3.461 \(\int x^2 \coth ^3(a+b x) \, dx\)

Optimal. Leaf size=114 \[ -\frac {\text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {\log (\sinh (a+b x))}{b^3}+\frac {x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^2}-\frac {x \coth (a+b x)}{b^2}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2}{2 b}-\frac {x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3720, 3475, 30, 3716, 2190, 2531, 2282, 6589} \[ \frac {x \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}-\frac {x \coth (a+b x)}{b^2}+\frac {\log (\sinh (a+b x))}{b^3}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2}{2 b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + b*x]^3,x]

[Out]

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

Rule 30

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

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 3475

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

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 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 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 ^3(a+b x) \, dx &=-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {\int x \coth ^2(a+b x) \, dx}{b}+\int x^2 \coth (a+b x) \, dx\\ &=-\frac {x^3}{3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \coth ^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 \coth (a+b x) \, dx}{b^2}+\frac {\int x \, dx}{b}\\ &=\frac {x^2}{2 b}-\frac {x^3}{3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {\log (\sinh (a+b x))}{b^3}-\frac {2 \int x \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac {x^2}{2 b}-\frac {x^3}{3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {\log (\sinh (a+b x))}{b^3}+\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^2}{2 b}-\frac {x^3}{3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {\log (\sinh (a+b x))}{b^3}+\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^2}{2 b}-\frac {x^3}{3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \coth ^2(a+b x)}{2 b}+\frac {x^2 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {\log (\sinh (a+b x))}{b^3}+\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*}

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Mathematica [B]  time = 2.21, size = 295, normalized size = 2.59 \[ \frac {x \text {csch}(a) \sinh (b x) \text {csch}(a+b x)}{b^2}-\frac {e^{2 a} \left (2 e^{-2 a} b^3 x^3-3 e^{-2 a} \left (e^{2 a}-1\right ) b^2 x^2 \log \left (1-e^{-a-b x}\right )-3 e^{-2 a} \left (e^{2 a}-1\right ) b^2 x^2 \log \left (e^{-a-b x}+1\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (-e^{-a-b x}\right )+\text {Li}_3\left (-e^{-a-b x}\right )\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (e^{-a-b x}\right )+\text {Li}_3\left (e^{-a-b x}\right )\right )+6 e^{-2 a} b x+3 \left (1-e^{-2 a}\right ) \left (b x-\log \left (1-e^{a+b x}\right )\right )+3 \left (1-e^{-2 a}\right ) \left (b x-\log \left (e^{a+b x}+1\right )\right )\right )}{3 \left (e^{2 a}-1\right ) b^3}-\frac {x^2 \text {csch}^2(a+b x)}{2 b}+\frac {1}{3} x^3 \coth (a) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Coth[a + b*x]^3,x]

[Out]

(x^3*Coth[a])/3 - (x^2*Csch[a + b*x]^2)/(2*b) - (E^(2*a)*((6*b*x)/E^(2*a) + (2*b^3*x^3)/E^(2*a) - (3*b^2*(-1 +
E^(2*a))*x^2*Log[1 - E^(-a - b*x)])/E^(2*a) - (3*b^2*(-1 + E^(2*a))*x^2*Log[1 + E^(-a - b*x)])/E^(2*a) + 3*(1
- E^(-2*a))*(b*x - Log[1 - E^(a + b*x)]) + 3*(1 - E^(-2*a))*(b*x - Log[1 + E^(a + b*x)]) + 6*(1 - E^(-2*a))*(
b*x*PolyLog[2, -E^(-a - b*x)] + PolyLog[3, -E^(-a - b*x)]) + 6*(1 - E^(-2*a))*(b*x*PolyLog[2, E^(-a - b*x)] +
PolyLog[3, E^(-a - b*x)])))/(3*b^3*(-1 + E^(2*a))) + (x*Csch[a]*Csch[a + b*x]*Sinh[b*x])/b^2

