3.216 \(\int \log (a \text {sech}^2(x)) \, dx\)

Optimal. Leaf size=35 \[ x \log \left (a \text {sech}^2(x)\right )+\text {Li}_2\left (-e^{2 x}\right )-x^2+2 x \log \left (e^{2 x}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, 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, 3718, 2190, 2279, 2391} \[ \text {PolyLog}\left (2,-e^{2 x}\right )+x \log \left (a \text {sech}^2(x)\right )-x^2+2 x \log \left (e^{2 x}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2 + 2*x*Log[1 + E^(2*x)] + x*Log[a*Sech[x]^2] + PolyLog[2, -E^(2*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 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 \left (a \text {sech}^2(x)\right ) \, dx &=x \log \left (a \text {sech}^2(x)\right )-\int -2 x \tanh (x) \, dx\\ &=x \log \left (a \text {sech}^2(x)\right )+2 \int x \tanh (x) \, dx\\ &=-x^2+x \log \left (a \text {sech}^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 \text {sech}^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 \text {sech}^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 \text {sech}^2(x)\right )+\text {Li}_2\left (-e^{2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.94 \[ x \left (\log \left (a \text {sech}^2(x)\right )+x+2 \log \left (e^{-2 x}+1\right )\right )-\text {Li}_2\left (-e^{-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.48, size = 106, normalized size = 3.03 \[ -x^{2} + x \log \left (\frac {4 \, {\left (a \cosh \relax (x) + a \sinh \relax (x)\right )}}{\cosh \relax (x)^{3} + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + \sinh \relax (x)^{3} + {\left (3 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x) + 3 \, \cosh \relax (x)}\right ) + 2 \, x \log \left (i \, \cosh \relax (x) + i \, \sinh \relax (x) + 1\right ) + 2 \, x \log \left (-i \, \cosh \relax (x) - i \, \sinh \relax (x) + 1\right ) + 2 \, {\rm Li}_2\left (i \, \cosh \relax (x) + i \, \sinh \relax (x)\right ) + 2 \, {\rm Li}_2\left (-i \, \cosh \relax (x) - i \, \sinh \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + x*log(4*(a*cosh(x) + a*sinh(x))/(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*cosh(x)^2 + 1)*sinh(x
) + 3*cosh(x))) + 2*x*log(I*cosh(x) + I*sinh(x) + 1) + 2*x*log(-I*cosh(x) - I*sinh(x) + 1) + 2*dilog(I*cosh(x)
+ I*sinh(x)) + 2*dilog(-I*cosh(x) - I*sinh(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \operatorname {sech}\relax (x)^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sech(x)^2), x)

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maple [C]  time = 1.41, size = 480, normalized size = 13.71 \[ -\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right )^{2}\right )}{2}-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 )^{2}\right )^{2}+\frac {i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+1\right )^{2}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )}{2}+i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-\frac {i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+1\right )^{2}}\right )^{3}}{2}-x^{2}+x \ln \relax (a )+2 x \ln \left ({\mathrm e}^{x}\right )+2 \ln \left (-i {\mathrm e}^{x}+1\right ) \ln \left ({\mathrm e}^{x}\right )+2 \ln \left (i {\mathrm e}^{x}+1\right ) \ln \left ({\mathrm e}^{x}\right )-2 \ln \left ({\mathrm e}^{2 x}+1\right ) \ln \left ({\mathrm e}^{x}\right )+2 \ln \relax (2) x +2 \dilog \left (-i {\mathrm e}^{x}+1\right )+2 \dilog \left (i {\mathrm e}^{x}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.15, size = 32, normalized size = 0.91 \[ -x^{2} + x \log \left (a \operatorname {sech}\relax (x)^{2}\right ) + 2 \, x \log \left (e^{\left (2 \, x\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ -\int 2\,\ln \left (\mathrm {cosh}\relax (x)\right )-\ln \relax (a) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a/cosh(x)^2),x)

[Out]

-int(2*log(cosh(x)) - log(a), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \operatorname {sech}^{2}{\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sech(x)**2), x)

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