Integrand size = 54, antiderivative size = 144 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=\frac {\sqrt {\left (1+x^2\right )^2} \left (2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )+\frac {5}{2} \text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4-\text {$\#$1}^6+\text {$\#$1}^8\&,\frac {-\log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^2-\log (x-\text {$\#$1}) \text {$\#$1}^4+\log (x-\text {$\#$1}) \text {$\#$1}^6}{-\text {$\#$1}+2 \text {$\#$1}^3-3 \text {$\#$1}^5+4 \text {$\#$1}^7}\&\right ]\right )}{1+x^2} \]
Time = 0.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=\frac {4 \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {\left (1+x^2\right )^2}}\right )-\sqrt {5-\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} x}{\sqrt {\left (1+x^2\right )^2}}\right )-\sqrt {5+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} x}{\sqrt {\left (1+x^2\right )^2}}\right )}{\sqrt {2}} \]
Integrate[((-1 + x^2)*(1 + x^2)^3*Sqrt[1 + 2*x^2 + x^4])/((1 + x^4)*(1 - x ^2 + x^4 - x^6 + x^8)),x]
(4*ArcTanh[(Sqrt[2]*x)/Sqrt[(1 + x^2)^2]] - Sqrt[5 - Sqrt[5]]*ArcTanh[(Sqr t[(5 - Sqrt[5])/2]*x)/Sqrt[(1 + x^2)^2]] - Sqrt[5 + Sqrt[5]]*ArcTanh[(Sqrt [(5 + Sqrt[5])/2]*x)/Sqrt[(1 + x^2)^2]])/Sqrt[2]
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.
| \(\displaystyle \int \frac {\left (x^2-1\right ) \left (x^2+1\right )^3 \sqrt {x^4+2 x^2+1}}{\left (x^4+1\right ) \left (x^8-x^6+x^4-x^2+1\right )} \, dx\) |
| \(\Big \downarrow \) 1382 |
| \(\displaystyle \int -\frac {\left (1-x^2\right ) \left (x^2+1\right ) \left (x^4+2 x^2+1\right )^{3/2}}{\left (x^4+1\right ) \left (x^8-x^6+x^4-x^2+1\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (1-x^2\right ) \left (x^2+1\right ) \left (x^4+2 x^2+1\right )^{3/2}}{\left (x^4+1\right ) \left (x^8-x^6+x^4-x^2+1\right )}dx\) |
| \(\Big \downarrow \) 1383 |
| \(\displaystyle -\frac {\left (x^4+2 x^2+1\right )^{3/2} \int \frac {\left (1-x^2\right ) \left (x^2+1\right )^4}{\left (x^4+1\right ) \left (x^8-x^6+x^4-x^2+1\right )}dx}{\left (x^2+1\right )^3}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle -\frac {\left (x^4+2 x^2+1\right )^{3/2} \int \left (\frac {4 \left (x^2-1\right )}{x^4+1}-\frac {5 \left (x^6-x^4+x^2-1\right )}{x^8-x^6+x^4-x^2+1}\right )dx}{\left (x^2+1\right )^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (x^4+2 x^2+1\right )^{3/2} \left (-5 \int \frac {1}{-x^8+x^6-x^4+x^2-1}dx-5 \int \frac {x^2}{x^8-x^6+x^4-x^2+1}dx+5 \int \frac {x^4}{x^8-x^6+x^4-x^2+1}dx-5 \int \frac {x^6}{x^8-x^6+x^4-x^2+1}dx+\sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )-\sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )\right )}{\left (x^2+1\right )^3}\) |
Int[((-1 + x^2)*(1 + x^2)^3*Sqrt[1 + 2*x^2 + x^4])/((1 + x^4)*(1 - x^2 + x ^4 - x^6 + x^8)),x]
