3.202 \(\int \log (a \sinh ^2(x)) \, dx\)

Optimal. Leaf size=35 \[ -\text{PolyLog}\left (2,e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )+x^2-2 x \log \left (1-e^{2 x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.060352, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {2548, 12, 3716, 2190, 2279, 2391} \[ -\text{PolyLog}\left (2,e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )+x^2-2 x \log \left (1-e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 12

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

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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 \left (a \sinh ^2(x)\right ) \, dx &=x \log \left (a \sinh ^2(x)\right )-\int 2 x \coth (x) \, dx\\ &=x \log \left (a \sinh ^2(x)\right )-2 \int x \coth (x) \, dx\\ &=x^2+x \log \left (a \sinh ^2(x)\right )+4 \int \frac{e^{2 x} x}{1-e^{2 x}} \, dx\\ &=x^2-2 x \log \left (1-e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )+2 \int \log \left (1-e^{2 x}\right ) \, dx\\ &=x^2-2 x \log \left (1-e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )+\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 x}\right )\\ &=x^2-2 x \log \left (1-e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )-\text{Li}_2\left (e^{2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0182854, size = 33, normalized size = 0.94 \[ \text{PolyLog}\left (2,e^{-2 x}\right )+x \left (\log \left (a \sinh ^2(x)\right )-x-2 \log \left (1-e^{-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.157, size = 454, normalized size = 13. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.2536, size = 58, normalized size = 1.66 \begin{align*} x^{2} + x \log \left (a \sinh \left (x\right )^{2}\right ) - 2 \, x \log \left (e^{x} + 1\right ) - 2 \, x \log \left (-e^{x} + 1\right ) - 2 \,{\rm Li}_2\left (-e^{x}\right ) - 2 \,{\rm Li}_2\left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.01435, size = 246, normalized size = 7.03 \begin{align*} x^{2} + x \log \left (\frac{1}{2} \, a \cosh \left (x\right )^{2} + \frac{1}{2} \, a \sinh \left (x\right )^{2} - \frac{1}{2} \, a\right ) - 2 \, x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - 2 \, x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) - 2 \,{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 2 \,{\rm Li}_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(a*sinh(x)^2),x, algorithm="fricas")

[Out]

x^2 + x*log(1/2*a*cosh(x)^2 + 1/2*a*sinh(x)^2 - 1/2*a) - 2*x*log(cosh(x) + sinh(x) + 1) - 2*x*log(-cosh(x) - s
inh(x) + 1) - 2*dilog(cosh(x) + sinh(x)) - 2*dilog(-cosh(x) - sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \sinh ^{2}{\left (x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sinh(x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \sinh \left (x\right )^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sinh(x)^2), x)