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

Optimal. Leaf size=19 \[ \frac{\tanh ^5(x)}{5}-\frac{2 \tanh ^3(x)}{3}+\tanh (x) \]

[Out]

Tanh[x] - (2*Tanh[x]^3)/3 + Tanh[x]^5/5

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Rubi [A]  time = 0.0201132, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3175, 3767} \[ \frac{\tanh ^5(x)}{5}-\frac{2 \tanh ^3(x)}{3}+\tanh (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^2)^(-3),x]

[Out]

Tanh[x] - (2*Tanh[x]^3)/3 + Tanh[x]^5/5

Rule 3175

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 3767

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*} \int \frac{1}{\left (1+\sinh ^2(x)\right )^3} \, dx &=\int \text{sech}^6(x) \, dx\\ &=i \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \tanh (x)\right )\\ &=\tanh (x)-\frac{2 \tanh ^3(x)}{3}+\frac{\tanh ^5(x)}{5}\\ \end{align*}

Mathematica [A]  time = 0.0029111, size = 27, normalized size = 1.42 \[ \frac{8 \tanh (x)}{15}+\frac{1}{5} \tanh (x) \text{sech}^4(x)+\frac{4}{15} \tanh (x) \text{sech}^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^2)^(-3),x]

[Out]

(8*Tanh[x])/15 + (4*Sech[x]^2*Tanh[x])/15 + (Sech[x]^4*Tanh[x])/5

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Maple [B]  time = 0.019, size = 52, normalized size = 2.7 \begin{align*} -2\,{\frac{1}{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{5}} \left ( - \left ( \tanh \left ( x/2 \right ) \right ) ^{9}-4/3\, \left ( \tanh \left ( x/2 \right ) \right ) ^{7}-{\frac{58\, \left ( \tanh \left ( x/2 \right ) \right ) ^{5}}{15}}-4/3\, \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-\tanh \left ( x/2 \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-tanh(1/2*x)^9-4/3*tanh(1/2*x)^7-58/15*tanh(1/2*x)^5-4/3*tanh(1/2*x)^3-tanh(1/2*x))/(tanh(1/2*x)^2+1)^5

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Maxima [B]  time = 1.04492, size = 150, normalized size = 7.89 \begin{align*} \frac{16 \, e^{\left (-2 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} + \frac{32 \, e^{\left (-4 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} + \frac{16}{15 \,{\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*e^(-2*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) + 32/3*e^(-4*x)/(5*e^(-2*x
) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) + 16/15/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) +
5*e^(-8*x) + e^(-10*x) + 1)

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Fricas [B]  time = 1.78008, size = 628, normalized size = 33.05 \begin{align*} -\frac{16 \,{\left (11 \, \cosh \left (x\right )^{2} + 18 \, \cosh \left (x\right ) \sinh \left (x\right ) + 11 \, \sinh \left (x\right )^{2} + 5\right )}}{15 \,{\left (\cosh \left (x\right )^{8} + 8 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + \sinh \left (x\right )^{8} +{\left (28 \, \cosh \left (x\right )^{2} + 5\right )} \sinh \left (x\right )^{6} + 5 \, \cosh \left (x\right )^{6} + 2 \,{\left (28 \, \cosh \left (x\right )^{3} + 15 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + 5 \,{\left (14 \, \cosh \left (x\right )^{4} + 15 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{4} + 10 \, \cosh \left (x\right )^{4} + 4 \,{\left (14 \, \cosh \left (x\right )^{5} + 25 \, \cosh \left (x\right )^{3} + 10 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} +{\left (28 \, \cosh \left (x\right )^{6} + 75 \, \cosh \left (x\right )^{4} + 60 \, \cosh \left (x\right )^{2} + 11\right )} \sinh \left (x\right )^{2} + 11 \, \cosh \left (x\right )^{2} + 2 \,{\left (4 \, \cosh \left (x\right )^{7} + 15 \, \cosh \left (x\right )^{5} + 20 \, \cosh \left (x\right )^{3} + 9 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*(11*cosh(x)^2 + 18*cosh(x)*sinh(x) + 11*sinh(x)^2 + 5)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (
28*cosh(x)^2 + 5)*sinh(x)^6 + 5*cosh(x)^6 + 2*(28*cosh(x)^3 + 15*cosh(x))*sinh(x)^5 + 5*(14*cosh(x)^4 + 15*cos
h(x)^2 + 2)*sinh(x)^4 + 10*cosh(x)^4 + 4*(14*cosh(x)^5 + 25*cosh(x)^3 + 10*cosh(x))*sinh(x)^3 + (28*cosh(x)^6
+ 75*cosh(x)^4 + 60*cosh(x)^2 + 11)*sinh(x)^2 + 11*cosh(x)^2 + 2*(4*cosh(x)^7 + 15*cosh(x)^5 + 20*cosh(x)^3 +
9*cosh(x))*sinh(x) + 5)

