### 3.406 $$\int x^2 \cosh (a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=115 $-\frac{2 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{2 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{2 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{2 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{2 x \sinh (a+b x)}{b^2}+\frac{2 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}-\frac{2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.123183, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.438, Rules used = {5450, 3296, 2638, 4182, 2531, 2282, 6589} $-\frac{2 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{2 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{2 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{2 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{2 x \sinh (a+b x)}{b^2}+\frac{2 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}-\frac{2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

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

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A]  time = 3.82008, size = 138, normalized size = 1.2 $\frac{-2 b x \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))+2 b x \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))+2 \text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))-2 \text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))+b^2 x^2 \cosh (a+b x)-2 b^2 x^2 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))-2 b x \sinh (a+b x)+2 \cosh (a+b x)}{b^3}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.059, size = 196, normalized size = 1.7 \begin{align*}{\frac{ \left ({x}^{2}{b}^{2}-2\,bx+2 \right ){{\rm e}^{bx+a}}}{2\,{b}^{3}}}+{\frac{ \left ({x}^{2}{b}^{2}+2\,bx+2 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{3}}}-2\,{\frac{{a}^{2}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{b}}+{\frac{{a}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+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}}-{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}+2\,{\frac{x{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-2\,{\frac{{\it polylog} \left ( 3,{{\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),x)

[Out]

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

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Maxima [A]  time = 1.34865, size = 205, normalized size = 1.78 \begin{align*} \frac{{\left ({\left (b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + 2 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} +{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x\right )}\right )} e^{\left (-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*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="maxima")

[Out]

1/2*((b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) + 2*e^(2*a))*e^(b*x) + (b^2*x^2 + 2*b*x + 2)*e^(-b*x))*e^(-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 = 2.20668, size = 1084, normalized size = 9.43 \begin{align*} \frac{b^{2} x^{2} +{\left (b^{2} x^{2} - 2 \, b x + 2\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b^{2} x^{2} - 2 \, b x + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b^{2} x^{2} - 2 \, b x + 2\right )} \sinh \left (b x + a\right )^{2} + 2 \, b x + 4 \,{\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 4 \,{\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 2 \,{\left (b^{2} x^{2} \cosh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \,{\left (a^{2} \cosh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left ({\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) +{\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 4 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 4 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 2}{2 \,{\left (b^{3} \cosh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )\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),x, algorithm="fricas")

[Out]

1/2*(b^2*x^2 + (b^2*x^2 - 2*b*x + 2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - 2*b*x + 2)*cosh(b*x + a)*sinh(b*x + a) + (
b^2*x^2 - 2*b*x + 2)*sinh(b*x + a)^2 + 2*b*x + 4*(b*x*cosh(b*x + a) + b*x*sinh(b*x + a))*dilog(cosh(b*x + a) +
sinh(b*x + a)) - 4*(b*x*cosh(b*x + a) + b*x*sinh(b*x + a))*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 2*(b^2*x^2
*cosh(b*x + a) + b^2*x^2*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 2*(a^2*cosh(b*x + a) + a^2*si
nh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 2*((b^2*x^2 - a^2)*cosh(b*x + a) + (b^2*x^2 - a^2)*sinh(
b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - 4*(cosh(b*x + a) + sinh(b*x + a))*polylog(3, cosh(b*x + a)
+ sinh(b*x + a)) + 4*(cosh(b*x + a) + sinh(b*x + a))*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + 2)/(b^3*cos
h(b*x + a) + b^3*sinh(b*x + a))

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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),x)

[Out]

Timed out

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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 )\,{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),x, algorithm="giac")

[Out]

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