3.847 \(\int \frac{x^3 \text{csch}(x) \text{sech}(x)}{\sqrt{a \text{sech}^4(x)}} \, dx\)

Optimal. Leaf size=129 \[ \frac{3 x^2 \text{sech}^2(x) \text{PolyLog}\left (2,e^{2 x}\right )}{2 \sqrt{a \text{sech}^4(x)}}-\frac{3 x \text{sech}^2(x) \text{PolyLog}\left (3,e^{2 x}\right )}{2 \sqrt{a \text{sech}^4(x)}}+\frac{3 \text{sech}^2(x) \text{PolyLog}\left (4,e^{2 x}\right )}{4 \sqrt{a \text{sech}^4(x)}}-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}} \]

[Out]

-(x^4*Sech[x]^2)/(4*Sqrt[a*Sech[x]^4]) + (x^3*Log[1 - E^(2*x)]*Sech[x]^2)/Sqrt[a*Sech[x]^4] + (3*x^2*PolyLog[2
, E^(2*x)]*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4]) - (3*x*PolyLog[3, E^(2*x)]*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4]) + (3*P
olyLog[4, E^(2*x)]*Sech[x]^2)/(4*Sqrt[a*Sech[x]^4])

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Rubi [A]  time = 0.553772, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {6720, 3716, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 x^2 \text{sech}^2(x) \text{PolyLog}\left (2,e^{2 x}\right )}{2 \sqrt{a \text{sech}^4(x)}}-\frac{3 x \text{sech}^2(x) \text{PolyLog}\left (3,e^{2 x}\right )}{2 \sqrt{a \text{sech}^4(x)}}+\frac{3 \text{sech}^2(x) \text{PolyLog}\left (4,e^{2 x}\right )}{4 \sqrt{a \text{sech}^4(x)}}-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^4],x]

[Out]

-(x^4*Sech[x]^2)/(4*Sqrt[a*Sech[x]^4]) + (x^3*Log[1 - E^(2*x)]*Sech[x]^2)/Sqrt[a*Sech[x]^4] + (3*x^2*PolyLog[2
, E^(2*x)]*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4]) - (3*x*PolyLog[3, E^(2*x)]*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4]) + (3*P
olyLog[4, E^(2*x)]*Sech[x]^2)/(4*Sqrt[a*Sech[x]^4])

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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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 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}^4(x)}} \, dx &=\frac{\text{sech}^2(x) \int x^3 \coth (x) \, dx}{\sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}-\frac{\left (2 \text{sech}^2(x)\right ) \int \frac{e^{2 x} x^3}{1-e^{2 x}} \, dx}{\sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}}-\frac{\left (3 \text{sech}^2(x)\right ) \int x^2 \log \left (1-e^{2 x}\right ) \, dx}{\sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}}+\frac{3 x^2 \text{Li}_2\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}-\frac{\left (3 \text{sech}^2(x)\right ) \int x \text{Li}_2\left (e^{2 x}\right ) \, dx}{\sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}}+\frac{3 x^2 \text{Li}_2\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}-\frac{3 x \text{Li}_3\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}+\frac{\left (3 \text{sech}^2(x)\right ) \int \text{Li}_3\left (e^{2 x}\right ) \, dx}{2 \sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}}+\frac{3 x^2 \text{Li}_2\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}-\frac{3 x \text{Li}_3\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}+\frac{\left (3 \text{sech}^2(x)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 x}\right )}{4 \sqrt{a \text{sech}^4(x)}}\\ &=-\frac{x^4 \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}+\frac{x^3 \log \left (1-e^{2 x}\right ) \text{sech}^2(x)}{\sqrt{a \text{sech}^4(x)}}+\frac{3 x^2 \text{Li}_2\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}-\frac{3 x \text{Li}_3\left (e^{2 x}\right ) \text{sech}^2(x)}{2 \sqrt{a \text{sech}^4(x)}}+\frac{3 \text{Li}_4\left (e^{2 x}\right ) \text{sech}^2(x)}{4 \sqrt{a \text{sech}^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0542898, size = 68, normalized size = 0.53 \[ -\frac{\text{sech}^2(x) \left (-6 x^2 \text{PolyLog}\left (2,e^{2 x}\right )+6 x \text{PolyLog}\left (3,e^{2 x}\right )-3 \text{PolyLog}\left (4,e^{2 x}\right )+x^4-4 x^3 \log \left (1-e^{2 x}\right )\right )}{4 \sqrt{a \text{sech}^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^4],x]

