Integrand size = 11, antiderivative size = 185 \[ \int \frac {x^3}{1+x^5} \, dx=\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}+2 \sqrt {\frac {2}{5+\sqrt {5}}} x\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}-\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} x\right )-\frac {1}{5} \log (1+x)+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1-\sqrt {5}\right ) x+x^2\right )+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1+\sqrt {5}\right ) x+x^2\right ) \]
-1/5*ln(1+x)+1/20*ln(1+x^2-1/2*x*(-5^(1/2)+1))*(-5^(1/2)+1)+1/20*ln(1+x^2- 1/2*x*(5^(1/2)+1))*(5^(1/2)+1)+1/10*arctan(1/5*x*(50+10*5^(1/2))^(1/2)-1/5 *(25+10*5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)+1/10*arctan(1/5*(25-10*5^(1/2 ))^(1/2)+2*x*2^(1/2)/(5+5^(1/2))^(1/2))*(10+2*5^(1/2))^(1/2)
Time = 0.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{1+x^5} \, dx=\frac {1}{20} \left (-2 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {1+\sqrt {5}-4 x}{\sqrt {10-2 \sqrt {5}}}\right )+2 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {-1+\sqrt {5}+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-4 \log (1+x)-\left (-1+\sqrt {5}\right ) \log \left (1+\frac {1}{2} \left (-1+\sqrt {5}\right ) x+x^2\right )+\left (1+\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1+\sqrt {5}\right ) x+x^2\right )\right ) \]
(-2*Sqrt[10 - 2*Sqrt[5]]*ArcTan[(1 + Sqrt[5] - 4*x)/Sqrt[10 - 2*Sqrt[5]]] + 2*Sqrt[2*(5 + Sqrt[5])]*ArcTan[(-1 + Sqrt[5] + 4*x)/Sqrt[2*(5 + Sqrt[5]) ]] - 4*Log[1 + x] - (-1 + Sqrt[5])*Log[1 + ((-1 + Sqrt[5])*x)/2 + x^2] + ( 1 + Sqrt[5])*Log[1 - ((1 + Sqrt[5])*x)/2 + x^2])/20
Time = 0.37 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {822, 16, 27, 1142, 25, 1083, 217, 1103}
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.
| \(\displaystyle \int \frac {x^3}{x^5+1} \, dx\) |
| \(\Big \downarrow \) 822 |
| \(\displaystyle \frac {2}{5} \int \frac {\left (1-\sqrt {5}\right ) x+\sqrt {5}+1}{2 \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )}dx+\frac {2}{5} \int \frac {\left (1+\sqrt {5}\right ) x-\sqrt {5}+1}{2 \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )}dx-\frac {1}{5} \int \frac {1}{x+1}dx\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {2}{5} \int \frac {\left (1-\sqrt {5}\right ) x+\sqrt {5}+1}{2 \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )}dx+\frac {2}{5} \int \frac {\left (1+\sqrt {5}\right ) x-\sqrt {5}+1}{2 \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )}dx-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (1-\sqrt {5}\right ) x+\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx+\frac {1}{5} \int \frac {\left (1+\sqrt {5}\right ) x-\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (5+\sqrt {5}\right ) \int \frac {1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx+\frac {1}{4} \left (1-\sqrt {5}\right ) \int -\frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \left (\frac {1}{2} \left (5-\sqrt {5}\right ) \int \frac {1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx+\frac {1}{4} \left (1+\sqrt {5}\right ) \int -\frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx\right )-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (5+\sqrt {5}\right ) \int \frac {1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx-\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \left (\frac {1}{2} \left (5-\sqrt {5}\right ) \int \frac {1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx-\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx\right )-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{5} \left (-\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx-\left (5-\sqrt {5}\right ) \int \frac {1}{-\left (4 x-\sqrt {5}-1\right )^2-2 \left (5-\sqrt {5}\right )}d\left (4 x-\sqrt {5}-1\right )\right )+\frac {1}{5} \left (-\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx-\left (5+\sqrt {5}\right ) \int \frac {1}{-\left (4 x+\sqrt {5}-1\right )^2-2 \left (5+\sqrt {5}\right )}d\left (4 x+\sqrt {5}-1\right )\right )-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{5} \left (\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {4 x+\sqrt {5}-1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \left (\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {4 x-\sqrt {5}-1}{\sqrt {2 \left (5-\sqrt {5}\right )}}\right )-\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx\right )-\frac {1}{5} \log (x+1)\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{5} \left (\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {4 x+\sqrt {5}-1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )+\frac {1}{4} \left (1-\sqrt {5}\right ) \log \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )\right )+\frac {1}{5} \left (\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {4 x-\sqrt {5}-1}{\sqrt {2 \left (5-\sqrt {5}\right )}}\right )+\frac {1}{4} \left (1+\sqrt {5}\right ) \log \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )\right )-\frac {1}{5} \log (x+1)\) |
