3.468 \(\int x^2 \text{csch}(a+b x) \text{sech}(a+b x) \, dx\)

Optimal. Leaf size=97 \[ -\frac{x \text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{b^2}+\frac{x \text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac{\text{PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac{2 x^2 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

[Out]

(-2*x^2*ArcTanh[E^(2*a + 2*b*x)])/b - (x*PolyLog[2, -E^(2*a + 2*b*x)])/b^2 + (x*PolyLog[2, E^(2*a + 2*b*x)])/b
^2 + PolyLog[3, -E^(2*a + 2*b*x)]/(2*b^3) - PolyLog[3, E^(2*a + 2*b*x)]/(2*b^3)

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Rubi [A]  time = 0.105626, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5461, 4182, 2531, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{b^2}+\frac{x \text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac{\text{PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac{2 x^2 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Csch[a + b*x]*Sech[a + b*x],x]

[Out]

(-2*x^2*ArcTanh[E^(2*a + 2*b*x)])/b - (x*PolyLog[2, -E^(2*a + 2*b*x)])/b^2 + (x*PolyLog[2, E^(2*a + 2*b*x)])/b
^2 + PolyLog[3, -E^(2*a + 2*b*x)]/(2*b^3) - PolyLog[3, E^(2*a + 2*b*x)]/(2*b^3)

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

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

Mathematica [A]  time = 4.11854, size = 108, normalized size = 1.11 \[ \frac{-2 b x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )+2 b x \text{PolyLog}\left (2,e^{2 (a+b x)}\right )+\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )-\text{PolyLog}\left (3,e^{2 (a+b x)}\right )+2 b^2 x^2 \log \left (1-e^{2 (a+b x)}\right )-2 b^2 x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Csch[a + b*x]*Sech[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.043, size = 186, normalized size = 1.9 \begin{align*} -2\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{b}}+2\,{\frac{x{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{b}}+2\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}-2\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}}+{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}}-{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*csch(b*x+a)*sech(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.1186, size = 200, normalized size = 2.06 \begin{align*} -\frac{2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} + \frac{b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} + \frac{b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*csch(b*x+a)*sech(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.52657, size = 1018, normalized size = 10.49 \begin{align*} \frac{b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, b x{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, b x{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 2 \, b x{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 2 \, b x{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) -{\left (b^{2} x^{2} - a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) -{\left (b^{2} x^{2} - a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) +{\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \,{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 2 \,{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 2 \,{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*csch(b*x+a)*sech(b*x+a),x, algorithm="fricas")

[Out]

(b^2*x^2*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 2*b*x*dilog(cosh(b*x + a) + sinh(b*x + a)) - 2*b*x*dilog(I*c
osh(b*x + a) + I*sinh(b*x + a)) - 2*b*x*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 2*b*x*dilog(-cosh(b*x + a)
- sinh(b*x + a)) - a^2*log(cosh(b*x + a) + sinh(b*x + a) + I) - a^2*log(cosh(b*x + a) + sinh(b*x + a) - I) +
a^2*log(cosh(b*x + a) + sinh(b*x + a) - 1) - (b^2*x^2 - a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - (b^2
*x^2 - a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (b^2*x^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) +
1) - 2*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 2*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 2*polylo
g(3, -I*cosh(b*x + a) - I*sinh(b*x + a)) - 2*polylog(3, -cosh(b*x + a) - sinh(b*x + a)))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*csch(b*x+a)*sech(b*x+a),x)

[Out]

Integral(x**2*csch(a + b*x)*sech(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*csch(b*x+a)*sech(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*csch(b*x + a)*sech(b*x + a), x)