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

Optimal. Leaf size=104 \[ -\frac {2 x \text {Li}_2\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 x \text {Li}_2\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 \text {Li}_3\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 \text {Li}_3\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \]

[Out]

-2*x^2*arctanh(exp(x))*sech(x)/(a*sech(x)^2)^(1/2)-2*x*polylog(2,-exp(x))*sech(x)/(a*sech(x)^2)^(1/2)+2*x*poly
log(2,exp(x))*sech(x)/(a*sech(x)^2)^(1/2)+2*polylog(3,-exp(x))*sech(x)/(a*sech(x)^2)^(1/2)-2*polylog(3,exp(x))
*sech(x)/(a*sech(x)^2)^(1/2)

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Rubi [A]  time = 0.81, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6720, 4182, 2531, 2282, 6589} \[ -\frac {2 x \text {sech}(x) \text {PolyLog}\left (2,-e^x\right )}{\sqrt {a \text {sech}^2(x)}}+\frac {2 x \text {sech}(x) \text {PolyLog}\left (2,e^x\right )}{\sqrt {a \text {sech}^2(x)}}+\frac {2 \text {sech}(x) \text {PolyLog}\left (3,-e^x\right )}{\sqrt {a \text {sech}^2(x)}}-\frac {2 \text {sech}(x) \text {PolyLog}\left (3,e^x\right )}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*ArcTanh[E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (2*x*PolyLog[2, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (2*x*Poly
Log[2, E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (2*PolyLog[3, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (2*PolyLog[3, E^x]*S
ech[x])/Sqrt[a*Sech[x]^2]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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^2 \text {csch}(x) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}} \, dx &=\frac {\text {sech}(x) \int x^2 \text {csch}(x) \, dx}{\sqrt {a \text {sech}^2(x)}}\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {(2 \text {sech}(x)) \int x \log \left (1-e^x\right ) \, dx}{\sqrt {a \text {sech}^2(x)}}+\frac {(2 \text {sech}(x)) \int x \log \left (1+e^x\right ) \, dx}{\sqrt {a \text {sech}^2(x)}}\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x \text {Li}_2\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 x \text {Li}_2\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {(2 \text {sech}(x)) \int \text {Li}_2\left (-e^x\right ) \, dx}{\sqrt {a \text {sech}^2(x)}}-\frac {(2 \text {sech}(x)) \int \text {Li}_2\left (e^x\right ) \, dx}{\sqrt {a \text {sech}^2(x)}}\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x \text {Li}_2\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 x \text {Li}_2\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {(2 \text {sech}(x)) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^x\right )}{\sqrt {a \text {sech}^2(x)}}-\frac {(2 \text {sech}(x)) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )}{\sqrt {a \text {sech}^2(x)}}\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 x \text {Li}_2\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 x \text {Li}_2\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}+\frac {2 \text {Li}_3\left (-e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}-\frac {2 \text {Li}_3\left (e^x\right ) \text {sech}(x)}{\sqrt {a \text {sech}^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 83, normalized size = 0.80 \[ \frac {\text {sech}(x) \left (2 x \text {Li}_2\left (-e^{-x}\right )-2 x \text {Li}_2\left (e^{-x}\right )+2 \text {Li}_3\left (-e^{-x}\right )-2 \text {Li}_3\left (e^{-x}\right )+x^2 \log \left (1-e^{-x}\right )-x^2 \log \left (e^{-x}+1\right )\right )}{\sqrt {a \text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*Log[1 - E^(-x)] - x^2*Log[1 + E^(-x)] + 2*x*PolyLog[2, -E^(-x)] - 2*x*PolyLog[2, E^(-x)] + 2*PolyLog[3,
-E^(-x)] - 2*PolyLog[3, E^(-x)])*Sech[x])/Sqrt[a*Sech[x]^2]

