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

Optimal. Leaf size=139 \[ -\frac {1}{3} \sqrt {2} \tanh ^{-1}\left (\frac {1-\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\frac {1}{3} \sqrt [6]{-1} \log \left (-\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )+(-1)^{5/6}+1\right )+\frac {1}{3} \sqrt [6]{-1} \log \left (\sqrt [3]{-1} \tanh \left (\frac {x}{2}\right )+\sqrt [6]{-1}+1\right )-\frac {2 \sqrt [6]{-1} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )+i}{\sqrt {1-\sqrt [3]{-1}}}\right )}{3 \sqrt {1-\sqrt [3]{-1}}} \]

[Out]

-1/3*(-1)^(1/6)*ln(1+(-1)^(5/6)-(-1)^(1/6)*tanh(1/2*x))+1/3*(-1)^(1/6)*ln(1+(-1)^(1/6)+(-1)^(1/3)*tanh(1/2*x))
-1/3*arctanh(1/2*(1-tanh(1/2*x))*2^(1/2))*2^(1/2)-2/3*(-1)^(1/6)*arctan((I+(-1)^(1/6)*tanh(1/2*x))/(1-(-1)^(1/
3))^(1/2))/(1-(-1)^(1/3))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3213, 2660, 618, 204, 617, 206, 616, 31} \[ -\frac {1}{3} \sqrt {2} \tanh ^{-1}\left (\frac {1-\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\frac {1}{3} \sqrt [6]{-1} \log \left (-\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )+(-1)^{5/6}+1\right )+\frac {1}{3} \sqrt [6]{-1} \log \left (\sqrt [3]{-1} \tanh \left (\frac {x}{2}\right )+\sqrt [6]{-1}+1\right )-\frac {2 \sqrt [6]{-1} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )+i}{\sqrt {1-\sqrt [3]{-1}}}\right )}{3 \sqrt {1-\sqrt [3]{-1}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^3)^(-1),x]

[Out]

(-2*(-1)^(1/6)*ArcTan[(I + (-1)^(1/6)*Tanh[x/2])/Sqrt[1 - (-1)^(1/3)]])/(3*Sqrt[1 - (-1)^(1/3)]) - (Sqrt[2]*Ar
cTanh[(1 - Tanh[x/2])/Sqrt[2]])/3 - ((-1)^(1/6)*Log[1 + (-1)^(5/6) - (-1)^(1/6)*Tanh[x/2]])/3 + ((-1)^(1/6)*Lo
g[1 + (-1)^(1/6) + (-1)^(1/3)*Tanh[x/2]])/3

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

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

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 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {1}{1+\sinh ^3(x)} \, dx &=\int \left (\frac {\sqrt [6]{-1}}{3 \left (\sqrt [6]{-1}-i \sinh (x)\right )}+\frac {\sqrt [6]{-1}}{3 \left (\sqrt [6]{-1}+\sqrt [6]{-1} \sinh (x)\right )}+\frac {\sqrt [6]{-1}}{3 \left (\sqrt [6]{-1}+(-1)^{5/6} \sinh (x)\right )}\right ) \, dx\\ &=\frac {1}{3} \sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1}-i \sinh (x)} \, dx+\frac {1}{3} \sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1}+\sqrt [6]{-1} \sinh (x)} \, dx+\frac {1}{3} \sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1}+(-1)^{5/6} \sinh (x)} \, dx\\ &=\frac {1}{3} \left (2 \sqrt [6]{-1}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1}-2 i x-\sqrt [6]{-1} x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{3} \left (2 \sqrt [6]{-1}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1}+2 \sqrt [6]{-1} x-\sqrt [6]{-1} x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{3} \left (2 \sqrt [6]{-1}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1}+2 (-1)^{5/6} x-\sqrt [6]{-1} x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\\ &=-\left (\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,1-\tanh \left (\frac {x}{2}\right )\right )\right )-\frac {1}{3} \left (4 \sqrt [6]{-1}\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (1-\sqrt [3]{-1}\right )-x^2} \, dx,x,-2 i-2 \sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )\right )-\frac {1}{3} \sqrt [3]{-1} \operatorname {Subst}\left (\int \frac {1}{-1+(-1)^{5/6}-\sqrt [6]{-1} x} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{3} \sqrt [3]{-1} \operatorname {Subst}\left (\int \frac {1}{1+(-1)^{5/6}-\sqrt [6]{-1} x} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\\ &=-\frac {2 \sqrt [6]{-1} \tan ^{-1}\left (\frac {i+\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )}{\sqrt {1-\sqrt [3]{-1}}}\right )}{3 \sqrt {1-\sqrt [3]{-1}}}-\frac {1}{3} \sqrt {2} \tanh ^{-1}\left (\frac {1-\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\frac {1}{3} \sqrt [6]{-1} \log \left (1+(-1)^{5/6}-\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )\right )+\frac {1}{3} \sqrt [6]{-1} \log \left (1+\sqrt [6]{-1}+\sqrt [3]{-1} \tanh \left (\frac {x}{2}\right )\right )\\ \end {align*}

