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

Optimal. Leaf size=87 \[ -\frac {3 \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \coth (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3720, 3716, 2190, 2531, 2282, 6589, 30} \[ \frac {3 x \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \coth (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Coth[a + b*x]^2,x]

[Out]

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

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 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^3 \coth ^2(a+b x) \, dx &=-\frac {x^3 \coth (a+b x)}{b}+\frac {3 \int x^2 \coth (a+b x) \, dx}{b}+\int x^3 \, dx\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}-\frac {6 \int \frac {e^{2 (a+b x)} x^2}{1-e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {6 \int x \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac {3 \int \text {Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^4}\\ \end {align*}

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Mathematica [B]  time = 0.60, size = 204, normalized size = 2.34 \[ -\frac {e^{2 a} \left (2 e^{-2 a} b^3 x^3-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (1-e^{-a-b x}\right )-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (e^{-a-b x}+1\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (-e^{-a-b x}\right )+\text {Li}_3\left (-e^{-a-b x}\right )\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (e^{-a-b x}\right )+\text {Li}_3\left (e^{-a-b x}\right )\right )\right )}{\left (e^{2 a}-1\right ) b^4}+\frac {x^3 \text {csch}(a) \sinh (b x) \text {csch}(a+b x)}{b}+\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Coth[a + b*x]^2,x]

[Out]

x^4/4 - (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)] - 3*b^2*(1 - E^(-2*a))*
x^2*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)])))/(b^4*(-1 + E^(2*a))) + (x^3*Csch[a]*
Csch[a + b*x]*Sinh[b*x])/b

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fricas [C]  time = 0.50, size = 632, normalized size = 7.26 \[ -\frac {b^{4} x^{4} - 8 \, a^{3} - {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \sinh \left (b x + a\right )^{2} - 24 \, {\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 )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 24 \, {\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 )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 12 \, {\left (b^{2} x^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right )^{2} - b^{2} x^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 12 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 12 \, {\left (b^{2} x^{2} - {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 24 \, {\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 polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 24 \, {\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 polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, {\left (b^{4} \cosh \left (b x + a\right )^{2} + 2 \, b^{4} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )^{2} - b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^4*x^4 - 8*a^3 - (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)^2 - 2*(b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b
*x + a)*sinh(b*x + a) - (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*sinh(b*x + a)^2 - 24*(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)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 24*(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)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 1
2*(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 - b^2*x^2)*log(co
sh(b*x + a) + sinh(b*x + a) + 1) - 12*(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 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 12*(b^2*x^2 - (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 - a^2)*log(-cosh(b*x + a) - sinh(b*x
+ a) + 1) + 24*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, cosh(b*x + a
) + sinh(b*x + a)) + 24*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, -co
sh(b*x + a) - sinh(b*x + a)))/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a)^2 -
b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.58, size = 198, normalized size = 2.28 \[ \frac {x^{4}}{4}-\frac {2 x^{3}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {3 a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 a^{2} x}{b^{3}}+\frac {4 a^{3}}{b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}+\frac {6 x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {6 x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4-2*x^3/b/(exp(2*b*x+2*a)-1)+3/b^4*a^2*ln(exp(b*x+a)-1)-6/b^4*a^2*ln(exp(b*x+a))-2*x^3/b+6/b^3*a^2*x+4/b
^4*a^3+3/b^2*ln(1-exp(b*x+a))*x^2-3/b^4*ln(1-exp(b*x+a))*a^2+6*x*polylog(2,exp(b*x+a))/b^3-6*polylog(3,exp(b*x
+a))/b^4+3/b^2*ln(1+exp(b*x+a))*x^2+6*x*polylog(2,-exp(b*x+a))/b^3-6*polylog(3,-exp(b*x+a))/b^4

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maxima [A]  time = 0.45, size = 146, normalized size = 1.68 \[ -\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} - b x^{4} - 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} - b\right )}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^3/b + 1/4*(b*x^4*e^(2*b*x + 2*a) - b*x^4 - 8*x^3)/(b*e^(2*b*x + 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^4 + 3*(b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*di
log(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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