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

Optimal. Leaf size=87 $\frac{3 x \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac{3 \text{PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{x^3 \coth (a+b x)}{b}-\frac{x^3}{b}+\frac{x^4}{4}$

[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)

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Rubi [A]  time = 0.190154, antiderivative size = 87, normalized size of antiderivative = 1., 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 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 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]

Rule 30

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

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*}

Mathematica [B]  time = 0.60123, size = 204, normalized size = 2.34 $-\frac{e^{2 a} \left (6 \left (1-e^{-2 a}\right ) \left (b x \text{PolyLog}\left (2,-e^{-a-b x}\right )+\text{PolyLog}\left (3,-e^{-a-b x}\right )\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text{PolyLog}\left (2,e^{-a-b x}\right )+\text{PolyLog}\left (3,e^{-a-b x}\right )\right )+2 e^{-2 a} b^3 x^3-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (1-e^{-a-b x}\right )-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (e^{-a-b x}+1\right )\right )}{\left (e^{2 a}-1\right ) b^4}+\frac{x^3 \text{csch}(a) \sinh (b x) \text{csch}(a+b x)}{b}+\frac{x^4}{4}$

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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Maple [B]  time = 0.056, size = 198, normalized size = 2.3 \begin{align*}{\frac{{x}^{4}}{4}}-2\,{\frac{{x}^{3}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+3\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}}-6\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-2\,{\frac{{x}^{3}}{b}}+6\,{\frac{{a}^{2}x}{{b}^{3}}}+4\,{\frac{{a}^{3}}{{b}^{4}}}+3\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{4}}}+6\,{\frac{x{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

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+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^2-3/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

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Maxima [A]  time = 1.38485, size = 197, normalized size = 2.26 \begin{align*} -\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}} \end{align*}

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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Fricas [C]  time = 2.16178, size = 1643, normalized size = 18.89 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cosh \left (b x + a\right )^{2} \operatorname{csch}\left (b x + a\right )^{2}\,{d x} \end{align*}

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)