3.266 \(\int \frac{1}{1+\sinh ^4(x)} \, dx\)

Optimal. Leaf size=176 \[ -\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (2 \tanh ^2(x)-2 \sqrt{1+\sqrt{2}} \tanh (x)+\sqrt{2}\right )+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (\sqrt{2} \tanh ^2(x)+\sqrt{2 \left (1+\sqrt{2}\right )} \tanh (x)+1\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{1+\sqrt{2}}-2 \tanh (x)}{\sqrt{\sqrt{2}-1}}\right )}{4 \sqrt{1+\sqrt{2}}}+\frac{\tan ^{-1}\left (\frac{2 \tanh (x)+\sqrt{1+\sqrt{2}}}{\sqrt{\sqrt{2}-1}}\right )}{4 \sqrt{1+\sqrt{2}}} \]

[Out]

-ArcTan[(Sqrt[1 + Sqrt[2]] - 2*Tanh[x])/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[1 + Sqrt[2]]) + ArcTan[(Sqrt[1 + Sqrt[2]]
+ 2*Tanh[x])/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[1 + Sqrt[2]]) - (Sqrt[1 + Sqrt[2]]*Log[Sqrt[2] - 2*Sqrt[1 + Sqrt[2]]*
Tanh[x] + 2*Tanh[x]^2])/8 + (Sqrt[1 + Sqrt[2]]*Log[1 + Sqrt[2*(1 + Sqrt[2])]*Tanh[x] + Sqrt[2]*Tanh[x]^2])/8

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Rubi [A]  time = 0.159089, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3209, 1169, 634, 618, 204, 628} \[ -\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (2 \tanh ^2(x)-2 \sqrt{1+\sqrt{2}} \tanh (x)+\sqrt{2}\right )+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (\sqrt{2} \tanh ^2(x)+\sqrt{2 \left (1+\sqrt{2}\right )} \tanh (x)+1\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{1+\sqrt{2}}-2 \tanh (x)}{\sqrt{\sqrt{2}-1}}\right )}{4 \sqrt{1+\sqrt{2}}}+\frac{\tan ^{-1}\left (\frac{2 \tanh (x)+\sqrt{1+\sqrt{2}}}{\sqrt{\sqrt{2}-1}}\right )}{4 \sqrt{1+\sqrt{2}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sinh[x]^4)^(-1),x]

[Out]

-ArcTan[(Sqrt[1 + Sqrt[2]] - 2*Tanh[x])/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[1 + Sqrt[2]]) + ArcTan[(Sqrt[1 + Sqrt[2]]
+ 2*Tanh[x])/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[1 + Sqrt[2]]) - (Sqrt[1 + Sqrt[2]]*Log[Sqrt[2] - 2*Sqrt[1 + Sqrt[2]]*
Tanh[x] + 2*Tanh[x]^2])/8 + (Sqrt[1 + Sqrt[2]]*Log[1 + Sqrt[2*(1 + Sqrt[2])]*Tanh[x] + Sqrt[2]*Tanh[x]^2])/8

