3.433 \(\int x^2 \coth ^2(a+b x) \, dx\)

Optimal. Leaf size=65 \[ \frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}+\frac{2 x \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{x^2 \coth (a+b x)}{b}-\frac{x^2}{b}+\frac{x^3}{3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.124658, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3720, 3716, 2190, 2279, 2391, 30} \[ \frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}+\frac{2 x \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{x^2 \coth (a+b x)}{b}-\frac{x^2}{b}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + b*x]^2,x]

[Out]

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

Rule 3720

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [C]  time = 4.88174, size = 163, normalized size = 2.51 \[ \frac{-\text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )-b^2 x^2 e^{-\tanh ^{-1}(\tanh (a))} \coth (a) \sqrt{\text{sech}^2(a)}+2 b x \log \left (1-e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )+2 \tanh ^{-1}(\tanh (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\tanh (a))+b x\right )\right )+b x\right )+i \pi b x-i \pi \log \left (e^{2 b x}+1\right )+i \pi \log (\cosh (b x))}{b^3}+\frac{x^2 \text{csch}(a) \sinh (b x) \text{csch}(a+b x)}{b}+\frac{x^3}{3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Coth[a + b*x]^2,x]

[Out]

x^3/3 + (I*b*Pi*x - I*Pi*Log[1 + E^(2*b*x)] + 2*b*x*Log[1 - E^(-2*(b*x + ArcTanh[Tanh[a]]))] + I*Pi*Log[Cosh[b
*x]] + 2*ArcTanh[Tanh[a]]*(b*x + Log[1 - E^(-2*(b*x + ArcTanh[Tanh[a]]))] - Log[I*Sinh[b*x + ArcTanh[Tanh[a]]]
]) - PolyLog[2, E^(-2*(b*x + ArcTanh[Tanh[a]]))] - (b^2*x^2*Coth[a]*Sqrt[Sech[a]^2])/E^ArcTanh[Tanh[a]])/b^3 +
(x^2*Csch[a]*Csch[a + b*x]*Sinh[b*x])/b

________________________________________________________________________________________

Maple [B]  time = 0.056, size = 156, normalized size = 2.4 \begin{align*}{\frac{{x}^{3}}{3}}-2\,{\frac{{x}^{2}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{{x}^{2}}{b}}-4\,{\frac{ax}{{b}^{2}}}-2\,{\frac{{a}^{2}}{{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{\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{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}}+4\,{\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*cosh(b*x+a)^2*csch(b*x+a)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.40041, size = 146, normalized size = 2.25 \begin{align*} -\frac{2 \, x^{2}}{b} + \frac{b x^{3} e^{\left (2 \, b x + 2 \, a\right )} - b x^{3} - 6 \, x^{2}}{3 \,{\left (b e^{\left (2 \, b x + 2 \, a\right )} - b\right )}} + \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*cosh(b*x+a)^2*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

-2*x^2/b + 1/3*(b*x^3*e^(2*b*x + 2*a) - b*x^3 - 6*x^2)/(b*e^(2*b*x + 2*a) - b) + 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

________________________________________________________________________________________

Fricas [B]  time = 2.17313, size = 1211, normalized size = 18.63 \begin{align*} -\frac{b^{3} x^{3} -{\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) -{\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 6 \, a^{2} - 6 \,{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \,{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 6 \,{\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 6 \,{\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} - a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \,{\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x + a\right )} \sinh \left (b x + a\right )^{2} - b x - a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )}{3 \,{\left (b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} - b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 - 2*(b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)*
sinh(b*x + a) - (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*sinh(b*x + a)^2 + 6*a^2 - 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*s
inh(b*x + a) + sinh(b*x + a)^2 - 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a
)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 6*(b*x*cosh(b*x + a)^2 + 2*b*x*
cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 - b*x)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 6*(a*cosh(b*
x + a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2 - a)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 6
*((b*x + a)*cosh(b*x + a)^2 + 2*(b*x + a)*cosh(b*x + a)*sinh(b*x + a) + (b*x + a)*sinh(b*x + a)^2 - b*x - a)*l
og(-cosh(b*x + a) - sinh(b*x + a) + 1))/(b^3*cosh(b*x + a)^2 + 2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*
x + a)^2 - b^3)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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