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Sympy [B]  time = 18.1921, size = 260, normalized size = 13.68 \begin{align*} \frac{30 \tanh ^{9}{\left (\frac{x}{2} \right )}}{15 \tanh ^{10}{\left (\frac{x}{2} \right )} + 75 \tanh ^{8}{\left (\frac{x}{2} \right )} + 150 \tanh ^{6}{\left (\frac{x}{2} \right )} + 150 \tanh ^{4}{\left (\frac{x}{2} \right )} + 75 \tanh ^{2}{\left (\frac{x}{2} \right )} + 15} + \frac{40 \tanh ^{7}{\left (\frac{x}{2} \right )}}{15 \tanh ^{10}{\left (\frac{x}{2} \right )} + 75 \tanh ^{8}{\left (\frac{x}{2} \right )} + 150 \tanh ^{6}{\left (\frac{x}{2} \right )} + 150 \tanh ^{4}{\left (\frac{x}{2} \right )} + 75 \tanh ^{2}{\left (\frac{x}{2} \right )} + 15} + \frac{116 \tanh ^{5}{\left (\frac{x}{2} \right )}}{15 \tanh ^{10}{\left (\frac{x}{2} \right )} + 75 \tanh ^{8}{\left (\frac{x}{2} \right )} + 150 \tanh ^{6}{\left (\frac{x}{2} \right )} + 150 \tanh ^{4}{\left (\frac{x}{2} \right )} + 75 \tanh ^{2}{\left (\frac{x}{2} \right )} + 15} + \frac{40 \tanh ^{3}{\left (\frac{x}{2} \right )}}{15 \tanh ^{10}{\left (\frac{x}{2} \right )} + 75 \tanh ^{8}{\left (\frac{x}{2} \right )} + 150 \tanh ^{6}{\left (\frac{x}{2} \right )} + 150 \tanh ^{4}{\left (\frac{x}{2} \right )} + 75 \tanh ^{2}{\left (\frac{x}{2} \right )} + 15} + \frac{30 \tanh{\left (\frac{x}{2} \right )}}{15 \tanh ^{10}{\left (\frac{x}{2} \right )} + 75 \tanh ^{8}{\left (\frac{x}{2} \right )} + 150 \tanh ^{6}{\left (\frac{x}{2} \right )} + 150 \tanh ^{4}{\left (\frac{x}{2} \right )} + 75 \tanh ^{2}{\left (\frac{x}{2} \right )} + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

30*tanh(x/2)**9/(15*tanh(x/2)**10 + 75*tanh(x/2)**8 + 150*tanh(x/2)**6 + 150*tanh(x/2)**4 + 75*tanh(x/2)**2 +
15) + 40*tanh(x/2)**7/(15*tanh(x/2)**10 + 75*tanh(x/2)**8 + 150*tanh(x/2)**6 + 150*tanh(x/2)**4 + 75*tanh(x/2)
**2 + 15) + 116*tanh(x/2)**5/(15*tanh(x/2)**10 + 75*tanh(x/2)**8 + 150*tanh(x/2)**6 + 150*tanh(x/2)**4 + 75*ta
nh(x/2)**2 + 15) + 40*tanh(x/2)**3/(15*tanh(x/2)**10 + 75*tanh(x/2)**8 + 150*tanh(x/2)**6 + 150*tanh(x/2)**4 +
75*tanh(x/2)**2 + 15) + 30*tanh(x/2)/(15*tanh(x/2)**10 + 75*tanh(x/2)**8 + 150*tanh(x/2)**6 + 150*tanh(x/2)**
4 + 75*tanh(x/2)**2 + 15)

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Giac [A]  time = 1.27453, size = 32, normalized size = 1.68 \begin{align*} -\frac{16 \,{\left (10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \,{\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*(10*e^(4*x) + 5*e^(2*x) + 1)/(e^(2*x) + 1)^5