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Mathematica [A]  time = 1.50, size = 156, normalized size = 1.12 \[ \frac {2 \tanh ^{-1}\left (\frac {\tanh \left (\frac {x}{2}\right )-1}{\sqrt {2}}\right )+i \sqrt {-1-i \sqrt {3}} \left (\sqrt {3}+i\right ) \tan ^{-1}\left (\frac {2+\left (1-i \sqrt {3}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}}\right )+\left (-1-i \sqrt {3}\right ) \sqrt {-1+i \sqrt {3}} \tan ^{-1}\left (\frac {2+\left (1+i \sqrt {3}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2-2 i \sqrt {3}}}\right )}{3 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^3)^(-1),x]

[Out]

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

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fricas [B]  time = 0.90, size = 185, normalized size = 1.33 \[ -\frac {1}{6} \, \sqrt {3} \log \left (-4 \, {\left (\sqrt {3} + 1\right )} e^{x} + 4 \, \sqrt {3} + 4 \, e^{\left (2 \, x\right )} + 8\right ) + \frac {1}{6} \, \sqrt {3} \log \left (4 \, {\left (\sqrt {3} - 1\right )} e^{x} - 4 \, \sqrt {3} + 4 \, e^{\left (2 \, x\right )} + 8\right ) + \frac {1}{6} \, \sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} - 1\right )} e^{x} + 2 \, \sqrt {2} - e^{\left (2 \, x\right )} - 3}{e^{\left (2 \, x\right )} + 2 \, e^{x} - 1}\right ) + \frac {2}{3} \, \arctan \left (-{\left (\sqrt {3} + 1\right )} e^{x} + \sqrt {{\left (\sqrt {3} - 1\right )} e^{x} - \sqrt {3} + e^{\left (2 \, x\right )} + 2} {\left (\sqrt {3} + 1\right )} - 1\right ) - \frac {2}{3} \, \arctan \left (-{\left (\sqrt {3} - 1\right )} e^{x} + \frac {1}{2} \, \sqrt {-4 \, {\left (\sqrt {3} + 1\right )} e^{x} + 4 \, \sqrt {3} + 4 \, e^{\left (2 \, x\right )} + 8} {\left (\sqrt {3} - 1\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(-4*(sqrt(3) + 1)*e^x + 4*sqrt(3) + 4*e^(2*x) + 8) + 1/6*sqrt(3)*log(4*(sqrt(3) - 1)*e^x - 4*s
qrt(3) + 4*e^(2*x) + 8) + 1/6*sqrt(2)*log(-(2*(sqrt(2) - 1)*e^x + 2*sqrt(2) - e^(2*x) - 3)/(e^(2*x) + 2*e^x -
1)) + 2/3*arctan(-(sqrt(3) + 1)*e^x + sqrt((sqrt(3) - 1)*e^x - sqrt(3) + e^(2*x) + 2)*(sqrt(3) + 1) - 1) - 2/3
*arctan(-(sqrt(3) - 1)*e^x + 1/2*sqrt(-4*(sqrt(3) + 1)*e^x + 4*sqrt(3) + 4*e^(2*x) + 8)*(sqrt(3) - 1) + 1)

