### 3.413 $$\int x^2 \cosh ^2(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=126 $\frac{x \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}-\frac{x^3}{3}$

[Out]

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

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Rubi [A]  time = 0.196835, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5450, 5372, 3310, 30, 3716, 2190, 2531, 2282, 6589} $\frac{x \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}-\frac{x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 3310

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[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 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 ^2(a+b x) \coth (a+b x) \, dx &=\int x^2 \coth (a+b x) \, dx+\int x^2 \cosh (a+b x) \sinh (a+b x) \, dx\\ &=-\frac{x^3}{3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x^2}{1-e^{2 (a+b x)}} \, dx-\frac{\int x \sinh ^2(a+b x) \, dx}{b}\\ &=-\frac{x^3}{3}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{\int x \, dx}{2 b}-\frac{2 \int x \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x^2}{4 b}-\frac{x^3}{3}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^2}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}-\frac{\int \text{Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{x^2}{4 b}-\frac{x^3}{3}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^2}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^3}\\ &=\frac{x^2}{4 b}-\frac{x^3}{3}+\frac{x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^2}-\frac{\text{Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 2.51488, size = 178, normalized size = 1.41 $\frac{\sinh (a) (\sinh (a)+\cosh (a)) \left (-48 b x \text{PolyLog}\left (2,-e^{-a-b x}\right )-48 b x \text{PolyLog}\left (2,e^{-a-b x}\right )-48 \text{PolyLog}\left (3,-e^{-a-b x}\right )-48 \text{PolyLog}\left (3,e^{-a-b x}\right )+24 b^2 x^2 \log \left (1-e^{-a-b x}\right )+24 b^2 x^2 \log \left (e^{-a-b x}+1\right )+6 b^2 x^2 \cosh (2 (a+b x))-6 b x \sinh (2 (a+b x))+3 \cosh (2 (a+b x))+8 b^3 x^3\right )}{12 \left (e^{2 a}-1\right ) b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sinh[a]*(Cosh[a] + Sinh[a])*(8*b^3*x^3 + 3*Cosh[2*(a + b*x)] + 6*b^2*x^2*Cosh[2*(a + b*x)] + 24*b^2*x^2*Log[1
- E^(-a - b*x)] + 24*b^2*x^2*Log[1 + E^(-a - b*x)] - 48*b*x*PolyLog[2, -E^(-a - b*x)] - 48*b*x*PolyLog[2, E^(
-a - b*x)] - 48*PolyLog[3, -E^(-a - b*x)] - 48*PolyLog[3, E^(-a - b*x)] - 6*b*x*Sinh[2*(a + b*x)]))/(12*b^3*(-
1 + E^(2*a)))

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Maple [A]  time = 0.072, size = 222, normalized size = 1.8 \begin{align*} -{\frac{{x}^{3}}{3}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}-2\,bx+1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{3}}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}+2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{3}}}+{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{{a}^{2}x}{{b}^{2}}}+{\frac{4\,{a}^{3}}{3\,{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}}}-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)^3*csch(b*x+a),x)

[Out]

-1/3*x^3+1/16*(2*b^2*x^2-2*b*x+1)/b^3*exp(2*b*x+2*a)+1/16*(2*b^2*x^2+2*b*x+1)/b^3*exp(-2*b*x-2*a)+1/b^3*a^2*ln
(exp(b*x+a)-1)-2/b^3*a^2*ln(exp(b*x+a))+2/b^2*a^2*x+4/3/b^3*a^3+1/b*ln(1+exp(b*x+a))*x^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 = 1.41126, size = 231, normalized size = 1.83 \begin{align*} -\frac{2}{3} \, x^{3} + \frac{{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} + 3 \,{\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 3 \,{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, 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)^3*csch(b*x+a),x, algorithm="maxima")

[Out]

-2/3*x^3 + 1/48*(16*b^3*x^3*e^(2*a) + 3*(2*b^2*x^2*e^(4*a) - 2*b*x*e^(4*a) + e^(4*a))*e^(2*b*x) + 3*(2*b^2*x^2
+ 2*b*x + 1)*e^(-2*b*x))*e^(-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 = 2.17355, size = 1858, normalized size = 14.75 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="fricas")

[Out]

1/48*(3*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^4 + 12*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + 3
*(2*b^2*x^2 - 2*b*x + 1)*sinh(b*x + a)^4 + 6*b^2*x^2 - 16*(b^3*x^3 + 2*a^3)*cosh(b*x + a)^2 - 2*(8*b^3*x^3 + 1
6*a^3 - 9*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 6*b*x + 96*(b*x*cosh(b*x + a)^2 + 2*b*x*c
osh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2)*dilog(cosh(b*x + a) + sinh(b*x + a)) + 96*(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)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + 48*(b^2*
x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a)^2)*log(cosh(b*x + a) + sin
h(b*x + a) + 1) + 48*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2)*log(cosh(
b*x + a) + sinh(b*x + a) - 1) + 48*((b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x
+ a) + (b^2*x^2 - a^2)*sinh(b*x + a)^2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - 96*(cosh(b*x + a)^2 + 2*cos
h(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) - 96*(cosh(b*x + a)^2 +
2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + 4*(3*(2*b^2*x^2
- 2*b*x + 1)*cosh(b*x + a)^3 - 8*(b^3*x^3 + 2*a^3)*cosh(b*x + a))*sinh(b*x + a) + 3)/(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)

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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)**3*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 )^{3} \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)^3*csch(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*cosh(b*x + a)^3*csch(b*x + a), x)