3.496 $$\int x^2 \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx$$

Optimal. Leaf size=64 $\frac{\text{PolyLog}\left (2,e^{4 (a+b x)}\right )}{2 b^3}+\frac{2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{2 x^2 \coth (2 a+2 b x)}{b}-\frac{2 x^2}{b}$

[Out]

(-2*x^2)/b - (2*x^2*Coth[2*a + 2*b*x])/b + (2*x*Log[1 - E^(4*(a + b*x))])/b^2 + PolyLog[2, E^(4*(a + b*x))]/(2
*b^3)

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Rubi [A]  time = 0.167392, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {5461, 4184, 3716, 2190, 2279, 2391} $\frac{\text{PolyLog}\left (2,e^{4 (a+b x)}\right )}{2 b^3}+\frac{2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{2 x^2 \coth (2 a+2 b x)}{b}-\frac{2 x^2}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*x^2)/b - (2*x^2*Coth[2*a + 2*b*x])/b + (2*x*Log[1 - E^(4*(a + b*x))])/b^2 + PolyLog[2, E^(4*(a + 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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 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]

Rubi steps

\begin{align*} \int x^2 \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx &=4 \int x^2 \text{csch}^2(2 a+2 b x) \, dx\\ &=-\frac{2 x^2 \coth (2 a+2 b x)}{b}+\frac{4 \int x \coth (2 a+2 b x) \, dx}{b}\\ &=-\frac{2 x^2}{b}-\frac{2 x^2 \coth (2 a+2 b x)}{b}-\frac{8 \int \frac{e^{2 (2 a+2 b x)} x}{1-e^{2 (2 a+2 b x)}} \, dx}{b}\\ &=-\frac{2 x^2}{b}-\frac{2 x^2 \coth (2 a+2 b x)}{b}+\frac{2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{2 \int \log \left (1-e^{2 (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 x^2}{b}-\frac{2 x^2 \coth (2 a+2 b x)}{b}+\frac{2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac{2 x^2}{b}-\frac{2 x^2 \coth (2 a+2 b x)}{b}+\frac{2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac{\text{Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}\\ \end{align*}

Mathematica [B]  time = 4.06738, size = 216, normalized size = 3.38 $4 \left (\frac{x^2 \text{csch}(2 a) \sinh (2 b x) \text{csch}(2 a+2 b x)}{2 b}-\frac{e^{4 a} \left (2 \left (1-e^{-4 a}\right ) \text{PolyLog}\left (2,-e^{-a-b x}\right )+2 \left (1-e^{-4 a}\right ) \text{PolyLog}\left (2,e^{-a-b x}\right )+\left (1-e^{-4 a}\right ) \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+4 e^{-4 a} b^2 x^2-2 \left (1-e^{-4 a}\right ) b x \log \left (1-e^{-a-b x}\right )-2 \left (1-e^{-4 a}\right ) b x \log \left (e^{-a-b x}+1\right )-2 \left (1-e^{-4 a}\right ) b x \log \left (e^{-2 (a+b x)}+1\right )\right )}{4 \left (e^{4 a}-1\right ) b^3}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

4*(-(E^(4*a)*((4*b^2*x^2)/E^(4*a) - 2*b*(1 - E^(-4*a))*x*Log[1 - E^(-a - b*x)] - 2*b*(1 - E^(-4*a))*x*Log[1 +
E^(-a - b*x)] - 2*b*(1 - E^(-4*a))*x*Log[1 + E^(-2*(a + b*x))] + 2*(1 - E^(-4*a))*PolyLog[2, -E^(-a - b*x)] +
2*(1 - E^(-4*a))*PolyLog[2, E^(-a - b*x)] + (1 - E^(-4*a))*PolyLog[2, -E^(-2*(a + b*x))]))/(4*b^3*(-1 + E^(4*a
))) + (x^2*Csch[2*a]*Csch[2*a + 2*b*x]*Sinh[2*b*x])/(2*b))

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Maple [B]  time = 0.041, size = 199, normalized size = 3.1 \begin{align*} -4\,{\frac{{x}^{2}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-4\,{\frac{{x}^{2}}{b}}-8\,{\frac{ax}{{b}^{2}}}-4\,{\frac{{a}^{2}}{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}}+8\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^2/b/(exp(2*b*x+2*a)-1)/(1+exp(2*b*x+2*a))-4*x^2/b-8/b^2*a*x-4/b^3*a^2+2/b^2*ln(1-exp(b*x+a))*x+2/b^3*ln(1
-exp(b*x+a))*a+2*polylog(2,exp(b*x+a))/b^3+2*x*ln(1+exp(2*b*x+2*a))/b^2+polylog(2,-exp(2*b*x+2*a))/b^3+2/b^2*l
n(1+exp(b*x+a))*x+2*polylog(2,-exp(b*x+a))/b^3-2/b^3*a*ln(exp(b*x+a)-1)+8/b^3*a*ln(exp(b*x+a))

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Maxima [A]  time = 1.22338, size = 159, normalized size = 2.48 \begin{align*} -\frac{4 \, x^{2}}{b e^{\left (4 \, b x + 4 \, a\right )} - b} - \frac{4 \, x^{2}}{b} + \frac{2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{3}} + \frac{2 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac{2 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^2/(b*e^(4*b*x + 4*a) - b) - 4*x^2/b + (2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*a)))/b^3 + 2*
(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^3 + 2*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^3

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Fricas [C]  time = 2.82204, size = 3605, normalized size = 56.33 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*(b^2*x^2 - a^2)*cosh(b*x + a)^4 + 8*(b^2*x^2 - a^2)*cosh(b*x + a)^3*sinh(b*x + a) + 12*(b^2*x^2 - a^2)*c
osh(b*x + a)^2*sinh(b*x + a)^2 + 8*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + 2*(b^2*x^2 - a^2)*sinh(b*x
+ a)^4 + 2*a^2 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*co
sh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - (cosh(b*x + a)^4 + 4
*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*
x + a)^4 - 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) +
6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(-I*cosh(b*x +
a) - I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^
2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b*x*cosh(b
*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b*x*cosh(b*x + a)*
sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 + 4*a
*cosh(b*x + a)^3*sinh(b*x + a) + 6*a*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*s
inh(b*x + a)^4 - a)*log(cosh(b*x + a) + sinh(b*x + a) + I) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)^3*sinh(b*x
+ a) + 6*a*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - a)*log(c
osh(b*x + a) + sinh(b*x + a) - I) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)^3*sinh(b*x + a) + 6*a*cosh(b*x + a)
^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - a)*log(cosh(b*x + a) + sinh(b*x +
a) - 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)^3*sinh(b*x + a) + 6*(b*x + a)*cosh(b*x + a)^
2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x + a)^4 - b*x - a)*log(I*cos
h(b*x + a) + I*sinh(b*x + a) + 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)^3*sinh(b*x + a) + 6
*(b*x + a)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x +
a)^4 - b*x - a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*
x + a)^3*sinh(b*x + a) + 6*(b*x + a)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)
^3 + (b*x + a)*sinh(b*x + a)^4 - b*x - a)*log(-cosh(b*x + a) - sinh(b*x + a) + 1))/(b^3*cosh(b*x + a)^4 + 4*b^
3*cosh(b*x + a)^3*sinh(b*x + a) + 6*b^3*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3
+ b^3*sinh(b*x + a)^4 - b^3)

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

[Out]

Integral(x**2*csch(a + b*x)**2*sech(a + b*x)**2, 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 )^{2} \operatorname{sech}\left (b x + a\right )^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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