3.818 \(\int \frac{1}{(\text{sech}^2(x)-\tanh ^2(x))^2} \, dx\)

Optimal. Leaf size=31 \[ x-\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{\sqrt{2}}+\frac{\tanh (x)}{1-2 \tanh ^2(x)} \]

[Out]

x - ArcTanh[Sqrt[2]*Tanh[x]]/Sqrt[2] + Tanh[x]/(1 - 2*Tanh[x]^2)

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Rubi [A]  time = 0.054305, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {414, 12, 481, 206} \[ x-\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{\sqrt{2}}+\frac{\tanh (x)}{1-2 \tanh ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x]^2 - Tanh[x]^2)^(-2),x]

[Out]

x - ArcTanh[Sqrt[2]*Tanh[x]]/Sqrt[2] + Tanh[x]/(1 - 2*Tanh[x]^2)

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 12

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

Rule 481

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> -Dist[(a*e^n)/(b*c -
a*d), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[(c*e^n)/(b*c - a*d), Int[(e*x)^(m - n)/(c + d*x^n), x], x]
/; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

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 (\text{sech}^2(x)-\tanh ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-2 x^2\right )^2 \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac{\tanh (x)}{1-2 \tanh ^2(x)}+\frac{1}{2} \operatorname{Subst}\left (\int -\frac{2 x^2}{\left (1-2 x^2\right ) \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac{\tanh (x)}{1-2 \tanh ^2(x)}-\operatorname{Subst}\left (\int \frac{x^2}{\left (1-2 x^2\right ) \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac{\tanh (x)}{1-2 \tanh ^2(x)}-\operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\tanh (x)\right )+\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (x)\right )\\ &=x-\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{\sqrt{2}}+\frac{\tanh (x)}{1-2 \tanh ^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.156086, size = 42, normalized size = 1.35 \[ \frac{-3 x-\sinh (2 x)+x \cosh (2 x)}{\cosh (2 x)-3}-\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x]^2 - Tanh[x]^2)^(-2),x]

[Out]

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

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Maple [B]  time = 0.046, size = 108, normalized size = 3.5 \begin{align*} \ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -{\frac{1}{2} \left ( 2-2\,\tanh \left ( x/2 \right ) \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) -1 \right ) ^{-1}}-{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) +2 \right ) } \right ) }+{\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) +2 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) -1 \right ) ^{-1}}-{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) -2 \right ) } \right ) }-\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)^2-tanh(x)^2)^2,x)

[Out]

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

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Maxima [B]  time = 1.76692, size = 119, normalized size = 3.84 \begin{align*} -\frac{1}{4} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} + 1}{\sqrt{2} + e^{\left (-x\right )} - 1}\right ) + \frac{1}{4} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} - 1}{\sqrt{2} + e^{\left (-x\right )} + 1}\right ) + x - \frac{2 \,{\left (3 \, e^{\left (-2 \, x\right )} - 1\right )}}{6 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.28395, size = 883, normalized size = 28.48 \begin{align*} \frac{4 \, x \cosh \left (x\right )^{4} + 16 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + 4 \, x \sinh \left (x\right )^{4} - 24 \,{\left (x + 1\right )} \cosh \left (x\right )^{2} + 24 \,{\left (x \cosh \left (x\right )^{2} - x - 1\right )} \sinh \left (x\right )^{2} +{\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 ) + 16 \,{\left (x \cosh \left (x\right )^{3} - 3 \,{\left (x + 1\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 4 \, x + 8}{4 \,{\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/(sech(x)^2-tanh(x)^2)^2,x, algorithm="fricas")

[Out]

1/4*(4*x*cosh(x)^4 + 16*x*cosh(x)*sinh(x)^3 + 4*x*sinh(x)^4 - 24*(x + 1)*cosh(x)^2 + 24*(x*cosh(x)^2 - x - 1)*
sinh(x)^2 + (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)/(cos
h(x)^2 + sinh(x)^2 - 3)) + 16*(x*cosh(x)^3 - 3*(x + 1)*cosh(x))*sinh(x) + 4*x + 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 (- \tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )}\right )^{2} \left (\tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)**2-tanh(x)**2)**2,x)

[Out]

Integral(1/((-tanh(x) + sech(x))**2*(tanh(x) + sech(x))**2), x)

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Giac [B]  time = 1.1434, size = 85, normalized size = 2.74 \begin{align*} \frac{1}{4} \, \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 ) + x - \frac{2 \,{\left (3 \, e^{\left (2 \, x\right )} - 1\right )}}{e^{\left (4 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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