-1/5*Log[1 + x] + (Sqrt[(5 + Sqrt[5])/2]*ArcTan[(-1 + Sqrt[5] + 4*x)/Sqrt[ 2*(5 + Sqrt[5])]] + ((1 - Sqrt[5])*Log[2 - (1 - Sqrt[5])*x + 2*x^2])/4)/5 + (Sqrt[(5 - Sqrt[5])/2]*ArcTan[(-1 - Sqrt[5] + 4*x)/Sqrt[2*(5 - Sqrt[5])] ] + ((1 + Sqrt[5])*Log[2 - (1 + Sqrt[5])*x + 2*x^2])/4)/5
3.14.1.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator [Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; -(-r)^(m + 1)/(a*n*s^m) Int[1/(r + s*x), x] + 2*(r^(m + 1)/(a*n*s^m)) Sum[u, {k, 1, (n - 1)/2}], x]] /; FreeQ[{a, b} , x] && IGtQ[(n - 1)/2, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.24
| method | result | size |
| risch | \(-\frac {\ln \left (1+x \right )}{5}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{3}-\textit {\_R}^{2}+\textit {\_R} +x -1\right )\right )}{5}\) | \(45\) |
| default | \(-\frac {\ln \left (1+x \right )}{5}-\frac {\left (-\sqrt {5}-1\right ) \ln \left (-x \sqrt {5}+2 x^{2}-x +2\right )}{20}-\frac {2 \left (\sqrt {5}-1-\frac {\left (-\sqrt {5}-1\right )^{2}}{4}\right ) \arctan \left (\frac {-\sqrt {5}+4 x -1}{\sqrt {10-2 \sqrt {5}}}\right )}{5 \sqrt {10-2 \sqrt {5}}}+\frac {\left (-\sqrt {5}+1\right ) \ln \left (x \sqrt {5}+2 x^{2}-x +2\right )}{20}+\frac {2 \left (\sqrt {5}+1-\frac {\left (-\sqrt {5}+1\right ) \left (\sqrt {5}-1\right )}{4}\right ) \arctan \left (\frac {\sqrt {5}+4 x -1}{\sqrt {10+2 \sqrt {5}}}\right )}{5 \sqrt {10+2 \sqrt {5}}}\) | \(156\) |
| meijerg | \(-\frac {x^{4} \ln \left (1+\left (x^{5}\right )^{\frac {1}{5}}\right )}{5 \left (x^{5}\right )^{\frac {4}{5}}}+\frac {x^{4} \cos \left (\frac {\pi }{5}\right ) \ln \left (1-2 \cos \left (\frac {\pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}+\left (x^{5}\right )^{\frac {2}{5}}\right )}{5 \left (x^{5}\right )^{\frac {4}{5}}}+\frac {2 x^{4} \sin \left (\frac {\pi }{5}\right ) \arctan \left (\frac {\sin \left (\frac {\pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}}{1-\cos \left (\frac {\pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}}\right )}{5 \left (x^{5}\right )^{\frac {4}{5}}}-\frac {x^{4} \cos \left (\frac {2 \pi }{5}\right ) \ln \left (1+2 \cos \left (\frac {2 \pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}+\left (x^{5}\right )^{\frac {2}{5}}\right )}{5 \left (x^{5}\right )^{\frac {4}{5}}}+\frac {2 x^{4} \sin \left (\frac {2 \pi }{5}\right ) \arctan \left (\frac {\sin \left (\frac {2 \pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}}{1+\cos \left (\frac {2 \pi }{5}\right ) \left (x^{5}\right )^{\frac {1}{5}}}\right )}{5 \left (x^{5}\right )^{\frac {4}{5}}}\) | \(165\) |
Leaf count of result is larger than twice the leaf count of optimal. 905 vs. \(2 (122) = 244\).
Time = 0.96 (sec) , antiderivative size = 905, normalized size of antiderivative = 4.89 \[ \int \frac {x^3}{1+x^5} \, dx=\text {Too large to display} \]
-1/20*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)*log(1/64*(sqrt(5) + 10 *sqrt(-1/50*sqrt(5) - 1/10) - 1)^3 + 1/64*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 + 1/16*(sqrt (5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 + 1/64*((sqrt(5) + 10*sqrt(-1/5 0*sqrt(5) - 1/10) - 1)^2 + 4*sqrt(5) + 40*sqrt(-1/50*sqrt(5) - 1/10) - 4)* (sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1) + x + 1/4*sqrt(5) + 5/2*sqrt (-1/50*sqrt(5) - 1/10) - 1/4) - 1/20*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/ 10) - 1)*log(-1/64*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^3 - 1/16* (sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 + x - 1/4*sqrt(5) - 5/2*sq rt(-1/50*sqrt(5) - 1/10) - 3/4) + 1/20*(sqrt(5) + 2*sqrt(-3/16*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 - 1/8*(sqrt(5) + 10*sqrt(-1/50*sqrt(5 ) - 1/10) + 3)*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1) - 3/16*(sqrt( 5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 - 1/2*sqrt(5) - 5*sqrt(-1/50*sqr t(5) - 1/10) - 5/2) + 1)*log(-1/64*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10 ) - 1)*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 + 1/16*sqrt(-3/16*( sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 - 1/8*(sqrt(5) + 10*sqrt(-1 /50*sqrt(5) - 1/10) + 3)*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1) - 3 /16*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1)^2 - 1/2*sqrt(5) - 5*sqrt (-1/50*sqrt(5) - 1/10) - 5/2)*(sqrt(5) + 10*sqrt(-1/50*sqrt(5) - 1/10) - 1 )*(sqrt(5) - 10*sqrt(-1/50*sqrt(5) - 1/10) - 1) - 1/64*((sqrt(5) + 10*s...
