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

Optimal. Leaf size=39 \[ x \log (a \cosh (x))-\frac {1}{2} \text {Li}_2\left (-e^{2 x}\right )+\frac {x^2}{2}-x \log \left (e^{2 x}+1\right ) \]

[Out]

1/2*x^2-x*ln(1+exp(2*x))+x*ln(a*cosh(x))-1/2*polylog(2,-exp(2*x))

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.92 \[ \frac {1}{2} \left (\text {Li}_2\left (-e^{-2 x}\right )-x \left (-2 \log (a \cosh (x))+x+2 \log \left (e^{-2 x}+1\right )\right )\right ) \]

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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fricas [C]  time = 0.48, size = 65, normalized size = 1.67 \[ \frac {1}{2} \, x^{2} + x \log \left (a \cosh \relax (x)\right ) - x \log \left (i \, \cosh \relax (x) + i \, \sinh \relax (x) + 1\right ) - x \log \left (-i \, \cosh \relax (x) - i \, \sinh \relax (x) + 1\right ) - {\rm Li}_2\left (i \, \cosh \relax (x) + i \, \sinh \relax (x)\right ) - {\rm Li}_2\left (-i \, \cosh \relax (x) - i \, \sinh \relax (x)\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \cosh \relax (x)\right )\,{d x} \]

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)

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maple [C]  time = 1.04, size = 321, normalized size = 8.23 \[ -\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) a \,{\mathrm e}^{-x}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) a \,{\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) a \,{\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right ) a \,{\mathrm e}^{-x}\right )^{3}}{2}+\frac {x^{2}}{2}+x \ln \relax (a )-x \ln \left ({\mathrm e}^{x}\right )-\ln \left (-i {\mathrm e}^{x}+1\right ) \ln \left ({\mathrm e}^{x}\right )-\ln \left (i {\mathrm e}^{x}+1\right ) \ln \left ({\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{2 x}+1\right ) \ln \left ({\mathrm e}^{x}\right )-\ln \relax (2) x -\dilog \left (-i {\mathrm e}^{x}+1\right )-\dilog \left (i {\mathrm e}^{x}+1\right ) \]

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*a*(exp(2*x)+1)*exp(-x))^3*x+1/2*I*Pi*csgn(I*(exp(2*x)+1))*csgn(I*exp(-x)*(exp(2*
x)+1))^2*x-1/2*I*Pi*csgn(I*a)*csgn(I*exp(-x)*(exp(2*x)+1))*csgn(I*a*(exp(2*x)+1)*exp(-x))*x+1/2*I*Pi*csgn(I*ex
p(-x)*(exp(2*x)+1))*csgn(I*a*(exp(2*x)+1)*exp(-x))^2*x+1/2*I*Pi*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(2*x)+1))^2
*x+1/2*I*Pi*csgn(I*a)*csgn(I*a*(exp(2*x)+1)*exp(-x))^2*x+x*ln(a)-ln(2)*x+1/2*x^2-1/2*I*Pi*csgn(I*exp(-x))*csgn
(I*(exp(2*x)+1))*csgn(I*exp(-x)*(exp(2*x)+1))*x-1/2*I*Pi*csgn(I*exp(-x)*(exp(2*x)+1))^3*x+ln(exp(x))*ln(exp(2*
x)+1)-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 = 1.15, size = 32, normalized size = 0.82 \[ \frac {1}{2} \, x^{2} + x \log \left (a \cosh \relax (x)\right ) - x \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, {\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \ln \left (a\,\mathrm {cosh}\relax (x)\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(log(a*cosh(x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \cosh {\relax (x )} \right )}\, dx \]

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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