ) - e^{\left (-x\right )} - e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cosh(x)^2)*sinh(x),x, algorithm="giac")

[Out]

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