\(\int \frac {1}{1+\sinh ^3(x)} \, dx\) [13]

Optimal result

Integrand size = 8, antiderivative size = 139 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=-\frac {2 \sqrt [6]{-1} \arctan \left (\frac {i+\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )}{\sqrt {1-\sqrt [3]{-1}}}\right )}{3 \sqrt {1-\sqrt [3]{-1}}}-\frac {1}{3} \sqrt {2} \text {arctanh}\left (\frac {1-\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\frac {1}{3} \sqrt [6]{-1} \log \left (1+(-1)^{5/6}-\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )\right )+\frac {1}{3} \sqrt [6]{-1} \log \left (1+\sqrt [6]{-1}+\sqrt [3]{-1} \tanh \left (\frac {x}{2}\right )\right ) \] Output:

-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)-1/3*2^(1/2)*arctanh(1/2*(1-tanh(1/2*x))*2^(1/2))-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))
 

Mathematica [A] (verified)

Time = 5.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.12 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\frac {i \sqrt {-1-i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {2+\left (1-i \sqrt {3}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}}\right )+\left (-1-i \sqrt {3}\right ) \sqrt {-1+i \sqrt {3}} \arctan \left (\frac {2+\left (1+i \sqrt {3}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2-2 i \sqrt {3}}}\right )+2 \text {arctanh}\left (\frac {-1+\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )}{3 \sqrt {2}} \] Input:

Integrate[(1 + Sinh[x]^3)^(-1),x]
 

Output:

(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]]*ArcTa 
n[(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])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3692, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{1+i \sin (i x)^3}dx\)

\(\Big \downarrow \) 3692

\(\displaystyle \int \left (\frac {\sqrt [6]{-1}}{3 \left (\sqrt [6]{-1} \sinh (x)+\sqrt [6]{-1}\right )}+\frac {\sqrt [6]{-1}}{3 \left ((-1)^{5/6} \sinh (x)+\sqrt [6]{-1}\right )}+\frac {\sqrt [6]{-1}}{3 \left (\sqrt [6]{-1}-i \sinh (x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt [6]{-1} \arctan \left (\frac {\sqrt [6]{-1} \tanh \left (\frac {x}{2}\right )+i}{\sqrt {1-\sqrt [3]{-1}}}\right )}{3 \sqrt {1-\sqrt [3]{-1}}}-\frac {1}{3} \sqrt {2} \text {arctanh}\left (\frac {1-\tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+\frac {1}{3} \sqrt [6]{-1} \log \left (\sqrt [3]{-1} \tanh \left (\frac {x}{2}\right )+\sqrt [6]{-1}+1\right )-\frac {1}{3} \sqrt [6]{-1} \log \left (-\left (\left (\sqrt {3}+i\right ) \tanh \left (\frac {x}{2}\right )\right )-\sqrt {3}+(2+i)\right )\)

Input:

Int[(1 + Sinh[x]^3)^(-1),x]
 

Output:

(-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]*ArcTanh[(1 - Tanh[x/2])/Sqrt[2]])/3 + (( 
-1)^(1/6)*Log[1 + (-1)^(1/6) + (-1)^(1/3)*Tanh[x/2]])/3 - ((-1)^(1/6)*Log[ 
(2 + I) - Sqrt[3] - (I + Sqrt[3])*Tanh[x/2]])/3
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3692

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]))
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.36 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{x}+1-\sqrt {2}\right )}{6}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{x}+1+\sqrt {2}\right )}{6}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (81 \textit {\_Z}^{4}-9 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-9 \textit {\_R}^{2}+3 \textit {\_R} +{\mathrm e}^{x}\right )\right )\)	\(60\)
default	\(\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}+2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )}{\sum }\frac {\left (-\textit {\_R}^{2}-\textit {\_R} +1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3}+3 \textit {\_R}^{2}+2 \textit {\_R} -1}\right )}{3}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )-2\right ) \sqrt {2}}{4}\right )}{3}\)	\(82\)

Input:

int(1/(1+sinh(x)^3),x,method=_RETURNVERBOSE)
 

Output:

