3.810 \(\int \frac{1}{(\cosh ^2(x)+\sinh ^2(x))^3} \, dx\)

Optimal. Leaf size=26 \[ \frac{1}{2} \tan ^{-1}(\tanh (x))+\frac{\tanh (x) \text{sech}^2(x)}{2 \left (\tanh ^2(x)+1\right )^2} \]

[Out]

ArcTan[Tanh[x]]/2 + (Sech[x]^2*Tanh[x])/(2*(1 + Tanh[x]^2)^2)

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Rubi [A]  time = 0.0287418, antiderivative size = 26, 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 = {413, 21, 203} \[ \frac{1}{2} \tan ^{-1}(\tanh (x))+\frac{\tanh (x) \text{sech}^2(x)}{2 \left (\tanh ^2(x)+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[x]^2 + Sinh[x]^2)^(-3),x]

[Out]

ArcTan[Tanh[x]]/2 + (Sech[x]^2*Tanh[x])/(2*(1 + Tanh[x]^2)^2)

Rule 413

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.006351, size = 22, normalized size = 0.85 \[ \frac{1}{4} \tan ^{-1}(\sinh (2 x))+\frac{1}{4} \tanh (2 x) \text{sech}(2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[x]^2 + Sinh[x]^2)^(-3),x]

[Out]

ArcTan[Sinh[2*x]]/4 + (Sech[2*x]*Tanh[2*x])/4

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Maple [B]  time = 0.042, size = 166, normalized size = 6.4 \begin{align*} -2\,{\frac{-1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{7}+1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{5}+1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-1/2\,\tanh \left ( x/2 \right ) }{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+6\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{\sqrt{2}}{2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{2+2\,\sqrt{2}}} \right ) }-{\frac{1}{2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{2+2\,\sqrt{2}}} \right ) }+{\frac{\sqrt{2}}{-2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{-2+2\,\sqrt{2}}} \right ) }-{\frac{1}{-2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{-2+2\,\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2+sinh(x)^2)^3,x)

[Out]

-2*(-1/2*tanh(1/2*x)^7+1/2*tanh(1/2*x)^5+1/2*tanh(1/2*x)^3-1/2*tanh(1/2*x))/(tanh(1/2*x)^4+6*tanh(1/2*x)^2+1)^
2-2^(1/2)/(2+2*2^(1/2))*arctan(2*tanh(1/2*x)/(2+2*2^(1/2)))-1/(2+2*2^(1/2))*arctan(2*tanh(1/2*x)/(2+2*2^(1/2))
)+2^(1/2)/(-2+2*2^(1/2))*arctan(2*tanh(1/2*x)/(-2+2*2^(1/2)))-1/(-2+2*2^(1/2))*arctan(2*tanh(1/2*x)/(-2+2*2^(1
/2)))

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Maxima [B]  time = 1.7561, size = 86, normalized size = 3.31 \begin{align*} \frac{e^{\left (-2 \, x\right )} - e^{\left (-6 \, x\right )}}{2 \,{\left (2 \, e^{\left (-4 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} + \frac{1}{2} \, \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{\left (-x\right )}\right )}\right ) - \frac{1}{2} \, \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{\left (-x\right )}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cosh(x)^2+sinh(x)^2)^3,x, algorithm="maxima")

[Out]

1/2*(e^(-2*x) - e^(-6*x))/(2*e^(-4*x) + e^(-8*x) + 1) + 1/2*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^(-x))) - 1/2*arc
tan(-1/2*sqrt(2)*(sqrt(2) - 2*e^(-x)))

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Fricas [B]  time = 2.24376, size = 1022, normalized size = 39.31 \begin{align*} \frac{\cosh \left (x\right )^{6} + 20 \, \cosh \left (x\right )^{3} \sinh \left (x\right )^{3} + 15 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{4} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} +{\left (15 \, \cosh \left (x\right )^{4} - 1\right )} \sinh \left (x\right )^{2} -{\left (\cosh \left (x\right )^{8} + 56 \, \cosh \left (x\right )^{3} \sinh \left (x\right )^{5} + 28 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{6} + 8 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + \sinh \left (x\right )^{8} + 2 \,{\left (35 \, \cosh \left (x\right )^{4} + 1\right )} \sinh \left (x\right )^{4} + 2 \, \cosh \left (x\right )^{4} + 8 \,{\left (7 \, \cosh \left (x\right )^{5} + \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 4 \,{\left (7 \, \cosh \left (x\right )^{6} + 3 \, \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 8 \,{\left (\cosh \left (x\right )^{7} + \cosh \left (x\right )^{3}\right )} \sinh \left (x\right ) + 1\right )} \arctan \left (-\frac{\cosh \left (x\right ) + \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - \cosh \left (x\right )^{2} + 2 \,{\left (3 \, \cosh \left (x\right )^{5} - \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left (\cosh \left (x\right )^{8} + 56 \, \cosh \left (x\right )^{3} \sinh \left (x\right )^{5} + 28 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{6} + 8 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + \sinh \left (x\right )^{8} + 2 \,{\left (35 \, \cosh \left (x\right )^{4} + 1\right )} \sinh \left (x\right )^{4} + 2 \, \cosh \left (x\right )^{4} + 8 \,{\left (7 \, \cosh \left (x\right )^{5} + \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 4 \,{\left (7 \, \cosh \left (x\right )^{6} + 3 \, \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 8 \,{\left (\cosh \left (x\right )^{7} + \cosh \left (x\right )^{3}\right )} \sinh \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cosh(x)^2+sinh(x)^2)^3,x, algorithm="fricas")

[Out]

1/2*(cosh(x)^6 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(
x)^4 - 1)*sinh(x)^2 - (cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sin
h(x)^8 + 2*(35*cosh(x)^4 + 1)*sinh(x)^4 + 2*cosh(x)^4 + 8*(7*cosh(x)^5 + cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 +
 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 + cosh(x)^3)*sinh(x) + 1)*arctan(-(cosh(x) + sinh(x))/(cosh(x) - sinh(x
))) - cosh(x)^2 + 2*(3*cosh(x)^5 - cosh(x))*sinh(x))/(cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x
)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 + 1)*sinh(x)^4 + 2*cosh(x)^4 + 8*(7*cosh(x)^5 + cosh(x
))*sinh(x)^3 + 4*(7*cosh(x)^6 + 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 + cosh(x)^3)*sinh(x) + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cosh(x)**2+sinh(x)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.16879, size = 62, normalized size = 2.38 \begin{align*} \frac{e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}}{2 \,{\left ({\left (e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )}^{2} + 4\right )}} + \frac{1}{4} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (4 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cosh(x)^2+sinh(x)^2)^3,x, algorithm="giac")

[Out]

1/2*(e^(2*x) - e^(-2*x))/((e^(2*x) - e^(-2*x))^2 + 4) + 1/4*arctan(1/2*(e^(4*x) - 1)*e^(-2*x))