Integrand size = 8, antiderivative size = 411 \[ \int \frac {1}{1+\sinh ^8(x)} \, dx=-\frac {1}{8} \sqrt {-1+\sqrt {4-2 \sqrt {2}}} \arctan \left (\frac {\sqrt {1+\sqrt {4-2 \sqrt {2}}}-2 \tanh (x)}{\sqrt {-1+\sqrt {4-2 \sqrt {2}}}}\right )-\frac {1}{8} \sqrt {-1+\sqrt {2 \left (2+\sqrt {2}\right )}} \arctan \left (\frac {\sqrt {1+\sqrt {2 \left (2+\sqrt {2}\right )}}-2 \tanh (x)}{\sqrt {-1+\sqrt {2 \left (2+\sqrt {2}\right )}}}\right )+\frac {1}{8} \sqrt {-1+\sqrt {4-2 \sqrt {2}}} \arctan \left (\frac {\sqrt {1+\sqrt {4-2 \sqrt {2}}}+2 \tanh (x)}{\sqrt {-1+\sqrt {4-2 \sqrt {2}}}}\right )+\frac {1}{8} \sqrt {-1+\sqrt {2 \left (2+\sqrt {2}\right )}} \arctan \left (\frac {\sqrt {1+\sqrt {2 \left (2+\sqrt {2}\right )}}+2 \tanh (x)}{\sqrt {-1+\sqrt {2 \left (2+\sqrt {2}\right )}}}\right )+\frac {1}{8} \sqrt {1+\sqrt {4-2 \sqrt {2}}} \text {arctanh}\left (\frac {\sqrt {1+\sqrt {4-2 \sqrt {2}}} \tanh (x)}{\sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )}+\tanh ^2(x)}\right )+\frac {1}{8} \sqrt {1+\sqrt {2 \left (2+\sqrt {2}\right )}} \text {arctanh}\left (\frac {\sqrt {1+\sqrt {2 \left (2+\sqrt {2}\right )}} \tanh (x)}{\sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )}+\tanh ^2(x)}\right ) \] Output:
-1/8*(-1+(4-2*2^(1/2))^(1/2))^(1/2)*arctan(((1+(4-2*2^(1/2))^(1/2))^(1/2)- 2*tanh(x))/(-1+(4-2*2^(1/2))^(1/2))^(1/2))-1/8*(-1+(4+2*2^(1/2))^(1/2))^(1 /2)*arctan(((1+(4+2*2^(1/2))^(1/2))^(1/2)-2*tanh(x))/(-1+(4+2*2^(1/2))^(1/ 2))^(1/2))+1/8*(-1+(4-2*2^(1/2))^(1/2))^(1/2)*arctan(((1+(4-2*2^(1/2))^(1/ 2))^(1/2)+2*tanh(x))/(-1+(4-2*2^(1/2))^(1/2))^(1/2))+1/8*(-1+(4+2*2^(1/2)) ^(1/2))^(1/2)*arctan(((1+(4+2*2^(1/2))^(1/2))^(1/2)+2*tanh(x))/(-1+(4+2*2^ (1/2))^(1/2))^(1/2))+1/8*(1+(4-2*2^(1/2))^(1/2))^(1/2)*arctanh((1+(4-2*2^( 1/2))^(1/2))^(1/2)*tanh(x)/(1/2*(4-2*2^(1/2))^(1/2)+tanh(x)^2))+1/8*(1+(4+ 2*2^(1/2))^(1/2))^(1/2)*arctanh((1+(4+2*2^(1/2))^(1/2))^(1/2)*tanh(x)/(1/2 *(4+2*2^(1/2))^(1/2)+tanh(x)^2))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 5.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.31 \[ \int \frac {1}{1+\sinh ^8(x)} \, dx=16 \text {RootSum}\left [1-8 \text {$\#$1}+28 \text {$\#$1}^2-56 \text {$\#$1}^3+326 \text {$\#$1}^4-56 \text {$\#$1}^5+28 \text {$\#$1}^6-8 \text {$\#$1}^7+\text {$\#$1}^8\&,\frac {x \text {$\#$1}^3+\log (-\cosh (x)-\sinh (x)+\cosh (x) \text {$\#$1}-\sinh (x) \text {$\#$1}) \text {$\#$1}^3}{-1+7 \text {$\#$1}-21 \text {$\#$1}^2+163 \text {$\#$1}^3-35 \text {$\#$1}^4+21 \text {$\#$1}^5-7 \text {$\#$1}^6+\text {$\#$1}^7}\&\right ] \] Input:
Integrate[(1 + Sinh[x]^8)^(-1),x]
Output:
16*RootSum[1 - 8*#1 + 28*#1^2 - 56*#1^3 + 326*#1^4 - 56*#1^5 + 28*#1^6 - 8 *#1^7 + #1^8 & , (x*#1^3 + Log[-Cosh[x] - Sinh[x] + Cosh[x]*#1 - Sinh[x]*# 1]*#1^3)/(-1 + 7*#1 - 21*#1^2 + 163*#1^3 - 35*#1^4 + 21*#1^5 - 7*#1^6 + #1 ^7) & ]
Time = 0.46 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.31, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 3690, 3042, 3660, 219}
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.
