### 3.204 $$\int \log (a \cosh (x)) \, dx$$

Optimal. Leaf size=39 $-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 x}\right )+x \log (a \cosh (x))+\frac{x^2}{2}-x \log \left (e^{2 x}+1\right )$

[Out]

x^2/2 - x*Log[1 + E^(2*x)] + x*Log[a*Cosh[x]] - PolyLog[2, -E^(2*x)]/2

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Rubi [A]  time = 0.0556963, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 5, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 1., Rules used = {2548, 3718, 2190, 2279, 2391} $-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 x}\right )+x \log (a \cosh (x))+\frac{x^2}{2}-x \log \left (e^{2 x}+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[a*Cosh[x]],x]

[Out]

x^2/2 - x*Log[1 + E^(2*x)] + x*Log[a*Cosh[x]] - PolyLog[2, -E^(2*x)]/2

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \log (a \cosh (x)) \, dx &=x \log (a \cosh (x))-\int x \tanh (x) \, dx\\ &=\frac{x^2}{2}+x \log (a \cosh (x))-2 \int \frac{e^{2 x} x}{1+e^{2 x}} \, dx\\ &=\frac{x^2}{2}-x \log \left (1+e^{2 x}\right )+x \log (a \cosh (x))+\int \log \left (1+e^{2 x}\right ) \, dx\\ &=\frac{x^2}{2}-x \log \left (1+e^{2 x}\right )+x \log (a \cosh (x))+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 x}\right )\\ &=\frac{x^2}{2}-x \log \left (1+e^{2 x}\right )+x \log (a \cosh (x))-\frac{1}{2} \text{Li}_2\left (-e^{2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0183128, size = 36, normalized size = 0.92 $\frac{1}{2} \left (\text{PolyLog}\left (2,-e^{-2 x}\right )-x \left (-2 \log (a \cosh (x))+x+2 \log \left (e^{-2 x}+1\right )\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[a*Cosh[x]],x]

[Out]

(-(x*(x + 2*Log[1 + E^(-2*x)] - 2*Log[a*Cosh[x]])) + PolyLog[2, -E^(-2*x)])/2

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Maple [C]  time = 0.109, size = 321, normalized size = 8.2 \begin{align*} -x\ln \left ({{\rm e}^{x}} \right ) +{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) \left ({\it csgn} \left ( ia \left ( 1+{{\rm e}^{2\,x}} \right ){{\rm e}^{-x}} \right ) \right ) ^{2}x+{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ia \left ( 1+{{\rm e}^{2\,x}} \right ){{\rm e}^{-x}} \right ) \right ) ^{2}{\it csgn} \left ( ia \right ) x-{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ){\it csgn} \left ( ia \left ( 1+{{\rm e}^{2\,x}} \right ){{\rm e}^{-x}} \right ){\it csgn} \left ( ia \right ) x+{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) \right ) ^{2}x-{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) \right ) ^{3}x+{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \right ) \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) \right ) ^{2}x+\ln \left ( a \right ) x-\ln \left ( 2 \right ) x+{\frac{{x}^{2}}{2}}-{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ia \left ( 1+{{\rm e}^{2\,x}} \right ){{\rm e}^{-x}} \right ) \right ) ^{3}x-{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \right ){\it csgn} \left ( i \left ( 1+{{\rm e}^{2\,x}} \right ) \right ){\it csgn} \left ( i{{\rm e}^{-x}} \left ( 1+{{\rm e}^{2\,x}} \right ) \right ) x+\ln \left ({{\rm e}^{x}} \right ) \ln \left ( 1+{{\rm e}^{2\,x}} \right ) -\ln \left ({{\rm e}^{x}} \right ) \ln \left ( 1+i{{\rm e}^{x}} \right ) -\ln \left ({{\rm e}^{x}} \right ) \ln \left ( 1-i{{\rm e}^{x}} \right ) -{\it dilog} \left ( 1+i{{\rm e}^{x}} \right ) -{\it dilog} \left ( 1-i{{\rm e}^{x}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*cosh(x)),x)

[Out]

-x*ln(exp(x))+1/2*I*Pi*csgn(I*exp(-x)*(1+exp(2*x)))*csgn(I*a*(1+exp(2*x))*exp(-x))^2*x+1/2*I*Pi*csgn(I*a*(1+ex
p(2*x))*exp(-x))^2*csgn(I*a)*x-1/2*I*Pi*csgn(I*exp(-x)*(1+exp(2*x)))*csgn(I*a*(1+exp(2*x))*exp(-x))*csgn(I*a)*
x+1/2*I*Pi*csgn(I*(1+exp(2*x)))*csgn(I*exp(-x)*(1+exp(2*x)))^2*x-1/2*I*Pi*csgn(I*exp(-x)*(1+exp(2*x)))^3*x+1/2
*I*Pi*csgn(I*exp(-x))*csgn(I*exp(-x)*(1+exp(2*x)))^2*x+ln(a)*x-ln(2)*x+1/2*x^2-1/2*I*Pi*csgn(I*a*(1+exp(2*x))*
exp(-x))^3*x-1/2*I*Pi*csgn(I*exp(-x))*csgn(I*(1+exp(2*x)))*csgn(I*exp(-x)*(1+exp(2*x)))*x+ln(exp(x))*ln(1+exp(
2*x))-ln(exp(x))*ln(1+I*exp(x))-ln(exp(x))*ln(1-I*exp(x))-dilog(1+I*exp(x))-dilog(1-I*exp(x))

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Maxima [A]  time = 1.70867, size = 43, normalized size = 1.1 \begin{align*} \frac{1}{2} \, x^{2} + x \log \left (a \cosh \left (x\right )\right ) - x \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{2} \,{\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.00195, size = 219, normalized size = 5.62 \begin{align*} \frac{1}{2} \, x^{2} + x \log \left (a \cosh \left (x\right )\right ) - x \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - x \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) -{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) -{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)),x, algorithm="fricas")

[Out]

1/2*x^2 + x*log(a*cosh(x)) - x*log(I*cosh(x) + I*sinh(x) + 1) - x*log(-I*cosh(x) - I*sinh(x) + 1) - dilog(I*co
sh(x) + I*sinh(x)) - dilog(-I*cosh(x) - I*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \cosh{\left (x \right )} \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*cosh(x)),x)

[Out]

Integral(log(a*cosh(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \cosh \left (x\right )\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)),x, algorithm="giac")

[Out]

integrate(log(a*cosh(x)), x)