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fricas [C]  time = 0.43, size = 1467, normalized size = 12.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 + (b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x + a)^4 + 4*(b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x +
a)*sinh(b*x + a)^3 + (b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*sinh(b*x + a)^4 + 2*a^3 - 2*(b^3*x^3 - 3*b^2*x^2 + 2*a^3
+ 3*b*x + 6*a)*cosh(b*x + a)^2 - 2*(b^3*x^3 - 3*b^2*x^2 + 2*a^3 - 3*(b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x
+ a)^2 + 3*b*x + 6*a)*sinh(b*x + a)^2 - 6*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sin
h(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x
+ a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(cosh(b*x + a) + sinh(b*x + a)) - 6*(b*x*cosh(b*x + a)^4 + 4*
b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b
*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(-cosh(b*x + a) -
sinh(b*x + a)) - 3*((b^2*x^2 + 1)*cosh(b*x + a)^4 + 4*(b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 +
1)*sinh(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 + 1)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3*(b^2*x^2 + 1)*cosh(b*x + a)^2
+ 1)*sinh(b*x + a)^2 + 4*((b^2*x^2 + 1)*cosh(b*x + a)^3 - (b^2*x^2 + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)*log
(cosh(b*x + a) + sinh(b*x + a) + 1) - 3*((a^2 + 1)*cosh(b*x + a)^4 + 4*(a^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3
+ (a^2 + 1)*sinh(b*x + a)^4 - 2*(a^2 + 1)*cosh(b*x + a)^2 + 2*(3*(a^2 + 1)*cosh(b*x + a)^2 - a^2 - 1)*sinh(b*
x + a)^2 + a^2 + 4*((a^2 + 1)*cosh(b*x + a)^3 - (a^2 + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a)
+ sinh(b*x + a) - 1) - 3*((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 +
(b^2*x^2 - a^2)*sinh(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3*(b^2*x^2 - a^2)
*cosh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 - (b^2*x^2 - a^2)*cosh(b*x
+ a))*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x +
a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 -
cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 6*(cosh(b*x + a)^4 + 4*cosh(b*x
+ a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(c
osh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + 4*((b^3*x^3 +
2*a^3 + 6*b*x + 6*a)*cosh(b*x + a)^3 - (b^3*x^3 - 3*b^2*x^2 + 2*a^3 + 3*b*x + 6*a)*cosh(b*x + a))*sinh(b*x + a
) + 6*a)/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3 + b^3*sinh(b*x + a)^4 - 2*b^3*cosh(b*x + a
)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 - b^3)*sinh(b*x + a)^2 + 4*(b^3*cosh(b*x + a)^3 - b^3*cosh(b*x + a))*sinh
(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.60, size = 246, normalized size = 2.16 \[ -\frac {x^{3}}{3}-\frac {2 x \left (b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}-1\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {4 a^{3}}{3 b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 \polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 a^{2} x}{b^{2}}-\frac {2 \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.72, size = 226, normalized size = 1.98 \[ -\frac {2}{3} \, x^{3} + \frac {b^{2} x^{3} e^{\left (4 \, b x + 4 \, a\right )} + b^{2} x^{3} - 2 \, {\left (b^{2} x^{3} e^{\left (2 \, a\right )} + 3 \, b x^{2} e^{\left (2 \, a\right )} + 3 \, x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 6 \, x}{3 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} - \frac {2 \, x}{b^{2}} + \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}} + \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b^{3}} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^3 + 1/3*(b^2*x^3*e^(4*b*x + 4*a) + b^2*x^3 - 2*(b^2*x^3*e^(2*a) + 3*b*x^2*e^(2*a) + 3*x*e^(2*a))*e^(2*b
*x) + 6*x)/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) + b^2) - 2*x/b^2 + (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 + log(e^(b*x + a) + 1)/b^3 + log(e^(b*x + a) - 1)/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*cosh(a + b*x)^3)/sinh(a + b*x)^3,x)

[Out]

int((x^2*cosh(a + b*x)^3)/sinh(a + b*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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