3.93 \(\int \frac{\cosh ^2(x)}{1+\tanh (x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0492305, 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} \[ \frac{3 x}{8}+\frac{1}{8 (1-\tanh (x))}-\frac{1}{4 (\tanh (x)+1)}-\frac{1}{8 (\tanh (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(1 + Tanh[x]),x]

[Out]

(3*x)/8 + 1/(8*(1 - Tanh[x])) - 1/(8*(1 + Tanh[x])^2) - 1/(4*(1 + Tanh[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{\cosh ^2(x)}{1+\tanh (x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{(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{3}{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{3}{8} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{3 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]  time = 0.0299549, 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[Cosh[x]^2/(1 + Tanh[x]),x]

[Out]

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

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Maple [B]  time = 0.032, size = 76, normalized size = 2. \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{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-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(cosh(x)^2/(1+tanh(x)),x)

[Out]

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

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Maxima [A]  time = 1.01783, 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(cosh(x)^2/(1+tanh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.14854, size = 178, normalized size = 4.68 \begin{align*} \frac{\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 3 \, \sinh \left (x\right )^{3} + 6 \,{\left (2 \, x - 1\right )} \cosh \left (x\right ) + 3 \,{\left (3 \, \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(cosh(x)^2/(1+tanh(x)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)**2/(tanh(x) + 1), x)

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Giac [A]  time = 1.20796, 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(cosh(x)^2/(1+tanh(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)