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

Optimal. Leaf size=179 \[ \frac {3 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \coth (a+b x)}{2 b^2}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3720, 3716, 2190, 2279, 2391, 30, 2531, 6609, 2282, 6589} \[ \frac {3 x^2 \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 \text {PolyLog}\left (4,e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Coth[a + b*x]^3,x]

[Out]

(-3*x^2)/(2*b^2) + x^3/(2*b) - x^4/4 - (3*x^2*Coth[a + b*x])/(2*b^2) - (x^3*Coth[a + b*x]^2)/(2*b) + (3*x*Log[
1 - E^(2*(a + b*x))])/b^3 + (x^3*Log[1 - E^(2*(a + b*x))])/b + (3*PolyLog[2, E^(2*(a + b*x))])/(2*b^4) + (3*x^
2*PolyLog[2, E^(2*(a + b*x))])/(2*b^2) - (3*x*PolyLog[3, E^(2*(a + b*x))])/(2*b^3) + (3*PolyLog[4, E^(2*(a + b
*x))])/(4*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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int x^3 \coth ^3(a+b x) \, dx &=-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 \int x^2 \coth ^2(a+b x) \, dx}{2 b}+\int x^3 \coth (a+b x) \, dx\\ &=-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}-2 \int \frac {e^{2 (a+b x)} x^3}{1-e^{2 (a+b x)}} \, dx+\frac {3 \int x \coth (a+b x) \, dx}{b^2}+\frac {3 \int x^2 \, dx}{2 b}\\ &=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {6 \int \frac {e^{2 (a+b x)} x}{1-e^{2 (a+b x)}} \, dx}{b^2}-\frac {3 \int x^2 \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 \int \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b^3}-\frac {3 \int x \text {Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 \int \text {Li}_3\left (e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {x^4}{4}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4}\\ \end {align*}

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Mathematica [B]  time = 3.34, size = 390, normalized size = 2.18 \[ \frac {1}{4} \left (\frac {6 x^2 \text {csch}(a) \sinh (b x) \text {csch}(a+b x)}{b^2}-\frac {2 e^{2 a} \left (e^{-2 a} b^4 x^4-2 e^{-2 a} \left (e^{2 a}-1\right ) b^3 x^3 \log \left (1-e^{-a-b x}\right )-2 e^{-2 a} \left (e^{2 a}-1\right ) b^3 x^3 \log \left (e^{-a-b x}+1\right )+6 \left (1-e^{-2 a}\right ) \left (b^2 x^2 \text {Li}_2\left (-e^{-a-b x}\right )+2 \left (b x \text {Li}_3\left (-e^{-a-b x}\right )+\text {Li}_4\left (-e^{-a-b x}\right )\right )\right )+6 \left (1-e^{-2 a}\right ) \left (b^2 x^2 \text {Li}_2\left (e^{-a-b x}\right )+2 \left (b x \text {Li}_3\left (e^{-a-b x}\right )+\text {Li}_4\left (e^{-a-b x}\right )\right )\right )+6 e^{-2 a} b^2 x^2+6 \left (1-e^{-2 a}\right ) \text {Li}_2\left (-e^{-a-b x}\right )+6 \left (1-e^{-2 a}\right ) \text {Li}_2\left (e^{-a-b x}\right )-6 \left (1-e^{-2 a}\right ) b x \log \left (1-e^{-a-b x}\right )-6 \left (1-e^{-2 a}\right ) b x \log \left (e^{-a-b x}+1\right )\right )}{\left (e^{2 a}-1\right ) b^4}-\frac {2 x^3 \text {csch}^2(a+b x)}{b}+x^4 \coth (a)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Coth[a + b*x]^3,x]

[Out]

(x^4*Coth[a] - (2*x^3*Csch[a + b*x]^2)/b - (2*E^(2*a)*((6*b^2*x^2)/E^(2*a) + (b^4*x^4)/E^(2*a) - 6*b*(1 - E^(-
2*a))*x*Log[1 - E^(-a - b*x)] - (2*b^3*(-1 + E^(2*a))*x^3*Log[1 - E^(-a - b*x)])/E^(2*a) - 6*b*(1 - E^(-2*a))*
x*Log[1 + E^(-a - b*x)] - (2*b^3*(-1 + E^(2*a))*x^3*Log[1 + E^(-a - b*x)])/E^(2*a) + 6*(1 - E^(-2*a))*PolyLog[
2, -E^(-a - b*x)] + 6*(1 - E^(-2*a))*PolyLog[2, E^(-a - b*x)] + 6*(1 - E^(-2*a))*(b^2*x^2*PolyLog[2, -E^(-a -
b*x)] + 2*(b*x*PolyLog[3, -E^(-a - b*x)] + PolyLog[4, -E^(-a - b*x)])) + 6*(1 - E^(-2*a))*(b^2*x^2*PolyLog[2,
E^(-a - b*x)] + 2*(b*x*PolyLog[3, E^(-a - b*x)] + PolyLog[4, E^(-a - b*x)]))))/(b^4*(-1 + E^(2*a))) + (6*x^2*C
sch[a]*Csch[a + b*x]*Sinh[b*x])/b^2)/4

