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]

1/2*arctan(tanh(x))+1/2*sech(x)^2*tanh(x)/(1+tanh(x)^2)^2

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 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])

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]

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 = 0.01, 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

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 304, normalized size = 11.69 \[ \frac {\cosh \relax (x)^{6} + 20 \, \cosh \relax (x)^{3} \sinh \relax (x)^{3} + 15 \, \cosh \relax (x)^{2} \sinh \relax (x)^{4} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + {\left (15 \, \cosh \relax (x)^{4} - 1\right )} \sinh \relax (x)^{2} - {\left (\cosh \relax (x)^{8} + 56 \, \cosh \relax (x)^{3} \sinh \relax (x)^{5} + 28 \, \cosh \relax (x)^{2} \sinh \relax (x)^{6} + 8 \, \cosh \relax (x) \sinh \relax (x)^{7} + \sinh \relax (x)^{8} + 2 \, {\left (35 \, \cosh \relax (x)^{4} + 1\right )} \sinh \relax (x)^{4} + 2 \, \cosh \relax (x)^{4} + 8 \, {\left (7 \, \cosh \relax (x)^{5} + \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 4 \, {\left (7 \, \cosh \relax (x)^{6} + 3 \, \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 8 \, {\left (\cosh \relax (x)^{7} + \cosh \relax (x)^{3}\right )} \sinh \relax (x) + 1\right )} \arctan \left (-\frac {\cosh \relax (x) + \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) - \cosh \relax (x)^{2} + 2 \, {\left (3 \, \cosh \relax (x)^{5} - \cosh \relax (x)\right )} \sinh \relax (x)}{2 \, {\left (\cosh \relax (x)^{8} + 56 \, \cosh \relax (x)^{3} \sinh \relax (x)^{5} + 28 \, \cosh \relax (x)^{2} \sinh \relax (x)^{6} + 8 \, \cosh \relax (x) \sinh \relax (x)^{7} + \sinh \relax (x)^{8} + 2 \, {\left (35 \, \cosh \relax (x)^{4} + 1\right )} \sinh \relax (x)^{4} + 2 \, \cosh \relax (x)^{4} + 8 \, {\left (7 \, \cosh \relax (x)^{5} + \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 4 \, {\left (7 \, \cosh \relax (x)^{6} + 3 \, \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 8 \, {\left (\cosh \relax (x)^{7} + \cosh \relax (x)^{3}\right )} \sinh \relax (x) + 1\right )}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.13, size = 46, normalized size = 1.77 \[ \frac {e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}}{2 \, {\left ({\left (e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )}^{2} + 4\right )}} + \frac {1}{4} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (4 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )}\right ) \]

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))

________________________________________________________________________________________

maple [B]  time = 0.23, size = 166, normalized size = 6.38 \[ -\frac {2 \left (-\frac {\left (\tanh ^{7}\left (\frac {x}{2}\right )\right )}{2}+\frac {\left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{2}+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{2}-\frac {\tanh \left (\frac {x}{2}\right )}{2}\right )}{\left (\tanh ^{4}\left (\frac {x}{2}\right )+6 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+1\right )^{2}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{-2+2 \sqrt {2}}\right )}{-2+2 \sqrt {2}}-\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{-2+2 \sqrt {2}}\right )}{-2+2 \sqrt {2}}-\frac {\sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{2+2 \sqrt {2}}-\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{2+2 \sqrt {2}} \]

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)))

________________________________________________________________________________________

maxima [B]  time = 0.41, size = 64, normalized size = 2.46 \[ \frac {e^{\left (-2 \, x\right )} - e^{\left (-6 \, x\right )}}{2 \, {\left (2 \, e^{\left (-4 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} + \frac {1}{2} \, \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (-x\right )}\right )}\right ) - \frac {1}{2} \, \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (-x\right )}\right )}\right ) \]

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)))

________________________________________________________________________________________

mupad [B]  time = 1.56, size = 28, normalized size = 1.08 \[ \frac {\mathrm {atan}\left ({\mathrm {e}}^{2\,x}\right )}{2}-\frac {{\mathrm {e}}^{-2\,x}}{4\,{\mathrm {cosh}\left (2\,x\right )}^2}+\frac {1}{4\,\mathrm {cosh}\left (2\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2 + sinh(x)^2)^3,x)

[Out]

atan(exp(2*x))/2 - exp(-2*x)/(4*cosh(2*x)^2) + 1/(4*cosh(2*x))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________