\(\int \log (a \sinh ^2(x)) \, dx\) [202]

Optimal result
Mathematica [A] (verified)
Rubi [C] (verified)
Maple [C] (warning: unable to verify)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 7, antiderivative size = 35 \[ \int \log \left (a \sinh ^2(x)\right ) \, dx=x^2-2 x \log \left (1-e^{2 x}\right )+x \log \left (a \sinh ^2(x)\right )-\operatorname {PolyLog}\left (2,e^{2 x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \log \left (a \sinh ^2(x)\right ) \, dx=x \left (-x-2 \log \left (1-e^{-2 x}\right )+\log \left (a \sinh ^2(x)\right )\right )+\operatorname {PolyLog}\left (2,e^{-2 x}\right ) \] Input:

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

Output:

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

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.286, Rules used = {3028, 27, 3042, 26, 4199, 25, 2620, 2715, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \log \left (a \sinh ^2(x)\right ) \, dx\)

\(\Big \downarrow \) 3028

\(\displaystyle x \log \left (a \sinh ^2(x)\right )-\int 2 x \coth (x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle x \log \left (a \sinh ^2(x)\right )-2 \int x \coth (x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle x \log \left (a \sinh ^2(x)\right )-2 \int -i x \tan \left (i x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 26

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \int x \tan \left (i x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 4199

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \left (2 i \int -\frac {e^{2 x} x}{1-e^{2 x}}dx-\frac {i x^2}{2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \left (-2 i \int \frac {e^{2 x} x}{1-e^{2 x}}dx-\frac {i x^2}{2}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \left (-2 i \left (\frac {1}{2} \int \log \left (1-e^{2 x}\right )dx-\frac {1}{2} x \log \left (1-e^{2 x}\right )\right )-\frac {i x^2}{2}\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \left (-2 i \left (\frac {1}{4} \int e^{-2 x} \log \left (1-e^{2 x}\right )de^{2 x}-\frac {1}{2} x \log \left (1-e^{2 x}\right )\right )-\frac {i x^2}{2}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle x \log \left (a \sinh ^2(x)\right )+2 i \left (-2 i \left (-\frac {\operatorname {PolyLog}\left (2,e^{2 x}\right )}{4}-\frac {1}{2} x \log \left (1-e^{2 x}\right )\right )-\frac {i x^2}{2}\right )\)

Input:

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

Output:

x*Log[a*Sinh[x]^2] + (2*I)*((-1/2*I)*x^2 - (2*I)*(-1/2*(x*Log[1 - E^(2*x)] 
) - PolyLog[2, E^(2*x)]/4))
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2715
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838
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 3028
Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, 
 x]/u), x], x] /; InverseFunctionFreeQ[u, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4199
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] + Simp 
[2*I   Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x 
))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && In 
tegerQ[4*k] && IGtQ[m, 0]
 
Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.05 (sec) , antiderivative size = 454, normalized size of antiderivative = 12.97

method result size
risch \(x^{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2}\right ) {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{2} x}{2}-\frac {i \pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{3} x}{2}+\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) x}{2}-\frac {i \pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2}\right )}^{3} x}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{2} x}{2}-\frac {i \pi {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{3} x}{2}+i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right ) {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2}\right )}^{2} x +\frac {i \pi \,\operatorname {csgn}\left (i a \right ) {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{2} x}{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} x +\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (i a \right ) \operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right ) x}{2}+2 \operatorname {dilog}\left ({\mathrm e}^{x}\right )-2 \operatorname {dilog}\left ({\mathrm e}^{x}+1\right )+\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right ) {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right )}^{2} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}\right ) x}{2}-\frac {i \pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right )}^{2} \operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )^{2}\right ) x}{2}+2 \ln \left ({\mathrm e}^{x}\right ) \ln \left (-1+{\mathrm e}^{2 x}\right )-2 \ln \left ({\mathrm e}^{x}\right ) \ln \left ({\mathrm e}^{x}+1\right )-2 x \ln \left ({\mathrm e}^{x}\right )+x \ln \left (a \right )-2 \ln \left (2\right ) x\) \(454\)

Input:

int(ln(a*sinh(x)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (32) = 64\).

Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.97 \[ \int \log \left (a \sinh ^2(x)\right ) \, dx=x^{2} + x \log \left (\frac {1}{2} \, a \cosh \left (x\right )^{2} + \frac {1}{2} \, a \sinh \left (x\right )^{2} - \frac {1}{2} \, a\right ) - 2 \, x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - 2 \, x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) - 2 \, {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 2 \, {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \] Input:

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

Output:

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) - sinh(x) + 1) - 2*dilog(cosh(x) + sinh(x 
)) - 2*dilog(-cosh(x) - sinh(x))
 

Sympy [F]

\[ \int \log \left (a \sinh ^2(x)\right ) \, dx=\int \log {\left (a \sinh ^{2}{\left (x \right )} \right )}\, dx \] Input:

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

Output:

Integral(log(a*sinh(x)**2), x)
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \log \left (a \sinh ^2(x)\right ) \, dx=x^{2} + x \log \left (a \sinh \left (x\right )^{2}\right ) - 2 \, x \log \left (e^{x} + 1\right ) - 2 \, x \log \left (-e^{x} + 1\right ) - 2 \, {\rm Li}_2\left (-e^{x}\right ) - 2 \, {\rm Li}_2\left (e^{x}\right ) \] Input:

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

Output:

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)
 

Giac [F]

\[ \int \log \left (a \sinh ^2(x)\right ) \, dx=\int { \log \left (a \sinh \left (x\right )^{2}\right ) \,d x } \] Input:

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

Output:

integrate(log(a*sinh(x)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \log \left (a \sinh ^2(x)\right ) \, dx=\int \ln \left (a\,{\mathrm {sinh}\left (x\right )}^2\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \log \left (a \sinh ^2(x)\right ) \, dx=\int \mathrm {log}\left (\sinh \left (x \right )^{2} a \right )d x \] Input:

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

Output:

int(log(sinh(x)**2*a),x)