3.3.15 \(\int \log (a \text {sech}(x)) \, dx\) [215]

Optimal. Leaf size=38 \[ -\frac {x^2}{2}+x \log \left (1+e^{2 x}\right )+x \log (a \text {sech}(x))+\frac {1}{2} \text {Li}_2\left (-e^{2 x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 38, 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 = {2628, 3799, 2221, 2317, 2438} \begin {gather*} \frac {1}{2} \text {PolyLog}\left (2,-e^{2 x}\right )+x \log (a \text {sech}(x))-\frac {x^2}{2}+x \log \left (e^{2 x}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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]
- 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 2317

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 2438

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 2628

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

Rule 3799

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

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Mathematica [A]
time = 0.01, size = 37, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (x \left (x+2 \log \left (1+e^{-2 x}\right )+2 \log (a \text {sech}(x))\right )-\text {Li}_2\left (-e^{-2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 314, normalized size = 8.26

method	result	size
risch	\(x \ln \left ({\mathrm e}^{x}\right )+\frac {i x \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{2}}{2}-\frac {i x \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{3}}{2}-\frac {i x \pi \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{3}}{2}-\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{2}}{2}-\frac {i x \pi \,\mathrm {csgn}\left (i a \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i a \,{\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )}{2}+x \ln \left (2\right )+x \ln \left (a \right )-\frac {x^{2}}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{2}}{2}+\frac {i x \pi \,\mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{1+{\mathrm e}^{2 x}}\right )^{2}}{2}-\ln \left ({\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{2 x}\right )+\ln \left ({\mathrm e}^{x}\right ) \ln \left (1+i {\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}\right ) \ln \left (1-i {\mathrm e}^{x}\right )+\dilog \left (1+i {\mathrm e}^{x}\right )+\dilog \left (1-i {\mathrm e}^{x}\right )\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*sech(x)),x,method=_RETURNVERBOSE)

[Out]

x*ln(exp(x))+1/2*I*x*Pi*csgn(I*exp(x)/(1+exp(2*x)))*csgn(I*a/(1+exp(2*x))*exp(x))^2-1/2*I*x*Pi*csgn(I*exp(x)/(
1+exp(2*x)))^3-1/2*I*x*Pi*csgn(I*a/(1+exp(2*x))*exp(x))^3-1/2*I*x*Pi*csgn(I*exp(x))*csgn(I/(1+exp(2*x)))*csgn(
I*exp(x)/(1+exp(2*x)))+1/2*I*x*Pi*csgn(I*a)*csgn(I*a/(1+exp(2*x))*exp(x))^2-1/2*I*x*Pi*csgn(I*a)*csgn(I*exp(x)
/(1+exp(2*x)))*csgn(I*a/(1+exp(2*x))*exp(x))+x*ln(2)+x*ln(a)-1/2*x^2+1/2*I*x*Pi*csgn(I*exp(x))*csgn(I*exp(x)/(
1+exp(2*x)))^2+1/2*I*x*Pi*csgn(I/(1+exp(2*x)))*csgn(I*exp(x)/(1+exp(2*x)))^2-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 = 0.54, size = 31, normalized size = 0.82 \begin {gather*} -\frac {1}{2} \, x^{2} + x \log \left (a \operatorname {sech}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.40, size = 84, normalized size = 2.21 \begin {gather*} -\frac {1}{2} \, x^{2} + x \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \log {\left (a \operatorname {sech}{\left (x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sech(x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sech(x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \ln \left (\mathrm {cosh}\left (x\right )\right )-\ln \left (a\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-int(log(cosh(x)) - log(a), x)

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