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

Optimal. Leaf size=63 \[ -\frac {\text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^2}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^3}{3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3716, 2190, 2531, 2282, 6589} \[ \frac {x \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + b*x],x]

[Out]

-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)

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 66, normalized size = 1.05 \[ -\frac {\text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}+\frac {x \text {Li}_2\left (e^{2 a+2 b x}\right )}{b^2}+\frac {x^2 \log \left (1-e^{2 a+2 b x}\right )}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Coth[a + b*x],x]

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 1.00, size = 168, normalized size = 2.67 \[ -\frac {b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 6 \, b x {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, b x {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - 3*b^2*x^2*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 6*b*x*dilog(cosh(b*x + a) + sinh(b*x + a))
- 6*b*x*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 3*a^2*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 3*(b^2*x^2 - a^
2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 6*polylog(3, -cosh(
b*x + a) - sinh(b*x + a)))/b^3

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh \left (b x + a\right ) \operatorname {csch}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.27, size = 166, normalized size = 2.63 \[ -\frac {x^{3}}{3}+\frac {a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 a^{2} x}{b^{2}}+\frac {4 a^{3}}{3 b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{3}}+\frac {2 \polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 96, normalized size = 1.52 \[ -\frac {1}{3} \, x^{3} + \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,\mathrm {cosh}\left (a+b\,x\right )}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*cosh(a + b*x))/sinh(a + b*x),x)

[Out]

int((x^2*cosh(a + b*x))/sinh(a + b*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh {\left (a + b x \right )} \operatorname {csch}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________