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fricas [C]  time = 0.43, size = 188, normalized size = 1.81 \[ -\frac {{\left (2 \, \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} {\left (e^{\left (2 \, x\right )} + 1\right )} e^{x} {\rm polylog}\left (3, \cosh \relax (x) + \sinh \relax (x)\right ) - 2 \, \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} {\left (e^{\left (2 \, x\right )} + 1\right )} e^{x} {\rm polylog}\left (3, -\cosh \relax (x) - \sinh \relax (x)\right ) - {\left (2 \, {\left (x e^{\left (2 \, x\right )} + x\right )} {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) - 2 \, {\left (x e^{\left (2 \, x\right )} + x\right )} {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) - {\left (x^{2} e^{\left (2 \, x\right )} + x^{2}\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (x^{2} e^{\left (2 \, x\right )} + x^{2}\right )} \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right )\right )} \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} e^{x}\right )} e^{\left (-x\right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*(e^(2*x) + 1)*e^x*polylog(3, cosh(x) + sinh(x)) - 2*sqrt(a/(e^(4*x) + 2*
e^(2*x) + 1))*(e^(2*x) + 1)*e^x*polylog(3, -cosh(x) - sinh(x)) - (2*(x*e^(2*x) + x)*dilog(cosh(x) + sinh(x)) -
2*(x*e^(2*x) + x)*dilog(-cosh(x) - sinh(x)) - (x^2*e^(2*x) + x^2)*log(cosh(x) + sinh(x) + 1) + (x^2*e^(2*x) +
x^2)*log(-cosh(x) - sinh(x) + 1))*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*e^x)*e^(-x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.35, size = 209, normalized size = 2.01 \[ -\frac {{\mathrm e}^{x} x^{2} \ln \left ({\mathrm e}^{x}+1\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}-\frac {2 \,{\mathrm e}^{x} x \polylog \left (2, -{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {2 \,{\mathrm e}^{x} \polylog \left (3, -{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {{\mathrm e}^{x} x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {2 \,{\mathrm e}^{x} x \polylog \left (2, {\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}-\frac {2 \,{\mathrm e}^{x} \polylog \left (3, {\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*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^2*ln(exp(x)+1)-2/(a*exp(2*x)/(1+exp(2*x))^2)^(1/2)/
(1+exp(2*x))*exp(x)*x*polylog(2,-exp(x))+2/(a*exp(2*x)/(1+exp(2*x))^2)^(1/2)/(1+exp(2*x))*exp(x)*polylog(3,-ex
p(x))+1/(a*exp(2*x)/(1+exp(2*x))^2)^(1/2)/(1+exp(2*x))*exp(x)*x^2*ln(1-exp(x))+2/(a*exp(2*x)/(1+exp(2*x))^2)^(
1/2)/(1+exp(2*x))*exp(x)*x*polylog(2,exp(x))-2/(a*exp(2*x)/(1+exp(2*x))^2)^(1/2)/(1+exp(2*x))*exp(x)*polylog(3
,exp(x))

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maxima [A]  time = 0.44, size = 60, normalized size = 0.58 \[ -\frac {x^{2} \log \left (e^{x} + 1\right ) + 2 \, x {\rm Li}_2\left (-e^{x}\right ) - 2 \, {\rm Li}_{3}(-e^{x})}{\sqrt {a}} + \frac {x^{2} \log \left (-e^{x} + 1\right ) + 2 \, x {\rm Li}_2\left (e^{x}\right ) - 2 \, {\rm Li}_{3}(e^{x})}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2*log(e^x + 1) + 2*x*dilog(-e^x) - 2*polylog(3, -e^x))/sqrt(a) + (x^2*log(-e^x + 1) + 2*x*dilog(e^x) - 2*p
olylog(3, e^x))/sqrt(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)\,\sqrt {\frac {a}{{\mathrm {cosh}\relax (x)}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/(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 \[ \int \frac {x^{2} \operatorname {csch}{\relax (x )} \operatorname {sech}{\relax (x )}}{\sqrt {a \operatorname {sech}^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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