3.842 \(\int \frac {x \text {csch}(x) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \, dx\)

Optimal. Leaf size=59 \[ -\frac {\text {Li}_2\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {\text {Li}_2\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.70, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6720, 4182, 2279, 2391} \[ -\frac {\text {sech}(x) \text {PolyLog}\left (2,-e^x\right )}{\sqrt {a \text {sech}^2(x)}}+\frac {\text {sech}(x) \text {PolyLog}\left (2,e^x\right )}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^2],x]

[Out]

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

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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.93 \[ \frac {\text {sech}(x) \left (\text {Li}_2\left (-e^{-x}\right )-\text {Li}_2\left (e^{-x}\right )+x \left (\log \left (1-e^{-x}\right )-\log \left (e^{-x}+1\right )\right )\right )}{\sqrt {a \text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^2],x]

[Out]

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

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fricas [A]  time = 0.42, size = 91, normalized size = 1.54 \[ \frac {{\left ({\left (e^{\left (2 \, x\right )} + 1\right )} {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) - {\left (e^{\left (2 \, x\right )} + 1\right )} {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) - {\left (x e^{\left (2 \, x\right )} + x\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (x e^{\left (2 \, x\right )} + x\right )} \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right )\right )} \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x*csch(x)*sech(x)/sqrt(a*sech(x)^2), x)

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maple [B]  time = 0.34, size = 136, normalized size = 2.31 \[ -\frac {{\mathrm e}^{x} x \ln \left ({\mathrm e}^{x}+1\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}-\frac {{\mathrm e}^{x} \polylog \left (2, -{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {{\mathrm e}^{x} x \ln \left (1-{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {{\mathrm e}^{x} \polylog \left (2, {\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.44, size = 36, normalized size = 0.61 \[ -\frac {x \log \left (e^{x} + 1\right ) + {\rm Li}_2\left (-e^{x}\right )}{\sqrt {a}} + \frac {x \log \left (-e^{x} + 1\right ) + {\rm Li}_2\left (e^{x}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)\,\sqrt {\frac {a}{{\mathrm {cosh}\relax (x)}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(cosh(x)*sinh(x)*(a/cosh(x)^2)^(1/2)),x)

[Out]

int(x/(cosh(x)*sinh(x)*(a/cosh(x)^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)**2)**(1/2),x)

[Out]

Integral(x*csch(x)*sech(x)/sqrt(a*sech(x)**2), x)

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