\(\displaystyle \int \frac {1}{\sinh ^8(x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{1+\sin (i x)^8}dx\) |
\(\Big \downarrow \) 3690 |
\(\displaystyle \frac {1}{4} \int \frac {1}{1-\sqrt [4]{-1} \sinh ^2(x)}dx+\frac {1}{4} \int \frac {1}{\sqrt [4]{-1} \sinh ^2(x)+1}dx+\frac {1}{4} \int \frac {1}{1-(-1)^{3/4} \sinh ^2(x)}dx+\frac {1}{4} \int \frac {1}{(-1)^{3/4} \sinh ^2(x)+1}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \frac {1}{1-\sqrt [4]{-1} \sin (i x)^2}dx+\frac {1}{4} \int \frac {1}{\sqrt [4]{-1} \sin (i x)^2+1}dx+\frac {1}{4} \int \frac {1}{1-(-1)^{3/4} \sin (i x)^2}dx+\frac {1}{4} \int \frac {1}{(-1)^{3/4} \sin (i x)^2+1}dx\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {1}{4} \int \frac {1}{1-\left (1-\sqrt [4]{-1}\right ) \tanh ^2(x)}d\tanh (x)+\frac {1}{4} \int \frac {1}{1-\left (1+\sqrt [4]{-1}\right ) \tanh ^2(x)}d\tanh (x)+\frac {1}{4} \int \frac {1}{1-\left (1-(-1)^{3/4}\right ) \tanh ^2(x)}d\tanh (x)+\frac {1}{4} \int \frac {1}{1-\left (1+(-1)^{3/4}\right ) \tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\sqrt {1-\sqrt [4]{-1}} \tanh (x)\right )}{4 \sqrt {1-\sqrt [4]{-1}}}+\frac {\text {arctanh}\left (\sqrt {1+\sqrt [4]{-1}} \tanh (x)\right )}{4 \sqrt {1+\sqrt [4]{-1}}}+\frac {\text {arctanh}\left (\sqrt {1-(-1)^{3/4}} \tanh (x)\right )}{4 \sqrt {1-(-1)^{3/4}}}+\frac {\text {arctanh}\left (\sqrt {1+(-1)^{3/4}} \tanh (x)\right )}{4 \sqrt {1+(-1)^{3/4}}}\) |
Input:
Int[(1 + Sinh[x]^8)^(-1),x]
Output:
ArcTanh[Sqrt[1 - (-1)^(1/4)]*Tanh[x]]/(4*Sqrt[1 - (-1)^(1/4)]) + ArcTanh[S qrt[1 + (-1)^(1/4)]*Tanh[x]]/(4*Sqrt[1 + (-1)^(1/4)]) + ArcTanh[Sqrt[1 - ( -1)^(3/4)]*Tanh[x]]/(4*Sqrt[1 - (-1)^(3/4)]) + ArcTanh[Sqrt[1 + (-1)^(3/4) ]*Tanh[x]]/(4*Sqrt[1 + (-1)^(3/4)])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{ k}, Simp[2/(a*n) Sum[Int[1/(1 - Sin[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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.53 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.16
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-4 \textit {\_Z}^{6}+6 \textit {\_Z}^{4}-4 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+\left (-4 \textit {\_R}^{7}+8 \textit {\_R}^{5}-12 \textit {\_R}^{3}+8 \textit {\_R} \right ) \tanh \left (\frac {x}{2}\right )+1\right )\right )}{8}\) | \(64\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (33554432 \textit {\_Z}^{8}-1048576 \textit {\_Z}^{6}+24576 \textit {\_Z}^{4}-256 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (8388608 \textit {\_R}^{7}-1048576 \textit {\_R}^{6}-131072 \textit {\_R}^{5}+16384 \textit {\_R}^{4}+4096 \textit {\_R}^{3}-512 \textit {\_R}^{2}+{\mathrm e}^{2 x}+1\right )\) | \(66\) |
Input:
int(1/(1+sinh(x)^8),x,method=_RETURNVERBOSE)
Output:
1/8*sum(_R*ln(tanh(1/2*x)^2+(-4*_R^7+8*_R^5-12*_R^3+8*_R)*tanh(1/2*x)+1),_ R=RootOf(2*_Z^8-4*_Z^6+6*_Z^4-4*_Z^2+1))
Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (295) = 590\).