[Out]

-((x^4 - 4*x^3*Log[1 - E^(2*x)] - 6*x^2*PolyLog[2, E^(2*x)] + 6*x*PolyLog[3, E^(2*x)] - 3*PolyLog[4, E^(2*x)])
*Sech[x]^2)/(4*Sqrt[a*Sech[x]^4])

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Maple [B]  time = 0.072, size = 329, normalized size = 2.6 \begin{align*} -{\frac{{{\rm e}^{2\,x}}{x}^{4}}{4\, \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+{\frac{{{\rm e}^{2\,x}}{x}^{3}\ln \left ({{\rm e}^{x}}+1 \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+3\,{\frac{{{\rm e}^{2\,x}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}-6\,{\frac{x{{\rm e}^{2\,x}}{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+6\,{\frac{{{\rm e}^{2\,x}}{\it polylog} \left ( 4,-{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+{\frac{{{\rm e}^{2\,x}}{x}^{3}\ln \left ( 1-{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+3\,{\frac{{{\rm e}^{2\,x}}{x}^{2}{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}-6\,{\frac{x{{\rm e}^{2\,x}}{\it polylog} \left ( 3,{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}}+6\,{\frac{{{\rm e}^{2\,x}}{\it polylog} \left ( 4,{{\rm e}^{x}} \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{\frac{1}{\sqrt{{\frac{{{\rm e}^{4\,x}}a}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{4}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x)

[Out]

-1/4/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*x^4+1/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*
x)+1)^2*exp(2*x)*x^3*ln(exp(x)+1)+3/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*x^2*polylog(2,-e
xp(x))-6/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*x*polylog(3,-exp(x))+6/(a*exp(4*x)/(exp(2*x
)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*polylog(4,-exp(x))+1/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*ex
p(2*x)*x^3*ln(1-exp(x))+3/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*x^2*polylog(2,exp(x))-6/(a
*exp(4*x)/(exp(2*x)+1)^4)^(1/2)/(exp(2*x)+1)^2*exp(2*x)*x*polylog(3,exp(x))+6/(a*exp(4*x)/(exp(2*x)+1)^4)^(1/2
)/(exp(2*x)+1)^2*exp(2*x)*polylog(4,exp(x))

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Maxima [A]  time = 1.91817, size = 117, normalized size = 0.91 \begin{align*} -\frac{x^{4}}{4 \, \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}} + \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)^4)^(1/2),x, algorithm="maxima")

[Out]

-1/4*x^4/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)
+ (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 = 2.17801, size = 1223, normalized size = 9.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x, algorithm="fricas")

[Out]

1/4*(24*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))*(e^(4*x) + 2*e^(2*x) + 1)*e^(2*x)*polylog(4,
cosh(x) + sinh(x)) + 24*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))*(e^(4*x) + 2*e^(2*x) + 1)*e
^(2*x)*polylog(4, -cosh(x) - sinh(x)) - 24*(x*e^(4*x) + 2*x*e^(2*x) + x)*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*
x) + 4*e^(2*x) + 1))*e^(2*x)*polylog(3, cosh(x) + sinh(x)) - 24*(x*e^(4*x) + 2*x*e^(2*x) + x)*sqrt(a/(e^(8*x)
+ 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1))*e^(2*x)*polylog(3, -cosh(x) - sinh(x)) - (x^4*e^(4*x) + 2*x^4*e^(2*x
) + x^4 - 12*(x^2*e^(4*x) + 2*x^2*e^(2*x) + x^2)*dilog(cosh(x) + sinh(x)) - 12*(x^2*e^(4*x) + 2*x^2*e^(2*x) +
x^2)*dilog(-cosh(x) - sinh(x)) - 4*(x^3*e^(4*x) + 2*x^3*e^(2*x) + x^3)*log(cosh(x) + sinh(x) + 1) - 4*(x^3*e^(
4*x) + 2*x^3*e^(2*x) + x^3)*log(-cosh(x) - sinh(x) + 1))*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) +
1))*e^(2*x))*e^(-2*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}^{4}{\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)**4)**(1/2),x)

[Out]

Integral(x**3*csch(x)*sech(x)/sqrt(a*sech(x)**4), 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 )^{4}}}\,{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)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3*csch(x)*sech(x)/sqrt(a*sech(x)^4), x)