3.64 \(\int \frac{1}{(1-\sinh ^2(x))^2} \, dx\)

Optimal. Leaf size=37 \[ \frac{3 \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{\sinh (x) \cosh (x)}{4 \left (1-\sinh ^2(x)\right )} \]

[Out]

(3*ArcTanh[Sqrt[2]*Tanh[x]])/(4*Sqrt[2]) + (Cosh[x]*Sinh[x])/(4*(1 - Sinh[x]^2))

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Rubi [A]  time = 0.0268138, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3184, 12, 3181, 206} \[ \frac{3 \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{\sinh (x) \cosh (x)}{4 \left (1-\sinh ^2(x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sinh[x]^2)^(-2),x]

[Out]

(3*ArcTanh[Sqrt[2]*Tanh[x]])/(4*Sqrt[2]) + (Cosh[x]*Sinh[x])/(4*(1 - Sinh[x]^2))

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\left (1-\sinh ^2(x)\right )^2} \, dx &=\frac{\cosh (x) \sinh (x)}{4 \left (1-\sinh ^2(x)\right )}-\frac{1}{4} \int -\frac{3}{1-\sinh ^2(x)} \, dx\\ &=\frac{\cosh (x) \sinh (x)}{4 \left (1-\sinh ^2(x)\right )}+\frac{3}{4} \int \frac{1}{1-\sinh ^2(x)} \, dx\\ &=\frac{\cosh (x) \sinh (x)}{4 \left (1-\sinh ^2(x)\right )}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{3 \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{\cosh (x) \sinh (x)}{4 \left (1-\sinh ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.128136, size = 35, normalized size = 0.95 \[ \frac{3 \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}-\frac{\sinh (2 x)}{4 (\cosh (2 x)-3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sinh[x]^2)^(-2),x]

[Out]

(3*ArcTanh[Sqrt[2]*Tanh[x]])/(4*Sqrt[2]) - Sinh[2*x]/(4*(-3 + Cosh[2*x]))

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Maple [B]  time = 0.02, size = 92, normalized size = 2.5 \begin{align*} -{ \left ( -{\frac{1}{4}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{4}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) -1 \right ) ^{-1}}+{\frac{3\,\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) -2 \right ) } \right ) }-{ \left ( -{\frac{1}{4}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{4}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) -1 \right ) ^{-1}}+{\frac{3\,\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) +2 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.54868, size = 117, normalized size = 3.16 \begin{align*} \frac{3}{16} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} + 1}{\sqrt{2} + e^{\left (-x\right )} - 1}\right ) - \frac{3}{16} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} - 1}{\sqrt{2} + e^{\left (-x\right )} + 1}\right ) - \frac{3 \, e^{\left (-2 \, x\right )} - 1}{2 \,{\left (6 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*sqrt(2)*log(-(sqrt(2) - e^(-x) + 1)/(sqrt(2) + e^(-x) - 1)) - 3/16*sqrt(2)*log(-(sqrt(2) - e^(-x) - 1)/(s
qrt(2) + e^(-x) + 1)) - 1/2*(3*e^(-2*x) - 1)/(6*e^(-2*x) - e^(-4*x) - 1)

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Fricas [B]  time = 1.8539, size = 729, normalized size = 19.7 \begin{align*} -\frac{24 \, \cosh \left (x\right )^{2} - 3 \,{\left (\sqrt{2} \cosh \left (x\right )^{4} + 4 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt{2} \sinh \left (x\right )^{4} + 6 \,{\left (\sqrt{2} \cosh \left (x\right )^{2} - \sqrt{2}\right )} \sinh \left (x\right )^{2} - 6 \, \sqrt{2} \cosh \left (x\right )^{2} + 4 \,{\left (\sqrt{2} \cosh \left (x\right )^{3} - 3 \, \sqrt{2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt{2}\right )} \log \left (-\frac{3 \,{\left (2 \, \sqrt{2} - 3\right )} \cosh \left (x\right )^{2} - 4 \,{\left (3 \, \sqrt{2} - 4\right )} \cosh \left (x\right ) \sinh \left (x\right ) + 3 \,{\left (2 \, \sqrt{2} - 3\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt{2} + 3}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} - 3}\right ) + 48 \, \cosh \left (x\right ) \sinh \left (x\right ) + 24 \, \sinh \left (x\right )^{2} - 8}{16 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 6 \,{\left (\cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 6 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - 3 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(24*cosh(x)^2 - 3*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + 6*(sqrt(2)*cosh
(x)^2 - sqrt(2))*sinh(x)^2 - 6*sqrt(2)*cosh(x)^2 + 4*(sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x) + sqrt(2)
)*log(-(3*(2*sqrt(2) - 3)*cosh(x)^2 - 4*(3*sqrt(2) - 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^2 - 2*sqrt
(2) + 3)/(cosh(x)^2 + sinh(x)^2 - 3)) + 48*cosh(x)*sinh(x) + 24*sinh(x)^2 - 8)/(cosh(x)^4 + 4*cosh(x)*sinh(x)^
3 + sinh(x)^4 + 6*(cosh(x)^2 - 1)*sinh(x)^2 - 6*cosh(x)^2 + 4*(cosh(x)^3 - 3*cosh(x))*sinh(x) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.2771, size = 84, normalized size = 2.27 \begin{align*} -\frac{3}{16} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}{{\left | 4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}\right ) - \frac{3 \, e^{\left (2 \, x\right )} - 1}{2 \,{\left (e^{\left (4 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*sqrt(2)*log(abs(-4*sqrt(2) + 2*e^(2*x) - 6)/abs(4*sqrt(2) + 2*e^(2*x) - 6)) - 1/2*(3*e^(2*x) - 1)/(e^(4*
x) - 6*e^(2*x) + 1)