3.271 \(\int \frac{1}{1+\sinh ^6(x)} \, dx\)

Optimal. Leaf size=71 \[ \frac{\tanh ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tanh (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tanh ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tanh (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}+\frac{\tanh (x)}{3} \]

[Out]

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

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Rubi [A]  time = 0.137859, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3211, 3181, 206, 3175, 3767, 8} \[ \frac{\tanh ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tanh (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tanh ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tanh (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}+\frac{\tanh (x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^6)^(-1),x]

[Out]

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

Rule 3211

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{k}, Dist[2/(a*n), Sum[Int[1/(1 - Si
n[e + f*x]^2/((-1)^((4*k)/n)*Rt[-(a/b), n/2])), x], {k, 1, n/2}], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[n/
2]

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])

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.201319, size = 87, normalized size = 1.23 \[ \frac{1}{18} \left (6 \tanh (x)+\sqrt [4]{-3} \left (\left (-3-i \sqrt{3}\right ) \tan ^{-1}\left (\frac{1}{2} \sqrt [4]{-3} \left (1+i \sqrt{3}\right ) \tanh (x)\right )-\left (\sqrt{3}+3 i\right ) \tan ^{-1}\left (\frac{1}{2} \sqrt [4]{-\frac{1}{3}} \left (3+i \sqrt{3}\right ) \tanh (x)\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^6)^(-1),x]

[Out]

((-3)^(1/4)*((-3 - I*Sqrt[3])*ArcTan[((-3)^(1/4)*(1 + I*Sqrt[3])*Tanh[x])/2] - (3*I + Sqrt[3])*ArcTan[((-1/3)^
(1/4)*(3 + I*Sqrt[3])*Tanh[x])/2]) + 6*Tanh[x])/18

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Maple [C]  time = 0.033, size = 61, normalized size = 0.9 \begin{align*}{\frac{1}{6}\sum _{{\it \_R}={\it RootOf} \left ( 3\,{{\it \_Z}}^{4}-3\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+ \left ( -6\,{{\it \_R}}^{3}+6\,{\it \_R} \right ) \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{2}{3}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sinh(x)^6),x)

[Out]

1/6*sum(_R*ln(tanh(1/2*x)^2+(-6*_R^3+6*_R)*tanh(1/2*x)+1),_R=RootOf(3*_Z^4-3*_Z^2+1))+2/3*tanh(1/2*x)/(tanh(1/
2*x)^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2}{3 \,{\left (e^{\left (2 \, x\right )} + 1\right )}} - \int \frac{4 \,{\left (e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}\right )}}{3 \,{\left (e^{\left (8 \, x\right )} - 8 \, e^{\left (6 \, x\right )} + 30 \, e^{\left (4 \, x\right )} - 8 \, e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sinh(x)^6),x, algorithm="maxima")

[Out]

-2/3/(e^(2*x) + 1) - integrate(4/3*(e^(6*x) - 10*e^(4*x) + e^(2*x))/(e^(8*x) - 8*e^(6*x) + 30*e^(4*x) - 8*e^(2
*x) + 1), x)

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Fricas [B]  time = 1.84705, size = 2303, normalized size = 32.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sinh(x)^6),x, algorithm="fricas")

[Out]

-1/144*(4*(12^(1/4)*sqrt(6)*e^(2*x) + 12^(1/4)*sqrt(6))*sqrt(-4*sqrt(3) + 8)*arctan((sqrt(3) + 2)*e^(2*x) + 1/
216*sqrt(-6*(12^(1/4)*sqrt(6)*(sqrt(3) + 3)*e^(2*x) - 12^(1/4)*sqrt(6)*(5*sqrt(3) + 9))*sqrt(-4*sqrt(3) + 8) +
144*sqrt(3) + 36*e^(4*x) - 144*e^(2*x) + 252)*((12^(3/4)*sqrt(6)*(sqrt(3) + 3) + 3*12^(1/4)*sqrt(6)*(sqrt(3)
+ 3))*sqrt(-4*sqrt(3) + 8) - 36*sqrt(3) - 72) - 2/3*sqrt(3)*(2*sqrt(3) - 3) - 1/36*(12^(3/4)*sqrt(6)*(sqrt(3)
- 3) + (12^(3/4)*sqrt(6)*(sqrt(3) + 3) + 3*12^(1/4)*sqrt(6)*(sqrt(3) + 3))*e^(2*x) + 3*12^(1/4)*sqrt(6)*(sqrt(
3) - 3))*sqrt(-4*sqrt(3) + 8) - 2*sqrt(3) + 4) + 4*(12^(1/4)*sqrt(6)*e^(2*x) + 12^(1/4)*sqrt(6))*sqrt(-4*sqrt(
3) + 8)*arctan(-(sqrt(3) + 2)*e^(2*x) + 1/216*sqrt(6*(12^(1/4)*sqrt(6)*(sqrt(3) + 3)*e^(2*x) - 12^(1/4)*sqrt(6
)*(5*sqrt(3) + 9))*sqrt(-4*sqrt(3) + 8) + 144*sqrt(3) + 36*e^(4*x) - 144*e^(2*x) + 252)*((12^(3/4)*sqrt(6)*(sq
rt(3) + 3) + 3*12^(1/4)*sqrt(6)*(sqrt(3) + 3))*sqrt(-4*sqrt(3) + 8) + 36*sqrt(3) + 72) + 2/3*sqrt(3)*(2*sqrt(3
) - 3) - 1/36*(12^(3/4)*sqrt(6)*(sqrt(3) - 3) + (12^(3/4)*sqrt(6)*(sqrt(3) + 3) + 3*12^(1/4)*sqrt(6)*(sqrt(3)
+ 3))*e^(2*x) + 3*12^(1/4)*sqrt(6)*(sqrt(3) - 3))*sqrt(-4*sqrt(3) + 8) + 2*sqrt(3) - 4) - (12^(1/4)*sqrt(6)*(s
qrt(3) + 2)*e^(2*x) + 12^(1/4)*sqrt(6)*(sqrt(3) + 2))*sqrt(-4*sqrt(3) + 8)*log(6*(12^(1/4)*sqrt(6)*(sqrt(3) +
3)*e^(2*x) - 12^(1/4)*sqrt(6)*(5*sqrt(3) + 9))*sqrt(-4*sqrt(3) + 8) + 144*sqrt(3) + 36*e^(4*x) - 144*e^(2*x) +
252) + (12^(1/4)*sqrt(6)*(sqrt(3) + 2)*e^(2*x) + 12^(1/4)*sqrt(6)*(sqrt(3) + 2))*sqrt(-4*sqrt(3) + 8)*log(-6*
(12^(1/4)*sqrt(6)*(sqrt(3) + 3)*e^(2*x) - 12^(1/4)*sqrt(6)*(5*sqrt(3) + 9))*sqrt(-4*sqrt(3) + 8) + 144*sqrt(3)
+ 36*e^(4*x) - 144*e^(2*x) + 252) + 96)/(e^(2*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/(1+sinh(x)**6),x)

[Out]

Timed out

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Giac [A]  time = 1.27431, size = 14, normalized size = 0.2 \begin{align*} -\frac{2}{3 \,{\left (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)^6),x, algorithm="giac")

[Out]

-2/3/(e^(2*x) + 1)