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giac [A]  time = 0.13, size = 102, normalized size = 0.73 \[ \frac {1}{6} \, \pi + \frac {1}{6} \, \sqrt {3} \log \left ({\left (\sqrt {3} + e^{x} - 1\right )}^{2} + e^{\left (2 \, x\right )}\right ) - \frac {1}{6} \, \sqrt {3} \log \left ({\left (\sqrt {3} - e^{x} + 1\right )}^{2} + e^{\left (2 \, x\right )}\right ) + \frac {1}{6} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, e^{x} + 2 \right |}}{2 \, {\left (\sqrt {2} + e^{x} + 1\right )}}\right ) + \frac {1}{3} \, \arctan \left (-{\left (\sqrt {3} + 1\right )} e^{x} - 1\right ) + \frac {1}{3} \, \arctan \left ({\left (\sqrt {3} - 1\right )} e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.06, size = 82, normalized size = 0.59 \[ \frac {2 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}+2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )}{\sum }\frac {\left (-\textit {\_R}^{2}-\textit {\_R} +1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3}+3 \textit {\_R}^{2}+2 \textit {\_R} -1}\right )}{3}+\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )-2\right ) \sqrt {2}}{4}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - e^{x} - 1}{\sqrt {2} + e^{x} + 1}\right ) - \int \frac {2 \, {\left (e^{\left (3 \, x\right )} - 4 \, e^{\left (2 \, x\right )} - e^{x}\right )}}{3 \, {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 2 \, e^{x} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.77, size = 203, normalized size = 1.46 \[ \frac {\mathrm {atan}\left (\frac {77824\,{\mathrm {e}}^x-32768\,\sqrt {3}-45056\,\sqrt {3}\,{\mathrm {e}}^x+57344}{77824\,{\mathrm {e}}^x-45056\,\sqrt {3}\,{\mathrm {e}}^x}\right )}{3}-\frac {\mathrm {atan}\left (\frac {77824\,{\mathrm {e}}^x+45056\,\sqrt {3}\,{\mathrm {e}}^x}{77824\,{\mathrm {e}}^x+32768\,\sqrt {3}+45056\,\sqrt {3}\,{\mathrm {e}}^x+57344}\right )}{3}-\frac {\sqrt {2}\,\ln \left (41984\,\sqrt {2}\,{\mathrm {e}}^x-17408\,\sqrt {2}-59392\,{\mathrm {e}}^x+24576\right )}{6}+\frac {\sqrt {2}\,\ln \left (17408\,\sqrt {2}-59392\,{\mathrm {e}}^x-41984\,\sqrt {2}\,{\mathrm {e}}^x+24576\right )}{6}-\frac {\sqrt {3}\,\ln \left ({\left (77824\,{\mathrm {e}}^x-32768\,\sqrt {3}-45056\,\sqrt {3}\,{\mathrm {e}}^x+57344\right )}^2+{\left (77824\,{\mathrm {e}}^x-45056\,\sqrt {3}\,{\mathrm {e}}^x\right )}^2\right )}{6}+\frac {\sqrt {3}\,\ln \left ({\left (77824\,{\mathrm {e}}^x+32768\,\sqrt {3}+45056\,\sqrt {3}\,{\mathrm {e}}^x+57344\right )}^2+{\left (77824\,{\mathrm {e}}^x+45056\,\sqrt {3}\,{\mathrm {e}}^x\right )}^2\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3 + 1),x)

[Out]

atan((77824*exp(x) - 32768*3^(1/2) - 45056*3^(1/2)*exp(x) + 57344)/(77824*exp(x) - 45056*3^(1/2)*exp(x)))/3 -
atan((77824*exp(x) + 45056*3^(1/2)*exp(x))/(77824*exp(x) + 32768*3^(1/2) + 45056*3^(1/2)*exp(x) + 57344))/3 -
(2^(1/2)*log(41984*2^(1/2)*exp(x) - 17408*2^(1/2) - 59392*exp(x) + 24576))/6 + (2^(1/2)*log(17408*2^(1/2) - 59
392*exp(x) - 41984*2^(1/2)*exp(x) + 24576))/6 - (3^(1/2)*log((77824*exp(x) - 32768*3^(1/2) - 45056*3^(1/2)*exp
(x) + 57344)^2 + (77824*exp(x) - 45056*3^(1/2)*exp(x))^2))/6 + (3^(1/2)*log((77824*exp(x) + 32768*3^(1/2) + 45
056*3^(1/2)*exp(x) + 57344)^2 + (77824*exp(x) + 45056*3^(1/2)*exp(x))^2))/6

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

[Out]

Timed out

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