3.91 \(\int \frac{\sinh ^2(x)}{1+\coth (x)} \, dx\)

Optimal. Leaf size=38 \[ -\frac{3 x}{8}-\frac{1}{8 (1-\coth (x))}+\frac{1}{4 (\coth (x)+1)}+\frac{1}{8 (\coth (x)+1)^2} \]

[Out]

(-3*x)/8 - 1/(8*(1 - Coth[x])) + 1/(8*(1 + Coth[x])^2) + 1/(4*(1 + Coth[x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0522239, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3487, 44, 207} \[ -\frac{3 x}{8}-\frac{1}{8 (1-\coth (x))}+\frac{1}{4 (\coth (x)+1)}+\frac{1}{8 (\coth (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 + Coth[x]),x]

[Out]

(-3*x)/8 - 1/(8*(1 - Coth[x])) + 1/(8*(1 + Coth[x])^2) + 1/(4*(1 + Coth[x]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sinh ^2(x)}{1+\coth (x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{(1-x)^2 (1+x)^3} \, dx,x,\coth (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{1}{8 (-1+x)^2}+\frac{1}{4 (1+x)^3}+\frac{1}{4 (1+x)^2}-\frac{3}{8 \left (-1+x^2\right )}\right ) \, dx,x,\coth (x)\right )\\ &=-\frac{1}{8 (1-\coth (x))}+\frac{1}{8 (1+\coth (x))^2}+\frac{1}{4 (1+\coth (x))}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\coth (x)\right )\\ &=-\frac{3 x}{8}-\frac{1}{8 (1-\coth (x))}+\frac{1}{8 (1+\coth (x))^2}+\frac{1}{4 (1+\coth (x))}\\ \end{align*}

Mathematica [A]  time = 0.0523216, size = 30, normalized size = 0.79 \[ \frac{1}{32} (-12 x+8 \sinh (2 x)-\sinh (4 x)-4 \cosh (2 x)+\cosh (4 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(1 + Coth[x]),x]

[Out]

(-12*x - 4*Cosh[2*x] + Cosh[4*x] + 8*Sinh[2*x] - Sinh[4*x])/32

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 70, normalized size = 1.8 \begin{align*}{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{3}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(1+coth(x)),x)

[Out]

1/2/(tanh(1/2*x)+1)^4-1/(tanh(1/2*x)+1)^3+1/2/(tanh(1/2*x)+1)-3/8*ln(tanh(1/2*x)+1)+1/4/(tanh(1/2*x)-1)^2+1/4/
(tanh(1/2*x)-1)+3/8*ln(tanh(1/2*x)-1)

________________________________________________________________________________________

Maxima [A]  time = 1.08251, size = 30, normalized size = 0.79 \begin{align*} -\frac{3}{8} \, x + \frac{1}{16} \, e^{\left (2 \, x\right )} - \frac{3}{16} \, e^{\left (-2 \, x\right )} + \frac{1}{32} \, e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+coth(x)),x, algorithm="maxima")

[Out]

-3/8*x + 1/16*e^(2*x) - 3/16*e^(-2*x) + 1/32*e^(-4*x)

________________________________________________________________________________________

Fricas [A]  time = 2.45374, size = 176, normalized size = 4.63 \begin{align*} \frac{3 \, \cosh \left (x\right )^{3} + 9 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} - 6 \,{\left (2 \, x + 1\right )} \cosh \left (x\right ) + 3 \,{\left (\cosh \left (x\right )^{2} - 4 \, x + 2\right )} \sinh \left (x\right )}{32 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+coth(x)),x, algorithm="fricas")

[Out]

1/32*(3*cosh(x)^3 + 9*cosh(x)*sinh(x)^2 + sinh(x)^3 - 6*(2*x + 1)*cosh(x) + 3*(cosh(x)^2 - 4*x + 2)*sinh(x))/(
cosh(x) + sinh(x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (x \right )}}{\coth{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**2/(1+coth(x)),x)

[Out]

Integral(sinh(x)**2/(coth(x) + 1), x)

________________________________________________________________________________________

Giac [A]  time = 1.12842, size = 41, normalized size = 1.08 \begin{align*} \frac{1}{32} \,{\left (9 \, e^{\left (4 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-4 \, x\right )} - \frac{3}{8} \, x + \frac{1}{16} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+coth(x)),x, algorithm="giac")

[Out]

1/32*(9*e^(4*x) - 6*e^(2*x) + 1)*e^(-4*x) - 3/8*x + 1/16*e^(2*x)