Rule 3209

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rule 1169

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

Rule 634

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

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 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{1+\sinh ^4(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1-x^2}{1-2 x^2+2 x^4} \, dx,x,\tanh (x)\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1+\sqrt{2}}-\left (1+\frac{1}{\sqrt{2}}\right ) x}{\frac{1}{\sqrt{2}}-\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1+\sqrt{2}}+\left (1+\frac{1}{\sqrt{2}}\right ) x}{\frac{1}{\sqrt{2}}+\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}\\ &=\frac{1}{8} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{\sqrt{2}}-\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )+\frac{1}{8} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{\sqrt{2}}+\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )-\frac{1}{8} \sqrt{1+\sqrt{2}} \operatorname{Subst}\left (\int \frac{-\sqrt{1+\sqrt{2}}+2 x}{\frac{1}{\sqrt{2}}-\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )+\frac{1}{8} \sqrt{1+\sqrt{2}} \operatorname{Subst}\left (\int \frac{\sqrt{1+\sqrt{2}}+2 x}{\frac{1}{\sqrt{2}}+\sqrt{1+\sqrt{2}} x+x^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (\sqrt{2}-2 \sqrt{1+\sqrt{2}} \tanh (x)+2 \tanh ^2(x)\right )+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (1+\sqrt{2 \left (1+\sqrt{2}\right )} \tanh (x)+\sqrt{2} \tanh ^2(x)\right )-\frac{1}{4} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2}-x^2} \, dx,x,-\sqrt{1+\sqrt{2}}+2 \tanh (x)\right )-\frac{1}{4} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2}-x^2} \, dx,x,\sqrt{1+\sqrt{2}}+2 \tanh (x)\right )\\ &=-\frac{1}{4} \sqrt{-1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{1+\sqrt{2}}-2 \tanh (x)}{\sqrt{-1+\sqrt{2}}}\right )+\frac{1}{4} \sqrt{-1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{1+\sqrt{2}}+2 \tanh (x)}{\sqrt{-1+\sqrt{2}}}\right )-\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (\sqrt{2}-2 \sqrt{1+\sqrt{2}} \tanh (x)+2 \tanh ^2(x)\right )+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left (1+\sqrt{2 \left (1+\sqrt{2}\right )} \tanh (x)+\sqrt{2} \tanh ^2(x)\right )\\ \end{align*}

Mathematica [C]  time = 0.072179, size = 45, normalized size = 0.26 \[ \frac{\tanh ^{-1}\left (\sqrt{1-i} \tanh (x)\right )}{2 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\sqrt{1+i} \tanh (x)\right )}{2 \sqrt{1+i}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sinh[x]^4)^(-1),x]

[Out]

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

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Maple [C]  time = 0.024, size = 44, normalized size = 0.3 \begin{align*}{\frac{1}{4}\sum _{{\it \_R}={\it RootOf} \left ( 2\,{{\it \_Z}}^{4}-2\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+ \left ( -4\,{{\it \_R}}^{3}+4\,{\it \_R} \right ) \tanh \left ({\frac{x}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sinh(x)^4),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sinh \left (x\right )^{4} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sinh(x)^4),x, algorithm="maxima")

[Out]

integrate(1/(sinh(x)^4 + 1), x)

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Fricas [B]  time = 2.257, size = 1874, normalized size = 10.65 \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)^4),x, algorithm="fricas")

[Out]

1/16*2^(1/4)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4)*log((2^(3/4)*e^(2*x) - 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2
) + 4) + 4*sqrt(2) + e^(4*x) - 2*e^(2*x) + 5) - 1/16*2^(1/4)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4)*log(-(2^(3/4)*
e^(2*x) - 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sqrt(2) + e^(4*x) - 2*e^(2*x) + 5) - 1/4*2^(1/4)*s
qrt(-2*sqrt(2) + 4)*arctan(1/14*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*e^(2*x) - 1/28*(2*sqrt(2)*(5*sqrt(2)
+ 6) - (2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*sqrt(-2*sqrt(2) + 4) + 16*sqrt(2) + 8)*sqrt(-(2
^(3/4)*e^(2*x) - 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sqrt(2) + e^(4*x) - 2*e^(2*x) + 5) - 1/14*s
qrt(2)*(3*sqrt(2) - 2) - 1/28*((2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*e^(2*x) - 2^(3/4)*(2*sqr
t(2) + 1) - 2*2^(1/4)*(3*sqrt(2) - 2))*sqrt(-2*sqrt(2) + 4) - 1/7*sqrt(2) + 3/7) - 1/4*2^(1/4)*sqrt(-2*sqrt(2)
+ 4)*arctan(-1/14*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*e^(2*x) + 1/28*(2*sqrt(2)*(5*sqrt(2) + 6) + (2^(3
/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*sqrt(-2*sqrt(2) + 4) + 16*sqrt(2) + 8)*sqrt((2^(3/4)*e^(2*x)
- 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sqrt(2) + e^(4*x) - 2*e^(2*x) + 5) + 1/14*sqrt(2)*(3*sqrt
(2) - 2) - 1/28*((2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*e^(2*x) - 2^(3/4)*(2*sqrt(2) + 1) - 2*
2^(1/4)*(3*sqrt(2) - 2))*sqrt(-2*sqrt(2) + 4) + 1/7*sqrt(2) - 3/7)

