### 3.405 $$\int x^3 \cosh (a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=165 $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{3 x^2 \sinh (a+b x)}{b^2}-\frac{6 \sinh (a+b x)}{b^4}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

(-2*x^3*ArcTanh[E^(a + b*x)])/b + (6*x*Cosh[a + b*x])/b^3 + (x^3*Cosh[a + b*x])/b - (3*x^2*PolyLog[2, -E^(a +
b*x)])/b^2 + (3*x^2*PolyLog[2, E^(a + b*x)])/b^2 + (6*x*PolyLog[3, -E^(a + b*x)])/b^3 - (6*x*PolyLog[3, E^(a +
b*x)])/b^3 - (6*PolyLog[4, -E^(a + b*x)])/b^4 + (6*PolyLog[4, E^(a + b*x)])/b^4 - (6*Sinh[a + b*x])/b^4 - (3*
x^2*Sinh[a + b*x])/b^2

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Rubi [A]  time = 0.177773, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5450, 3296, 2637, 4182, 2531, 6609, 2282, 6589} $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{3 x^2 \sinh (a+b x)}{b^2}-\frac{6 \sinh (a+b x)}{b^4}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

(-2*x^3*ArcTanh[E^(a + b*x)])/b + (6*x*Cosh[a + b*x])/b^3 + (x^3*Cosh[a + b*x])/b - (3*x^2*PolyLog[2, -E^(a +
b*x)])/b^2 + (3*x^2*PolyLog[2, E^(a + b*x)])/b^2 + (6*x*PolyLog[3, -E^(a + b*x)])/b^3 - (6*x*PolyLog[3, E^(a +
b*x)])/b^3 - (6*PolyLog[4, -E^(a + b*x)])/b^4 + (6*PolyLog[4, E^(a + b*x)])/b^4 - (6*Sinh[a + b*x])/b^4 - (3*
x^2*Sinh[a + b*x])/b^2

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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 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]

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]

Rubi steps

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

Mathematica [A]  time = 3.90936, size = 202, normalized size = 1.22 $\frac{-3 b^2 x^2 \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))+3 b^2 x^2 \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))+6 b x \text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))-6 b x \text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))-6 \text{PolyLog}(4,-\sinh (a+b x)-\cosh (a+b x))+6 \text{PolyLog}(4,\sinh (a+b x)+\cosh (a+b x))-3 b^2 x^2 \sinh (a+b x)+b^3 x^3 \cosh (a+b x)-2 b^3 x^3 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))-6 \sinh (a+b x)+6 b x \cosh (a+b x)}{b^4}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

(-2*b^3*x^3*ArcTanh[Cosh[a + b*x] + Sinh[a + b*x]] + 6*b*x*Cosh[a + b*x] + b^3*x^3*Cosh[a + b*x] - 3*b^2*x^2*P
olyLog[2, -Cosh[a + b*x] - Sinh[a + b*x]] + 3*b^2*x^2*PolyLog[2, Cosh[a + b*x] + Sinh[a + b*x]] + 6*b*x*PolyLo
g[3, -Cosh[a + b*x] - Sinh[a + b*x]] - 6*b*x*PolyLog[3, Cosh[a + b*x] + Sinh[a + b*x]] - 6*PolyLog[4, -Cosh[a
+ b*x] - Sinh[a + b*x]] + 6*PolyLog[4, Cosh[a + b*x] + Sinh[a + b*x]] - 6*Sinh[a + b*x] - 3*b^2*x^2*Sinh[a + b
*x])/b^4

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Maple [A]  time = 0.092, size = 246, normalized size = 1.5 \begin{align*}{\frac{ \left ({x}^{3}{b}^{3}-3\,{x}^{2}{b}^{2}+6\,bx-6 \right ){{\rm e}^{bx+a}}}{2\,{b}^{4}}}+{\frac{ \left ({x}^{3}{b}^{3}+3\,{x}^{2}{b}^{2}+6\,bx+6 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 4,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}-{\frac{{a}^{3}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{3}}{{b}^{4}}}+2\,{\frac{{a}^{3}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-6\,{\frac{{\it polylog} \left ( 4,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-3\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}+3\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}} \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),x)

[Out]

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

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Maxima [A]  time = 1.4724, size = 278, normalized size = 1.68 \begin{align*} \frac{{\left ({\left (b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 6 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} +{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{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}} \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),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.18018, size = 1409, normalized size = 8.54 \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),x, algorithm="fricas")

[Out]

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

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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),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 )\,{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),x, algorithm="giac")

[Out]

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