3.6.76 \(\int \frac {1+\cosh ^2(x)}{1-\cosh ^2(x)} \, dx\) [576]

Optimal. Leaf size=8 \[ -x+2 \coth (x) \]

[Out]

-x+2*coth(x)

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Rubi [A]
time = 0.03, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3250, 3254, 3852, 8} \begin {gather*} 2 \coth (x)-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Cosh[x]^2)/(1 - Cosh[x]^2),x]

[Out]

-x + 2*Coth[x]

Rule 8

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

Rule 3250

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[B*(x
/b), x] + Dist[(A*b - a*B)/b, Int[1/(a + b*Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {1+\cosh ^2(x)}{1-\cosh ^2(x)} \, dx &=-x+2 \int \frac {1}{1-\cosh ^2(x)} \, dx\\ &=-x-2 \int \text {csch}^2(x) \, dx\\ &=-x+2 i \text {Subst}(\int 1 \, dx,x,-i \coth (x))\\ &=-x+2 \coth (x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 8, normalized size = 1.00 \begin {gather*} -x+2 \coth (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cosh[x]^2)/(1 - Cosh[x]^2),x]

[Out]

-x + 2*Coth[x]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(27\) vs. \(2(8)=16\).
time = 0.75, size = 28, normalized size = 3.50

method	result	size
risch	\(-x +\frac {4}{{\mathrm e}^{2 x}-1}\)	\(15\)
default	\(\tanh \left (\frac {x}{2}\right )-\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\frac {1}{\tanh \left (\frac {x}{2}\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)^2+1)/(1-cosh(x)^2),x,method=_RETURNVERBOSE)

[Out]

tanh(1/2*x)-ln(tanh(1/2*x)+1)+ln(tanh(1/2*x)-1)+1/tanh(1/2*x)

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Maxima [A]
time = 0.26, size = 14, normalized size = 1.75 \begin {gather*} -x - \frac {4}{e^{\left (-2 \, x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cosh(x)^2)/(1-cosh(x)^2),x, algorithm="maxima")

[Out]

-x - 4/(e^(-2*x) - 1)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (8) = 16\).
time = 0.37, size = 17, normalized size = 2.12 \begin {gather*} -\frac {{\left (x + 2\right )} \sinh \left (x\right ) - 2 \, \cosh \left (x\right )}{\sinh \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cosh(x)^2)/(1-cosh(x)^2),x, algorithm="fricas")

[Out]

-((x + 2)*sinh(x) - 2*cosh(x))/sinh(x)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 12 vs. \(2 (5) = 10\).
time = 0.34, size = 12, normalized size = 1.50 \begin {gather*} - x + \tanh {\left (\frac {x}{2} \right )} + \frac {1}{\tanh {\left (\frac {x}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + tanh(x/2) + 1/tanh(x/2)

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Giac [A]
time = 0.40, size = 14, normalized size = 1.75 \begin {gather*} -x + \frac {4}{e^{\left (2 \, x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+cosh(x)^2)/(1-cosh(x)^2),x, algorithm="giac")

[Out]

-x + 4/(e^(2*x) - 1)

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Mupad [B]
time = 0.05, size = 14, normalized size = 1.75 \begin {gather*} \frac {4}{{\mathrm {e}}^{2\,x}-1}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(cosh(x)^2 + 1)/(cosh(x)^2 - 1),x)

[Out]

4/(exp(2*x) - 1) - x

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