3.21.27.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_), x_Symbol] :> Simp[e^(q - 1)/c^((q - 1)/2) Int[u*(d + e*x^ n)*(a + b*x^n + c*x^(2*n))^(p + (q - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && !IntegerQ[p] && I ntegerQ[(q - 1)/2]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^p/(d + e*x^n)^(2 *p) Int[u*(d + e*x^n)^(q + 2*p), x], x] /; FreeQ[{a, b, c, d, e, n, p, q} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && !Int egerQ[p]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.42 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (x \sqrt {2}+x^{2}+1\right )}{x^{2}+1}-\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (-x \sqrt {2}+x^{2}+1\right )}{x^{2}+1}+\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-5 \textit {\_Z}^{2}+5\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R} x +x^{2}+1\right )\right )}{2 x^{2}+2}\) | \(110\) |
| pseudoelliptic | \(\frac {\left (\sqrt {10+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 x^{2}+2}{x \sqrt {10-2 \sqrt {5}}}\right )-\sqrt {10-2 \sqrt {5}}\, \left (5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 x^{2}+2}{x \sqrt {10+2 \sqrt {5}}}\right )+8 \sqrt {5}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2 x}\right )\right ) \operatorname {csgn}\left (\frac {x^{2}+1}{x}\right ) \sqrt {5}}{20}\) | \(110\) |
| default | \(\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \left (\sqrt {2}\, \ln \left (-\frac {x \sqrt {2}+x^{2}+1}{x \sqrt {2}-x^{2}-1}\right )-\sqrt {2}\, \ln \left (-\frac {x \sqrt {2}-x^{2}-1}{x \sqrt {2}+x^{2}+1}\right )+5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (125 \textit {\_Z}^{4}-25 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-5 \textit {\_R} x +x^{2}+1\right )\right )\right )}{2 x^{2}+2}\) | \(113\) |
((x^2+1)^2)^(1/2)/(x^2+1)*2^(1/2)*ln(x*2^(1/2)+x^2+1)-((x^2+1)^2)^(1/2)/(x ^2+1)*2^(1/2)*ln(-x*2^(1/2)+x^2+1)+1/2*((x^2+1)^2)^(1/2)/(x^2+1)*sum(_R*ln (-_R*x+x^2+1),_R=RootOf(_Z^4-5*_Z^2+5))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=-\frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {5} + 5} \log \left (2 \, x^{2} + \sqrt {2} x \sqrt {\sqrt {5} + 5} + 2\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {5} + 5} \log \left (2 \, x^{2} - \sqrt {2} x \sqrt {\sqrt {5} + 5} + 2\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {-\sqrt {5} + 5} \log \left (2 \, x^{2} + \sqrt {2} x \sqrt {-\sqrt {5} + 5} + 2\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {-\sqrt {5} + 5} \log \left (2 \, x^{2} - \sqrt {2} x \sqrt {-\sqrt {5} + 5} + 2\right ) + \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} + 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \]
integrate((x^2-1)*(x^2+1)^3*((x^2+1)^2)^(1/2)/(x^4+1)/(x^8-x^6+x^4-x^2+1), x, algorithm="fricas")
-1/4*sqrt(2)*sqrt(sqrt(5) + 5)*log(2*x^2 + sqrt(2)*x*sqrt(sqrt(5) + 5) + 2 ) + 1/4*sqrt(2)*sqrt(sqrt(5) + 5)*log(2*x^2 - sqrt(2)*x*sqrt(sqrt(5) + 5) + 2) - 1/4*sqrt(2)*sqrt(-sqrt(5) + 5)*log(2*x^2 + sqrt(2)*x*sqrt(-sqrt(5) + 5) + 2) + 1/4*sqrt(2)*sqrt(-sqrt(5) + 5)*log(2*x^2 - sqrt(2)*x*sqrt(-sqr t(5) + 5) + 2) + sqrt(2)*log((x^4 + 4*x^2 + 2*sqrt(2)*(x^3 + x) + 1)/(x^4 + 1))
Timed out. \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=\text {Timed out} \]
Not integrable
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.49 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=\int { \frac {\sqrt {{\left (x^{2} + 1\right )}^{2}} {\left (x^{2} + 1\right )}^{3} {\left (x^{2} - 1\right )}}{{\left (x^{8} - x^{6} + x^{4} - x^{2} + 1\right )} {\left (x^{4} + 1\right )}} \,d x } \]
