\(\int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx\) [71]

Optimal result

Integrand size = 11, antiderivative size = 38 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3597, 862, 90, 213} \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}+\frac {1}{4 (\tanh (x)+1)}-\frac {1}{8 (\tanh (x)+1)^2} \]

[In]

Int[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

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

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 213

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

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3597

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^2}{(1+x) \left (-1+x^2\right )^2} \, dx,x,\tanh (x)\right ) \\ & = \text {Subst}\left (\int \frac {x^2}{(-1+x)^2 (1+x)^3} \, dx,x,\tanh (x)\right ) \\ & = \text {Subst}\left (\int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}-\frac {1}{4 (1+x)^2}+\frac {1}{8 \left (-1+x^2\right )}\right ) \, dx,x,\tanh (x)\right ) \\ & = \frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (x)\right ) \\ & = -\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{32} (-4 x+4 \cosh (2 x)-\cosh (4 x)+\sinh (4 x)) \]

[In]

Integrate[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {x}{8}+\frac {{\mathrm e}^{2 x}}{16}+\frac {{\mathrm e}^{-2 x}}{16}-\frac {{\mathrm e}^{-4 x}}{32}\)	\(23\)
parallelrisch	\(-\frac {x}{8}+\frac {\sinh \left (4 x \right )}{32}-\frac {\cosh \left (4 x \right )}{32}+\frac {\cosh \left (2 x \right )}{8}-\frac {3}{32}\)	\(24\)
default	\(-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{4 \tanh \left (\frac {x}{2}\right )-4}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}\)	\(68\)

[In]

int(sinh(x)^2/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-1/8*x+1/16*exp(2*x)+1/16*exp(-2*x)-1/32*exp(-4*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 3 \, \sinh \left (x\right )^{3} - 2 \, {\left (2 \, x - 1\right )} \cosh \left (x\right ) + {\left (9 \, \cosh \left (x\right )^{2} - 4 \, x - 2\right )} \sinh \left (x\right )}{32 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\int \frac {\sinh ^{2}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \]

[In]

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

[Out]

Integral(sinh(x)**2/(tanh(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.58 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (-2 \, x\right )} - \frac {1}{32} \, e^{\left (-4 \, x\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{32} \, {\left (3 \, e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-4 \, x\right )} - \frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.58 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}}{16}-\frac {x}{8}+\frac {{\mathrm {e}}^{2\,x}}{16}-\frac {{\mathrm {e}}^{-4\,x}}{32} \]

[In]

int(sinh(x)^2/(tanh(x) + 1),x)

[Out]

exp(-2*x)/16 - x/8 + exp(2*x)/16 - exp(-4*x)/32