1/6*2^(1/2)*ln(exp(x)+1-2^(1/2))-1/6*2^(1/2)*ln(exp(x)+1+2^(1/2))+sum(_R*l 
n(-9*_R^2+3*_R+exp(x)),_R=RootOf(81*_Z^4-9*_Z^2+1))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.15 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=-\frac {1}{6} \, \sqrt {3} \log \left (\frac {{\left (\sqrt {3} + 3\right )} \cosh \left (x\right ) - {\left (\sqrt {3} + 1\right )} \sinh \left (x\right ) - \sqrt {3} - 1}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\frac {{\left (\sqrt {3} - 3\right )} \cosh \left (x\right ) - {\left (\sqrt {3} - 1\right )} \sinh \left (x\right ) - \sqrt {3} + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \frac {1}{6} \, \sqrt {2} \log \left (-\frac {{\left (\sqrt {2} - 2\right )} \cosh \left (x\right ) - {\left (\sqrt {2} - 1\right )} \sinh \left (x\right ) + \sqrt {2} - 1}{\sinh \left (x\right ) + 1}\right ) - \frac {1}{3} \, \arctan \left ({\left (\sqrt {3} + 1\right )} \cosh \left (x\right ) + {\left (\sqrt {3} + 1\right )} \sinh \left (x\right ) + 1\right ) + \frac {1}{3} \, \arctan \left ({\left (\sqrt {3} - 1\right )} \cosh \left (x\right ) + {\left (\sqrt {3} - 1\right )} \sinh \left (x\right ) - 1\right ) \] Input:

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

Output:

-1/6*sqrt(3)*log(((sqrt(3) + 3)*cosh(x) - (sqrt(3) + 1)*sinh(x) - sqrt(3) 
- 1)/(cosh(x) - sinh(x))) + 1/6*sqrt(3)*log(-((sqrt(3) - 3)*cosh(x) - (sqr 
t(3) - 1)*sinh(x) - sqrt(3) + 1)/(cosh(x) - sinh(x))) + 1/6*sqrt(2)*log(-( 
(sqrt(2) - 2)*cosh(x) - (sqrt(2) - 1)*sinh(x) + sqrt(2) - 1)/(sinh(x) + 1) 
) - 1/3*arctan((sqrt(3) + 1)*cosh(x) + (sqrt(3) + 1)*sinh(x) + 1) + 1/3*ar 
ctan((sqrt(3) - 1)*cosh(x) + (sqrt(3) - 1)*sinh(x) - 1)
 

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5697 vs. \(2 (133) = 266\).

Time = 35.32 (sec) , antiderivative size = 5697, normalized size of antiderivative = 40.99 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-2730935734518397297302171298494629987458472657311991598275400*sqrt(3)*I*s 
qrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + sqrt(2))/(5676143333269554149506871 
1547884217702675164237235096719281544 + 4013639441941715127028842203182095 
0068139251222039410549591998*sqrt(2) - 11586379061175742792768711092190588 
955898140233336817863353300*sqrt(3)*I*sqrt(1 + sqrt(3)*I) - 81928072035551 
91891906513895483889962375417971935974794826200*sqrt(6)*I*sqrt(1 + sqrt(3) 
*I) + 34759137183527228378306133276571766867694420700010453590059900*sqrt( 
1 + sqrt(3)*I) + 245784216106655756757195416864516698871262539158079243844 
78600*sqrt(2)*sqrt(1 + sqrt(3)*I)) - 1931063176862623798794785182031764825 
983023372222802977225550*sqrt(6)*I*sqrt(1 + sqrt(3)*I)*log(tanh(x/2) - 1 + 
 sqrt(2))/(56761433332695541495068711547884217702675164237235096719281544 
+ 40136394419417151270288422031820950068139251222039410549591998*sqrt(2) - 
 11586379061175742792768711092190588955898140233336817863353300*sqrt(3)*I* 
sqrt(1 + sqrt(3)*I) - 8192807203555191891906513895483889962375417971935974 
794826200*sqrt(6)*I*sqrt(1 + sqrt(3)*I) + 34759137183527228378306133276571 
766867694420700010453590059900*sqrt(1 + sqrt(3)*I) + 245784216106655756757 
19541686451669887126253915807924384478600*sqrt(2)*sqrt(1 + sqrt(3)*I)) + 1 
3378798139805717090096140677273650022713083740679803516530666*log(tanh(x/2 
) - 1 + sqrt(2))/(56761433332695541495068711547884217702675164237235096719 
281544 + 40136394419417151270288422031820950068139251222039410549591998...
 