integrate((x^2-1)*(x^2+1)^3*((x^2+1)^2)^(1/2)/(x^4+1)/(x^8-x^6+x^4-x^2+1), x, algorithm="maxima")
sqrt(2)*log(x^2 + sqrt(2)*x + 1) - sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 5*in tegrate((x^6 - x^4 + x^2 - 1)/(x^8 - x^6 + x^4 - x^2 + 1), x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.34 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=-\sqrt {2} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {2} + \frac {2}{x} \right |}}{{\left | 2 \, x + 2 \, \sqrt {2} + \frac {2}{x} \right |}}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} + 10} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} + \frac {1}{x} \right |}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} + 10} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} + \frac {1}{x} \right |}\right ) - \frac {1}{4} \, \sqrt {-2 \, \sqrt {5} + 10} \log \left ({\left | x + \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} + \frac {1}{x} \right |}\right ) + \frac {1}{4} \, \sqrt {-2 \, \sqrt {5} + 10} \log \left ({\left | x - \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} + \frac {1}{x} \right |}\right ) \]
integrate((x^2-1)*(x^2+1)^3*((x^2+1)^2)^(1/2)/(x^4+1)/(x^8-x^6+x^4-x^2+1), x, algorithm="giac")
-sqrt(2)*log(abs(2*x - 2*sqrt(2) + 2/x)/abs(2*x + 2*sqrt(2) + 2/x)) - 1/4* sqrt(2*sqrt(5) + 10)*log(abs(x + sqrt(1/2*sqrt(5) + 5/2) + 1/x)) + 1/4*sqr t(2*sqrt(5) + 10)*log(abs(x - sqrt(1/2*sqrt(5) + 5/2) + 1/x)) - 1/4*sqrt(- 2*sqrt(5) + 10)*log(abs(x + sqrt(-1/2*sqrt(5) + 5/2) + 1/x)) + 1/4*sqrt(-2 *sqrt(5) + 10)*log(abs(x - sqrt(-1/2*sqrt(5) + 5/2) + 1/x))
Time = 6.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-1+x^2\right ) \left (1+x^2\right )^3 \sqrt {1+2 x^2+x^4}}{\left (1+x^4\right ) \left (1-x^2+x^4-x^6+x^8\right )} \, dx=2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {420500000\,\sqrt {2}\,x}{420500000\,x^2+420500000}\right )+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {2125000\,\sqrt {2}\,x\,\sqrt {\sqrt {5}+5}}{1810000\,\sqrt {5}+1810000\,\sqrt {5}\,x^2-4250000\,x^2-4250000}-\frac {905000\,\sqrt {2}\,\sqrt {5}\,x\,\sqrt {\sqrt {5}+5}}{1810000\,\sqrt {5}+1810000\,\sqrt {5}\,x^2-4250000\,x^2-4250000}\right )\,\sqrt {\sqrt {5}+5}}{2}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {2125000\,\sqrt {2}\,x\,\sqrt {5-\sqrt {5}}}{1810000\,\sqrt {5}+1810000\,\sqrt {5}\,x^2+4250000\,x^2+4250000}+\frac {905000\,\sqrt {2}\,\sqrt {5}\,x\,\sqrt {5-\sqrt {5}}}{1810000\,\sqrt {5}+1810000\,\sqrt {5}\,x^2+4250000\,x^2+4250000}\right )\,\sqrt {5-\sqrt {5}}}{2} \]
2*2^(1/2)*atanh((420500000*2^(1/2)*x)/(420500000*x^2 + 420500000)) + (2^(1 /2)*atanh((2125000*2^(1/2)*x*(5^(1/2) + 5)^(1/2))/(1810000*5^(1/2) + 18100 00*5^(1/2)*x^2 - 4250000*x^2 - 4250000) - (905000*2^(1/2)*5^(1/2)*x*(5^(1/ 2) + 5)^(1/2))/(1810000*5^(1/2) + 1810000*5^(1/2)*x^2 - 4250000*x^2 - 4250 000))*(5^(1/2) + 5)^(1/2))/2 - (2^(1/2)*atanh((2125000*2^(1/2)*x*(5 - 5^(1 /2))^(1/2))/(1810000*5^(1/2) + 1810000*5^(1/2)*x^2 + 4250000*x^2 + 4250000 ) + (905000*2^(1/2)*5^(1/2)*x*(5 - 5^(1/2))^(1/2))/(1810000*5^(1/2) + 1810 000*5^(1/2)*x^2 + 4250000*x^2 + 4250000))*(5 - 5^(1/2))^(1/2))/2