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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)**4),x)

[Out]

Timed out

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Giac [C]  time = 1.24228, size = 293, normalized size = 1.66 \begin{align*} \left (\frac{1}{16} i + \frac{1}{16}\right ) \, \sqrt{2 \, \sqrt{2} - 2}{\left (-\frac{i}{\sqrt{2} - 1} + 1\right )} \log \left (2 \, \sqrt{10 \, \sqrt{2} + 14}{\left (-\frac{i}{5 \, \sqrt{2} + 7} + 1\right )} + \left (4 i + 2\right ) \, e^{\left (2 \, x\right )} - 10\right ) - \left (\frac{1}{16} i + \frac{1}{16}\right ) \, \sqrt{2 \, \sqrt{2} - 2}{\left (-\frac{i}{\sqrt{2} - 1} + 1\right )} \log \left (-2 \, \sqrt{10 \, \sqrt{2} + 14}{\left (-\frac{i}{5 \, \sqrt{2} + 7} + 1\right )} + \left (4 i + 2\right ) \, e^{\left (2 \, x\right )} - 10\right ) + \left (\frac{1}{16} i + \frac{1}{16}\right ) \, \sqrt{2 \, \sqrt{2} + 2}{\left (-\frac{i}{\sqrt{2} + 1} + 1\right )} \log \left (2 \, \sqrt{2 \, \sqrt{2} - 2}{\left (\frac{i}{\sqrt{2} - 1} + 1\right )} + 2 \, e^{\left (2 \, x\right )} - 4 i - 2\right ) - \left (\frac{1}{16} i + \frac{1}{16}\right ) \, \sqrt{2 \, \sqrt{2} + 2}{\left (-\frac{i}{\sqrt{2} + 1} + 1\right )} \log \left (-2 \, \sqrt{2 \, \sqrt{2} - 2}{\left (\frac{i}{\sqrt{2} - 1} + 1\right )} + 2 \, e^{\left (2 \, x\right )} - 4 i - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sinh(x)^4),x, algorithm="giac")

[Out]

(1/16*I + 1/16)*sqrt(2*sqrt(2) - 2)*(-I/(sqrt(2) - 1) + 1)*log(2*sqrt(10*sqrt(2) + 14)*(-I/(5*sqrt(2) + 7) + 1
) + (4*I + 2)*e^(2*x) - 10) - (1/16*I + 1/16)*sqrt(2*sqrt(2) - 2)*(-I/(sqrt(2) - 1) + 1)*log(-2*sqrt(10*sqrt(2
) + 14)*(-I/(5*sqrt(2) + 7) + 1) + (4*I + 2)*e^(2*x) - 10) + (1/16*I + 1/16)*sqrt(2*sqrt(2) + 2)*(-I/(sqrt(2)
+ 1) + 1)*log(2*sqrt(2*sqrt(2) - 2)*(I/(sqrt(2) - 1) + 1) + 2*e^(2*x) - 4*I - 2) - (1/16*I + 1/16)*sqrt(2*sqrt
(2) + 2)*(-I/(sqrt(2) + 1) + 1)*log(-2*sqrt(2*sqrt(2) - 2)*(I/(sqrt(2) - 1) + 1) + 2*e^(2*x) - 4*I - 2)