3.421 \(\int x \cosh ^2(x) \coth ^3(x) \, dx\)

Optimal. Leaf size=63 \[ \text{PolyLog}\left (2,e^{2 x}\right )-x^2+\frac{3 x}{4}+2 x \log \left (1-e^{2 x}\right )+\frac{1}{2} x \sinh ^2(x)-\frac{1}{2} x \coth ^2(x)-\frac{\coth (x)}{2}-\frac{1}{4} \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.162843, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5450, 5372, 2635, 8, 3716, 2190, 2279, 2391, 3720, 3473} \[ \text{PolyLog}\left (2,e^{2 x}\right )-x^2+\frac{3 x}{4}+2 x \log \left (1-e^{2 x}\right )+\frac{1}{2} x \sinh ^2(x)-\frac{1}{2} x \coth ^2(x)-\frac{\coth (x)}{2}-\frac{1}{4} \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[x]^2*Coth[x]^3,x]

[Out]

(3*x)/4 - x^2 - Coth[x]/2 - (x*Coth[x]^2)/2 + 2*x*Log[1 - E^(2*x)] + PolyLog[2, E^(2*x)] - (Cosh[x]*Sinh[x])/4
+ (x*Sinh[x]^2)/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 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

Mathematica [A]  time = 0.0765283, size = 56, normalized size = 0.89 \[ \frac{1}{8} \left (-8 \text{PolyLog}\left (2,e^{-2 x}\right )+8 x^2+16 x \log \left (1-e^{-2 x}\right )-\sinh (2 x)+2 x \cosh (2 x)-4 \coth (x)-4 x \text{csch}^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[x]^2*Coth[x]^3,x]

[Out]

(8*x^2 + 2*x*Cosh[2*x] - 4*Coth[x] - 4*x*Csch[x]^2 + 16*x*Log[1 - E^(-2*x)] - 8*PolyLog[2, E^(-2*x)] - Sinh[2*
x])/8

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Maple [A]  time = 0.047, size = 82, normalized size = 1.3 \begin{align*} -{x}^{2}+ \left ( -{\frac{1}{16}}+{\frac{x}{8}} \right ){{\rm e}^{2\,x}}+ \left ({\frac{1}{16}}+{\frac{x}{8}} \right ){{\rm e}^{-2\,x}}-{\frac{2\,x{{\rm e}^{2\,x}}+{{\rm e}^{2\,x}}-1}{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}}+2\,x\ln \left ({{\rm e}^{x}}+1 \right ) +2\,{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) +2\,x\ln \left ( 1-{{\rm e}^{x}} \right ) +2\,{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(x)^2*coth(x)^3,x)

[Out]

-x^2+(-1/16+1/8*x)*exp(2*x)+(1/16+1/8*x)*exp(-2*x)-(2*x*exp(2*x)+exp(2*x)-1)/(exp(2*x)-1)^2+2*x*ln(exp(x)+1)+2
*polylog(2,-exp(x))+2*x*ln(1-exp(x))+2*polylog(2,exp(x))

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Maxima [B]  time = 1.36158, size = 197, normalized size = 3.13 \begin{align*} -2 \, x^{2} + 2 \, x \log \left (e^{x} + 1\right ) + 2 \, x \log \left (-e^{x} + 1\right ) + \frac{5}{8} \, x + \frac{16 \, x^{2} +{\left (2 \, x - 1\right )} e^{\left (6 \, x\right )} + 2 \,{\left (8 \, x^{2} - 2 \, x + 1\right )} e^{\left (4 \, x\right )} -{\left (32 \, x^{2} + 8 \, x + 11\right )} e^{\left (2 \, x\right )} +{\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} - 14 \, x + 9}{16 \,{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )}} - \frac{5 \,{\left (2 \, x e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )} - 1\right )}}{16 \,{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )}} + 2 \,{\rm Li}_2\left (-e^{x}\right ) + 2 \,{\rm Li}_2\left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(x)^2*coth(x)^3,x, algorithm="maxima")

[Out]