Time = 0.13 (sec) , antiderivative size = 729, normalized size of antiderivative = 1.77 \[ \int \frac {1}{1+\sinh ^8(x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(1+sinh(x)^8),x, algorithm="fricas")
Output:
-1/8*sqrt(-1/2*sqrt(2*sqrt(2) - 3) + 1/2)*log(cosh(x)^2 + 2*cosh(x)*sinh(x ) + sinh(x)^2 + sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) + (sqrt(2*sqrt(2) - 3)*( sqrt(2) + 2) - sqrt(2) - 2)*sqrt(-1/2*sqrt(2*sqrt(2) - 3) + 1/2) - sqrt(2) - 1) + 1/8*sqrt(-1/2*sqrt(2*sqrt(2) - 3) + 1/2)*log(cosh(x)^2 + 2*cosh(x) *sinh(x) + sinh(x)^2 + sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) - (sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) - sqrt(2) - 2)*sqrt(-1/2*sqrt(2*sqrt(2) - 3) + 1/2) - sqrt(2) - 1) + 1/8*sqrt(1/2*sqrt(2*sqrt(2) - 3) + 1/2)*log(cosh(x)^2 + 2*c osh(x)*sinh(x) + sinh(x)^2 - sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) + (sqrt(2*s qrt(2) - 3)*(sqrt(2) + 2) + sqrt(2) + 2)*sqrt(1/2*sqrt(2*sqrt(2) - 3) + 1/ 2) - sqrt(2) - 1) - 1/8*sqrt(1/2*sqrt(2*sqrt(2) - 3) + 1/2)*log(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) - (sqr t(2*sqrt(2) - 3)*(sqrt(2) + 2) + sqrt(2) + 2)*sqrt(1/2*sqrt(2*sqrt(2) - 3) + 1/2) - sqrt(2) - 1) - 1/8*sqrt(1/2*sqrt(-2*sqrt(2) - 3) + 1/2)*log(cosh (x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + (sqrt(2) - 2)*sqrt(-2*sqrt(2) - 3) + ((sqrt(2) - 2)*sqrt(-2*sqrt(2) - 3) + sqrt(2) - 2)*sqrt(1/2*sqrt(-2*sqr t(2) - 3) + 1/2) + sqrt(2) - 1) + 1/8*sqrt(1/2*sqrt(-2*sqrt(2) - 3) + 1/2) *log(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + (sqrt(2) - 2)*sqrt(-2*sqr t(2) - 3) - ((sqrt(2) - 2)*sqrt(-2*sqrt(2) - 3) + sqrt(2) - 2)*sqrt(1/2*sq rt(-2*sqrt(2) - 3) + 1/2) + sqrt(2) - 1) + 1/8*sqrt(-1/2*sqrt(-2*sqrt(2) - 3) + 1/2)*log(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - (sqrt(2) - 2...
\[ \int \frac {1}{1+\sinh ^8(x)} \, dx=\int \frac {1}{\sinh ^{8}{\left (x \right )} + 1}\, dx \] Input:
integrate(1/(1+sinh(x)**8),x)
Output:
Integral(1/(sinh(x)**8 + 1), x)
\[ \int \frac {1}{1+\sinh ^8(x)} \, dx=\int { \frac {1}{\sinh \left (x\right )^{8} + 1} \,d x } \] Input:
integrate(1/(1+sinh(x)^8),x, algorithm="maxima")
Output:
integrate(1/(sinh(x)^8 + 1), x)
Time = 0.12 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {1}{1+\sinh ^8(x)} \, dx=0 \] Input:
integrate(1/(1+sinh(x)^8),x, algorithm="giac")
Output:
0
Timed out. \[ \int \frac {1}{1+\sinh ^8(x)} \, dx=\text {Hanged} \] Input:
int(1/(sinh(x)^8 + 1),x)
Output:
\text{Hanged}
\[ \int \frac {1}{1+\sinh ^8(x)} \, dx=\int \frac {1}{\sinh \left (x \right )^{8}+1}d x \] Input:
int(1/(1+sinh(x)^8),x)
Output:
int(1/(sinh(x)**8 + 1),x)