Time = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.19 \[ \int \frac {x^3}{1+x^5} \, dx=- \frac {\log {\left (x + 1 \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} - 125 t^{3} + 25 t^{2} - 5 t + 1, \left ( t \mapsto t \log {\left (625 t^{4} + x \right )} \right )\right )} \]
-log(x + 1)/5 + RootSum(625*_t**4 - 125*_t**3 + 25*_t**2 - 5*_t + 1, Lambd a(_t, _t*log(625*_t**4 + x)))
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{1+x^5} \, dx=\frac {\sqrt {5} {\left (\sqrt {5} + 1\right )} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {2 \, \sqrt {5} + 10}} + \frac {\sqrt {5} {\left (\sqrt {5} - 1\right )} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {-2 \, \sqrt {5} + 10}} + \frac {{\left (\sqrt {5} + 3\right )} \log \left (2 \, x^{2} - x {\left (\sqrt {5} + 1\right )} + 2\right )}{10 \, {\left (\sqrt {5} + 1\right )}} + \frac {{\left (\sqrt {5} - 3\right )} \log \left (2 \, x^{2} + x {\left (\sqrt {5} - 1\right )} + 2\right )}{10 \, {\left (\sqrt {5} - 1\right )}} - \frac {1}{5} \, \log \left (x + 1\right ) \]
1/5*sqrt(5)*(sqrt(5) + 1)*arctan((4*x + sqrt(5) - 1)/sqrt(2*sqrt(5) + 10)) /sqrt(2*sqrt(5) + 10) + 1/5*sqrt(5)*(sqrt(5) - 1)*arctan((4*x - sqrt(5) - 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) + 1/10*(sqrt(5) + 3)*log(2 *x^2 - x*(sqrt(5) + 1) + 2)/(sqrt(5) + 1) + 1/10*(sqrt(5) - 3)*log(2*x^2 + x*(sqrt(5) - 1) + 2)/(sqrt(5) - 1) - 1/5*log(x + 1)
Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.61 \[ \int \frac {x^3}{1+x^5} \, dx=\frac {1}{20} \, {\left (\sqrt {5} + 1\right )} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {5} + 1\right )} + 1\right ) - \frac {1}{20} \, {\left (\sqrt {5} - 1\right )} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {5} - 1\right )} + 1\right ) + \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) + \frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) - \frac {1}{5} \, \log \left ({\left | x + 1 \right |}\right ) \]
1/20*(sqrt(5) + 1)*log(x^2 - 1/2*x*(sqrt(5) + 1) + 1) - 1/20*(sqrt(5) - 1) *log(x^2 + 1/2*x*(sqrt(5) - 1) + 1) + 1/10*sqrt(2*sqrt(5) + 10)*arctan((4* x + sqrt(5) - 1)/sqrt(2*sqrt(5) + 10)) + 1/10*sqrt(-2*sqrt(5) + 10)*arctan ((4*x - sqrt(5) - 1)/sqrt(-2*sqrt(5) + 10)) - 1/5*log(abs(x + 1))
Time = 5.82 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{1+x^5} \, dx=\ln \left (25\,x\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}-\frac {\sqrt {5}}{20}+\frac {1}{20}\right )-5\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}-\frac {\sqrt {5}}{20}+\frac {1}{20}\right )-\frac {\ln \left (x+1\right )}{5}-\ln \left (25\,x\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}+\frac {\sqrt {5}}{20}-\frac {1}{20}\right )+5\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}+\frac {\sqrt {5}}{20}-\frac {1}{20}\right )+\ln \left (25\,x\,\left (\frac {\sqrt {5}}{20}-\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )-5\right )\,\left (\frac {\sqrt {5}}{20}-\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )+\ln \left (25\,x\,\left (\frac {\sqrt {5}}{20}+\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )-5\right )\,\left (\frac {\sqrt {5}}{20}+\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right ) \]
log(25*x*((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 - 5^(1/2)/20 + 1/20) - 5)*((2 ^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 - 5^(1/2)/20 + 1/20) - log(x + 1)/5 - log (25*x*((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 + 5^(1/2)/20 - 1/20) + 5)*((2^(1 /2)*(- 5^(1/2) - 5)^(1/2))/20 + 5^(1/2)/20 - 1/20) + log(25*x*(5^(1/2)/20 - (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20) - 5)*(5^(1/2)/20 - (2^(1/2)*(5^ (1/2) - 5)^(1/2))/20 + 1/20) + log(25*x*(5^(1/2)/20 + (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20) - 5)*(5^(1/2)/20 + (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20)