### 3.844 $$\int \frac{x^3 \text{csch}(x) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}} \, dx$$

Optimal. Leaf size=150 $-\frac{3 x^2 \text{sech}(x) \text{PolyLog}\left (2,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{sech}(x) \text{PolyLog}\left (2,e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{6 x \text{sech}(x) \text{PolyLog}\left (3,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{6 x \text{sech}(x) \text{PolyLog}\left (3,e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{6 \text{sech}(x) \text{PolyLog}\left (4,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{6 \text{sech}(x) \text{PolyLog}\left (4,e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}$

[Out]

(-2*x^3*ArcTanh[E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (3*x^2*PolyLog[2, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (3*x^2*
PolyLog[2, E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (6*x*PolyLog[3, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (6*x*PolyLog[3
, E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (6*PolyLog[4, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (6*PolyLog[4, E^x]*Sech[x
])/Sqrt[a*Sech[x]^2]

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Rubi [A]  time = 0.833424, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6720, 4182, 2531, 6609, 2282, 6589} $-\frac{3 x^2 \text{sech}(x) \text{PolyLog}\left (2,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{sech}(x) \text{PolyLog}\left (2,e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{6 x \text{sech}(x) \text{PolyLog}\left (3,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{6 x \text{sech}(x) \text{PolyLog}\left (3,e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{6 \text{sech}(x) \text{PolyLog}\left (4,-e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{6 \text{sech}(x) \text{PolyLog}\left (4,e^x\right )}{\sqrt{a \text{sech}^2(x)}}-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x^3*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^2],x]

[Out]

(-2*x^3*ArcTanh[E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (3*x^2*PolyLog[2, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (3*x^2*
PolyLog[2, E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (6*x*PolyLog[3, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (6*x*PolyLog[3
, E^x]*Sech[x])/Sqrt[a*Sech[x]^2] - (6*PolyLog[4, -E^x]*Sech[x])/Sqrt[a*Sech[x]^2] + (6*PolyLog[4, E^x]*Sech[x
])/Sqrt[a*Sech[x]^2]

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])

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 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 6609

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

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 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]

Rubi steps

\begin{align*} \int \frac{x^3 \text{csch}(x) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}} \, dx &=\frac{\text{sech}(x) \int x^3 \text{csch}(x) \, dx}{\sqrt{a \text{sech}^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{(3 \text{sech}(x)) \int x^2 \log \left (1-e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}+\frac{(3 \text{sech}(x)) \int x^2 \log \left (1+e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{3 x^2 \text{Li}_2\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{Li}_2\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{(6 \text{sech}(x)) \int x \text{Li}_2\left (-e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}-\frac{(6 \text{sech}(x)) \int x \text{Li}_2\left (e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{3 x^2 \text{Li}_2\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{Li}_2\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{6 x \text{Li}_3\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{6 x \text{Li}_3\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{(6 \text{sech}(x)) \int \text{Li}_3\left (-e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}+\frac{(6 \text{sech}(x)) \int \text{Li}_3\left (e^x\right ) \, dx}{\sqrt{a \text{sech}^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{3 x^2 \text{Li}_2\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{Li}_2\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{6 x \text{Li}_3\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{6 x \text{Li}_3\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{(6 \text{sech}(x)) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^x\right )}{\sqrt{a \text{sech}^2(x)}}+\frac{(6 \text{sech}(x)) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^x\right )}{\sqrt{a \text{sech}^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{3 x^2 \text{Li}_2\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{3 x^2 \text{Li}_2\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{6 x \text{Li}_3\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{6 x \text{Li}_3\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}-\frac{6 \text{Li}_4\left (-e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}+\frac{6 \text{Li}_4\left (e^x\right ) \text{sech}(x)}{\sqrt{a \text{sech}^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0803852, size = 113, normalized size = 0.75 $\frac{\text{sech}(x) \left (24 x^2 \text{PolyLog}\left (2,-e^{-x}\right )+24 x^2 \text{PolyLog}\left (2,e^x\right )+48 x \text{PolyLog}\left (3,-e^{-x}\right )-48 x \text{PolyLog}\left (3,e^x\right )+48 \text{PolyLog}\left (4,-e^{-x}\right )+48 \text{PolyLog}\left (4,e^x\right )-2 x^4-8 x^3 \log \left (e^{-x}+1\right )+8 x^3 \log \left (1-e^x\right )+\pi ^4\right )}{8 \sqrt{a \text{sech}^2(x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x^3*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^2],x]

[Out]

((Pi^4 - 2*x^4 - 8*x^3*Log[1 + E^(-x)] + 8*x^3*Log[1 - E^x] + 24*x^2*PolyLog[2, -E^(-x)] + 24*x^2*PolyLog[2, E
^x] + 48*x*PolyLog[3, -E^(-x)] - 48*x*PolyLog[3, E^x] + 48*PolyLog[4, -E^(-x)] + 48*PolyLog[4, E^x])*Sech[x])/
(8*Sqrt[a*Sech[x]^2])