Maxima [F]

\[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\int { \frac {1}{\sinh \left (x\right )^{3} + 1} \,d x } \] Input:

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

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\frac {1}{6} \, \pi + \frac {1}{6} \, \sqrt {3} \log \left ({\left (\sqrt {3} + e^{x} - 1\right )}^{2} + e^{\left (2 \, x\right )}\right ) - \frac {1}{6} \, \sqrt {3} \log \left ({\left (\sqrt {3} - e^{x} + 1\right )}^{2} + e^{\left (2 \, x\right )}\right ) + \frac {1}{6} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, e^{x} + 2 \right |}}{2 \, {\left (\sqrt {2} + e^{x} + 1\right )}}\right ) + \frac {1}{3} \, \arctan \left (-{\left (\sqrt {3} + 1\right )} e^{x} - 1\right ) + \frac {1}{3} \, \arctan \left ({\left (\sqrt {3} - 1\right )} e^{x} - 1\right ) \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.46 \[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\frac {\mathrm {atan}\left (\frac {77824\,{\mathrm {e}}^x-32768\,\sqrt {3}-45056\,\sqrt {3}\,{\mathrm {e}}^x+57344}{77824\,{\mathrm {e}}^x-45056\,\sqrt {3}\,{\mathrm {e}}^x}\right )}{3}-\frac {\mathrm {atan}\left (\frac {77824\,{\mathrm {e}}^x+45056\,\sqrt {3}\,{\mathrm {e}}^x}{77824\,{\mathrm {e}}^x+32768\,\sqrt {3}+45056\,\sqrt {3}\,{\mathrm {e}}^x+57344}\right )}{3}-\frac {\sqrt {2}\,\ln \left (41984\,\sqrt {2}\,{\mathrm {e}}^x-17408\,\sqrt {2}-59392\,{\mathrm {e}}^x+24576\right )}{6}+\frac {\sqrt {2}\,\ln \left (17408\,\sqrt {2}-59392\,{\mathrm {e}}^x-41984\,\sqrt {2}\,{\mathrm {e}}^x+24576\right )}{6}-\frac {\sqrt {3}\,\ln \left ({\left (77824\,{\mathrm {e}}^x-32768\,\sqrt {3}-45056\,\sqrt {3}\,{\mathrm {e}}^x+57344\right )}^2+{\left (77824\,{\mathrm {e}}^x-45056\,\sqrt {3}\,{\mathrm {e}}^x\right )}^2\right )}{6}+\frac {\sqrt {3}\,\ln \left ({\left (77824\,{\mathrm {e}}^x+32768\,\sqrt {3}+45056\,\sqrt {3}\,{\mathrm {e}}^x+57344\right )}^2+{\left (77824\,{\mathrm {e}}^x+45056\,\sqrt {3}\,{\mathrm {e}}^x\right )}^2\right )}{6} \] Input:

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

Output:

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) - 59392*exp(x) - 41984*2^(1/2)*exp(x) + 2457 
6))/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((7 
7824*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
 

Reduce [F]

\[ \int \frac {1}{1+\sinh ^3(x)} \, dx=\sqrt {2}\, \mathrm {log}\left (e^{x}-\sqrt {2}+1\right )-\sqrt {2}\, \mathrm {log}\left (e^{x}+\sqrt {2}+1\right )-16 \left (\int \frac {e^{2 x}}{e^{6 x}-3 e^{4 x}+8 e^{3 x}+3 e^{2 x}-1}d x \right )-8 \left (\int \frac {e^{x}}{e^{6 x}-3 e^{4 x}+8 e^{3 x}+3 e^{2 x}-1}d x \right )-\frac {2 \,\mathrm {log}\left (e^{4 x}-2 e^{3 x}+2 e^{2 x}+2 e^{x}+1\right )}{3}+\frac {4 \,\mathrm {log}\left (e^{x}-\sqrt {2}+1\right )}{3}+\frac {4 \,\mathrm {log}\left (e^{x}+\sqrt {2}+1\right )}{3} \] Input:

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

Output:

(3*sqrt(2)*log(e**x - sqrt(2) + 1) - 3*sqrt(2)*log(e**x + sqrt(2) + 1) - 4 
8*int(e**(2*x)/(e**(6*x) - 3*e**(4*x) + 8*e**(3*x) + 3*e**(2*x) - 1),x) - 
24*int(e**x/(e**(6*x) - 3*e**(4*x) + 8*e**(3*x) + 3*e**(2*x) - 1),x) - 2*l 
og(e**(4*x) - 2*e**(3*x) + 2*e**(2*x) + 2*e**x + 1) + 4*log(e**x - sqrt(2) 
 + 1) + 4*log(e**x + sqrt(2) + 1))/3