\(\int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx\) [575]

Optimal result

Integrand size = 17, antiderivative size = 8 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-x+2 \tanh (x) \]

[Out]

-x+2*tanh(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3250, 3254, 3852, 8} \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=2 \tanh (x)-x \]

[In]

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

[Out]

-x + 2*Tanh[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*} \text {integral}& = -x+2 \int \frac {1}{1+\sinh ^2(x)} \, dx \\ & = -x+2 \int \text {sech}^2(x) \, dx \\ & = -x+2 i \text {Subst}(\int 1 \, dx,x,-i \tanh (x)) \\ & = -x+2 \tanh (x) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(17\) vs. \(2(8)=16\).

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-\frac {x}{2}-\frac {1}{2} \text {arctanh}(\tanh (x))+2 \tanh (x) \]

[In]

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

[Out]

-1/2*x - ArcTanh[Tanh[x]]/2 + 2*Tanh[x]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-x +2 \tanh \left (x \right )\)	\(9\)
risch	\(-x -\frac {4}{1+{\mathrm e}^{2 x}}\)	\(15\)
default	\(-\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+\frac {4 \tanh \left (\frac {x}{2}\right )}{1+\tanh \left (\frac {x}{2}\right )^{2}}+\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )\)	\(34\)

[In]

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

[Out]

-x+2*tanh(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (8) = 16\).

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-\frac {{\left (x + 2\right )} \cosh \left (x\right ) - 2 \, \sinh \left (x\right )}{\cosh \left (x\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (5) = 10\).

Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 5.12 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=- \frac {x \tanh ^{2}{\left (\frac {x}{2} \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} + 1} - \frac {x}{\tanh ^{2}{\left (\frac {x}{2} \right )} + 1} + \frac {4 \tanh {\left (\frac {x}{2} \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} + 1} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-x + \frac {4}{e^{\left (-2 \, x\right )} + 1} \]

[In]

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

[Out]

-x + 4/(e^(-2*x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-x - \frac {4}{e^{\left (2 \, x\right )} + 1} \]

[In]

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

[Out]

-x - 4/(e^(2*x) + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \frac {1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx=-x-\frac {4}{{\mathrm {e}}^{2\,x}+1} \]

[In]

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

[Out]

- x - 4/(exp(2*x) + 1)