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Maple [B]  time = 0.072, size = 281, normalized size = 1.9 \begin{align*} -{\frac{{x}^{3}{{\rm e}^{x}}\ln \left ({{\rm e}^{x}}+1 \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-3\,{\frac{{x}^{2}{{\rm e}^{x}}{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+6\,{\frac{x{{\rm e}^{x}}{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-6\,{\frac{{{\rm e}^{x}}{\it polylog} \left ( 4,-{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{{x}^{3}{{\rm e}^{x}}\ln \left ( 1-{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+3\,{\frac{{x}^{2}{{\rm e}^{x}}{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-6\,{\frac{x{{\rm e}^{x}}{\it polylog} \left ( 3,{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+6\,{\frac{{{\rm e}^{x}}{\it polylog} \left ( 4,{{\rm e}^{x}} \right ) }{{{\rm e}^{2\,x}}+1}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x)

[Out]

-1/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*x^3*ln(exp(x)+1)-3/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/
(exp(2*x)+1)*exp(x)*x^2*polylog(2,-exp(x))+6/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*x*polylog(3
,-exp(x))-6/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*polylog(4,-exp(x))+1/(a*exp(2*x)/(exp(2*x)+1
)^2)^(1/2)/(exp(2*x)+1)*exp(x)*x^3*ln(1-exp(x))+3/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*x^2*po
lylog(2,exp(x))-6/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*x*polylog(3,exp(x))+6/(a*exp(2*x)/(exp
(2*x)+1)^2)^(1/2)/(exp(2*x)+1)*exp(x)*polylog(4,exp(x))

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Maxima [A]  time = 1.81711, size = 108, normalized size = 0.72 \begin{align*} -\frac{x^{3} \log \left (e^{x} + 1\right ) + 3 \, x^{2}{\rm Li}_2\left (-e^{x}\right ) - 6 \, x{\rm Li}_{3}(-e^{x}) + 6 \,{\rm Li}_{4}(-e^{x})}{\sqrt{a}} + \frac{x^{3} \log \left (-e^{x} + 1\right ) + 3 \, x^{2}{\rm Li}_2\left (e^{x}\right ) - 6 \, x{\rm Li}_{3}(e^{x}) + 6 \,{\rm Li}_{4}(e^{x})}{\sqrt{a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-(x^3*log(e^x + 1) + 3*x^2*dilog(-e^x) - 6*x*polylog(3, -e^x) + 6*polylog(4, -e^x))/sqrt(a) + (x^3*log(-e^x +
1) + 3*x^2*dilog(e^x) - 6*x*polylog(3, e^x) + 6*polylog(4, e^x))/sqrt(a)

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Fricas [C]  time = 2.10084, size = 807, normalized size = 5.38 \begin{align*} \frac{{\left (6 \, \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 (4, \cosh \left (x\right ) + \sinh \left (x\right )\right ) - 6 \, \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 (4, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) - 6 \,{\left (x e^{\left (2 \, x\right )} + x\right )} \sqrt{\frac{a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} e^{x}{\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + 6 \,{\left (x e^{\left (2 \, x\right )} + x\right )} \sqrt{\frac{a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} e^{x}{\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) +{\left (3 \,{\left (x^{2} e^{\left (2 \, x\right )} + x^{2}\right )}{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 3 \,{\left (x^{2} e^{\left (2 \, x\right )} + x^{2}\right )}{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) -{\left (x^{3} e^{\left (2 \, x\right )} + x^{3}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\left (x^{3} e^{\left (2 \, x\right )} + x^{3}\right )} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="fricas")

[Out]

(6*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*(e^(2*x) + 1)*e^x*polylog(4, cosh(x) + sinh(x)) - 6*sqrt(a/(e^(4*x) + 2*e
^(2*x) + 1))*(e^(2*x) + 1)*e^x*polylog(4, -cosh(x) - sinh(x)) - 6*(x*e^(2*x) + x)*sqrt(a/(e^(4*x) + 2*e^(2*x)
+ 1))*e^x*polylog(3, cosh(x) + sinh(x)) + 6*(x*e^(2*x) + x)*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*e^x*polylog(3, -
cosh(x) - sinh(x)) + (3*(x^2*e^(2*x) + x^2)*dilog(cosh(x) + sinh(x)) - 3*(x^2*e^(2*x) + x^2)*dilog(-cosh(x) -
sinh(x)) - (x^3*e^(2*x) + x^3)*log(cosh(x) + sinh(x) + 1) + (x^3*e^(2*x) + x^3)*log(-cosh(x) - sinh(x) + 1))*s
qrt(a/(e^(4*x) + 2*e^(2*x) + 1))*e^x)*e^(-x)/a

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{csch}{\left (x \right )} \operatorname{sech}{\left (x \right )}}{\sqrt{a \operatorname{sech}^{2}{\left (x \right )}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*csch(x)*sech(x)/(a*sech(x)**2)**(1/2),x)

[Out]

Integral(x**3*csch(x)*sech(x)/sqrt(a*sech(x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{csch}\left (x\right ) \operatorname{sech}\left (x\right )}{\sqrt{a \operatorname{sech}\left (x\right )^{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(x)*sech(x)/(a*sech(x)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3*csch(x)*sech(x)/sqrt(a*sech(x)^2), x)