-2*x^2 + 2*x*log(e^x + 1) + 2*x*log(-e^x + 1) + 5/8*x + 1/16*(16*x^2 + (2*x - 1)*e^(6*x) + 2*(8*x^2 - 2*x + 1)
*e^(4*x) - (32*x^2 + 8*x + 11)*e^(2*x) + (2*x + 1)*e^(-2*x) - 14*x + 9)/(e^(4*x) - 2*e^(2*x) + 1) - 5/16*(2*x*
e^(4*x) + e^(2*x) - 1)/(e^(4*x) - 2*e^(2*x) + 1) + 2*dilog(-e^x) + 2*dilog(e^x)

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Fricas [B]  time = 2.25806, size = 2881, normalized size = 45.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(x)^2*coth(x)^3,x, algorithm="fricas")

[Out]

1/16*((2*x - 1)*cosh(x)^8 + 8*(2*x - 1)*cosh(x)*sinh(x)^7 + (2*x - 1)*sinh(x)^8 - 2*(8*x^2 + 2*x - 1)*cosh(x)^
6 + 2*(14*(2*x - 1)*cosh(x)^2 - 8*x^2 - 2*x + 1)*sinh(x)^6 + 4*(14*(2*x - 1)*cosh(x)^3 - 3*(8*x^2 + 2*x - 1)*c
osh(x))*sinh(x)^5 + 4*(8*x^2 - 7*x - 4)*cosh(x)^4 + 2*(35*(2*x - 1)*cosh(x)^4 - 15*(8*x^2 + 2*x - 1)*cosh(x)^2
+ 16*x^2 - 14*x - 8)*sinh(x)^4 + 8*(7*(2*x - 1)*cosh(x)^5 - 5*(8*x^2 + 2*x - 1)*cosh(x)^3 + 2*(8*x^2 - 7*x -
4)*cosh(x))*sinh(x)^3 - 2*(8*x^2 + 2*x - 7)*cosh(x)^2 + 2*(14*(2*x - 1)*cosh(x)^6 - 15*(8*x^2 + 2*x - 1)*cosh(
x)^4 + 12*(8*x^2 - 7*x - 4)*cosh(x)^2 - 8*x^2 - 2*x + 7)*sinh(x)^2 + 32*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sin
h(x)^6 + (15*cosh(x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 -
12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))*dilog(cosh(x) + sin
h(x)) + 32*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*co
sh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 - 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*
cosh(x)^3 + cosh(x))*sinh(x))*dilog(-cosh(x) - sinh(x)) + 32*(x*cosh(x)^6 + 6*x*cosh(x)*sinh(x)^5 + x*sinh(x)^
6 - 2*x*cosh(x)^4 + (15*x*cosh(x)^2 - 2*x)*sinh(x)^4 + 4*(5*x*cosh(x)^3 - 2*x*cosh(x))*sinh(x)^3 + x*cosh(x)^2
+ (15*x*cosh(x)^4 - 12*x*cosh(x)^2 + x)*sinh(x)^2 + 2*(3*x*cosh(x)^5 - 4*x*cosh(x)^3 + x*cosh(x))*sinh(x))*lo
g(cosh(x) + sinh(x) + 1) + 32*(x*cosh(x)^6 + 6*x*cosh(x)*sinh(x)^5 + x*sinh(x)^6 - 2*x*cosh(x)^4 + (15*x*cosh(
x)^2 - 2*x)*sinh(x)^4 + 4*(5*x*cosh(x)^3 - 2*x*cosh(x))*sinh(x)^3 + x*cosh(x)^2 + (15*x*cosh(x)^4 - 12*x*cosh(
x)^2 + x)*sinh(x)^2 + 2*(3*x*cosh(x)^5 - 4*x*cosh(x)^3 + x*cosh(x))*sinh(x))*log(-cosh(x) - sinh(x) + 1) + 4*(
2*(2*x - 1)*cosh(x)^7 - 3*(8*x^2 + 2*x - 1)*cosh(x)^5 + 4*(8*x^2 - 7*x - 4)*cosh(x)^3 - (8*x^2 + 2*x - 7)*cosh
(x))*sinh(x) + 2*x + 1)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(x)^2 - 2)*sinh(x)^4 - 2*cosh(x
)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 - 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*c
osh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))

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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*cosh(x)**2*coth(x)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(x)^2*coth(x)^3,x, algorithm="giac")

[Out]

integrate(x*cosh(x)^2*coth(x)^3, x)