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fricas [C]  time = 0.45, size = 1985, normalized size = 11.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^4*x^4 + (b^4*x^4 - 2*a^4 + 12*b^2*x^2 - 12*a^2)*cosh(b*x + a)^4 + 4*(b^4*x^4 - 2*a^4 + 12*b^2*x^2 - 12
*a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^4*x^4 - 2*a^4 + 12*b^2*x^2 - 12*a^2)*sinh(b*x + a)^4 - 2*a^4 - 2*(b^4
*x^4 - 4*b^3*x^3 - 2*a^4 + 6*b^2*x^2 - 12*a^2)*cosh(b*x + a)^2 - 2*(b^4*x^4 - 4*b^3*x^3 - 2*a^4 + 6*b^2*x^2 -
3*(b^4*x^4 - 2*a^4 + 12*b^2*x^2 - 12*a^2)*cosh(b*x + a)^2 - 12*a^2)*sinh(b*x + a)^2 - 12*a^2 - 12*((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)*dilog(cosh(b*x + a) + sinh(b*x + a
)) - 12*((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)*dilog(-cosh(b*
x + a) - sinh(b*x + a)) - 4*(b^3*x^3 + (b^3*x^3 + 3*b*x)*cosh(b*x + a)^4 + 4*(b^3*x^3 + 3*b*x)*cosh(b*x + a)*s
inh(b*x + a)^3 + (b^3*x^3 + 3*b*x)*sinh(b*x + a)^4 - 2*(b^3*x^3 + 3*b*x)*cosh(b*x + a)^2 - 2*(b^3*x^3 - 3*(b^3
*x^3 + 3*b*x)*cosh(b*x + a)^2 + 3*b*x)*sinh(b*x + a)^2 + 3*b*x + 4*((b^3*x^3 + 3*b*x)*cosh(b*x + a)^3 - (b^3*x
^3 + 3*b*x)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 4*((a^3 + 3*a)*cosh(b*x + a
)^4 + 4*(a^3 + 3*a)*cosh(b*x + a)*sinh(b*x + a)^3 + (a^3 + 3*a)*sinh(b*x + a)^4 + a^3 - 2*(a^3 + 3*a)*cosh(b*x
+ a)^2 - 2*(a^3 - 3*(a^3 + 3*a)*cosh(b*x + a)^2 + 3*a)*sinh(b*x + a)^2 + 4*((a^3 + 3*a)*cosh(b*x + a)^3 - (a^
3 + 3*a)*cosh(b*x + a))*sinh(b*x + a) + 3*a)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 4*(b^3*x^3 + (b^3*x^3 +
a^3 + 3*b*x + 3*a)*cosh(b*x + a)^4 + 4*(b^3*x^3 + a^3 + 3*b*x + 3*a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^3*x^3
+ a^3 + 3*b*x + 3*a)*sinh(b*x + a)^4 + a^3 - 2*(b^3*x^3 + a^3 + 3*b*x + 3*a)*cosh(b*x + a)^2 - 2*(b^3*x^3 + a^
3 - 3*(b^3*x^3 + a^3 + 3*b*x + 3*a)*cosh(b*x + a)^2 + 3*b*x + 3*a)*sinh(b*x + a)^2 + 3*b*x + 4*((b^3*x^3 + a^3
+ 3*b*x + 3*a)*cosh(b*x + a)^3 - (b^3*x^3 + a^3 + 3*b*x + 3*a)*cosh(b*x + a))*sinh(b*x + a) + 3*a)*log(-cosh(
b*x + a) - sinh(b*x + a) + 1) - 24*(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(4, cosh(b*x + a) + sinh(b*x + a)) - 24*(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(4, -cosh(b*x + a) - sinh(b*x + a)) + 24*(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))*polylog(3, cosh(b*x + a) + sinh(b*x +
a)) + 24*(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))*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + 4*((b^4*x^4 - 2*a^4 + 12*b^2*x^2 - 12*a^2)*cosh(b*x +
a)^3 - (b^4*x^4 - 4*b^3*x^3 - 2*a^4 + 6*b^2*x^2 - 12*a^2)*cosh(b*x + a))*sinh(b*x + a))/(b^4*cosh(b*x + a)^4 +
4*b^4*cosh(b*x + a)*sinh(b*x + a)^3 + b^4*sinh(b*x + a)^4 - 2*b^4*cosh(b*x + a)^2 + b^4 + 2*(3*b^4*cosh(b*x +
a)^2 - b^4)*sinh(b*x + a)^2 + 4*(b^4*cosh(b*x + a)^3 - b^4*cosh(b*x + a))*sinh(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.60, size = 375, normalized size = 2.09 \[ -\frac {x^{2} \left (2 b x \,{\mathrm e}^{2 b x +2 a}+3 \,{\mathrm e}^{2 b x +2 a}-3\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}+\frac {3 x^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {2 a^{3} x}{b^{3}}+\frac {3 x^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 x \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {x^{4}}{4}-\frac {3 x^{2}}{b^{2}}+\frac {6 \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 a^{4}}{2 b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{4}}+\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {6 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}-\frac {3 a^{2}}{b^{4}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{3}}{b}-\frac {6 a x}{b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 302, normalized size = 1.69 \[ \frac {b^{2} x^{4} e^{\left (4 \, b x + 4 \, a\right )} + b^{2} x^{4} + 12 \, x^{2} - 2 \, {\left (b^{2} x^{4} e^{\left (2 \, a\right )} + 4 \, b x^{3} e^{\left (2 \, a\right )} + 6 \, x^{2} e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{4 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} - \frac {b^{4} x^{4} + 6 \, b^{2} x^{2}}{2 \, b^{4}} + \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} + \frac {3 \, {\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{4}} + \frac {3 \, {\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^4*e^(4*b*x + 4*a) + b^2*x^4 + 12*x^2 - 2*(b^2*x^4*e^(2*a) + 4*b*x^3*e^(2*a) + 6*x^2*e^(2*a))*e^(2*b
*x))/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) + b^2) - 1/2*(b^4*x^4 + 6*b^2*x^2)/b^4 + (b^3*x^3*log(e^(b*x
+ a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*b*x*polylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^4
+ (b^3*x^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*dilog(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4,
e^(b*x + a)))/b^4 + 3*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^4 + 3*(b*x*log(-e^(b*x + a) + 1) + d
ilog(e^(b*x + a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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