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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3516, 848, 88, 207} \[ -\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}+\frac {1}{4 (\tanh (x)+1)}-\frac {1}{8 (\tanh (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

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

Rule 88

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 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])

Rule 848

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*x)/e)^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 3516

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*} \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx &=\operatorname {Subst}\left (\int \frac {x^2}{(1+x) \left (-1+x^2\right )^2} \, dx,x,\tanh (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {x^2}{(-1+x)^2 (1+x)^3} \, dx,x,\tanh (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 {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} \operatorname {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*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 0.63 \[ \frac {1}{32} (-4 x+\sinh (4 x)+4 \cosh (2 x)-\cosh (4 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

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

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fricas [A]  time = 0.56, size = 51, normalized size = 1.34 \[ \frac {\cosh \relax (x)^{3} + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + 3 \, \sinh \relax (x)^{3} - 2 \, {\left (2 \, x - 1\right )} \cosh \relax (x) + {\left (9 \, \cosh \relax (x)^{2} - 4 \, x - 2\right )} \sinh \relax (x)}{32 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.13, size = 30, normalized size = 0.79 \[ \frac {1}{32} \, {\left (3 \, e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-4 \, x\right )} - \frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.08, size = 68, normalized size = 1.79 \[ \frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{4 \tanh \left (\frac {x}{2}\right )-4}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 22, normalized size = 0.58 \[ -\frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (-2 \, x\right )} - \frac {1}{32} \, e^{\left (-4 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.14, size = 22, normalized size = 0.58 \[ \frac {{\mathrm {e}}^{-2\,x}}{16}-\frac {x}{8}+\frac {{\mathrm {e}}^{2\,x}}{16}-\frac {{\mathrm {e}}^{-4\,x}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(tanh(x) + 1),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\relax (x )}}{\tanh {\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(x)**2/(tanh(x) + 1), x)

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