Integrand size = 13, antiderivative size = 211 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=-\frac {1}{2 a^5 x^2}-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^7}+\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^7}-\frac {\log (a+x)}{5 a^7}+\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^7}+\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^7} \]
-1/2/a^5/x^2-1/5*ln(a+x)/a^7+1/20*ln(a^2+x^2-1/2*a*x*(5^(1/2)+1))*(-5^(1/2 )+1)/a^7+1/20*ln(a^2+x^2-1/2*a*x*(-5^(1/2)+1))*(5^(1/2)+1)/a^7-1/10*arctan ((-4*x+a*(-5^(1/2)+1))/a/(10+2*5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)/a^7+1/ 10*arctan(1/20*(-4*x+a*(5^(1/2)+1))*(50+10*5^(1/2))^(1/2)/a)*(10+2*5^(1/2) )^(1/2)/a^7
Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=-\frac {\frac {10 a^2}{x^2}-2 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {\left (-1+\sqrt {5}\right ) a+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )+2 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {-\left (\left (1+\sqrt {5}\right ) a\right )+4 x}{\sqrt {10-2 \sqrt {5}} a}\right )+4 \log (a+x)-\left (1+\sqrt {5}\right ) \log \left (a^2+\frac {1}{2} \left (-1+\sqrt {5}\right ) a x+x^2\right )+\left (-1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^7} \]
-1/20*((10*a^2)/x^2 - 2*Sqrt[10 - 2*Sqrt[5]]*ArcTan[((-1 + Sqrt[5])*a + 4* x)/(Sqrt[2*(5 + Sqrt[5])]*a)] + 2*Sqrt[2*(5 + Sqrt[5])]*ArcTan[(-((1 + Sqr t[5])*a) + 4*x)/(Sqrt[10 - 2*Sqrt[5]]*a)] + 4*Log[a + x] - (1 + Sqrt[5])*L og[a^2 + ((-1 + Sqrt[5])*a*x)/2 + x^2] + (-1 + Sqrt[5])*Log[a^2 - ((1 + Sq rt[5])*a*x)/2 + x^2])/a^7
Time = 0.38 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {847, 822, 16, 27, 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.
| \(\displaystyle \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle -\frac {\int \frac {x^2}{a^5+x^5}dx}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 822 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\left (1+\sqrt {5}\right ) a+\left (1+\sqrt {5}\right ) x}{2 \left (2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2\right )}dx}{5 a^2}+\frac {2 \int -\frac {\left (1-\sqrt {5}\right ) (a+x)}{2 \left (2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2\right )}dx}{5 a^2}+\frac {\int \frac {1}{a+x}dx}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\left (1+\sqrt {5}\right ) a+\left (1+\sqrt {5}\right ) x}{2 \left (2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2\right )}dx}{5 a^2}+\frac {2 \int -\frac {\left (1-\sqrt {5}\right ) (a+x)}{2 \left (2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2\right )}dx}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\left (1+\sqrt {5}\right ) (a+x)}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \int \frac {a+x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\left (1+\sqrt {5}\right ) \int \frac {a+x}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \int \frac {a+x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle -\frac {-\frac {\left (1+\sqrt {5}\right ) \left (\frac {1}{4} \left (5-\sqrt {5}\right ) a \int \frac {1}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx+\frac {1}{4} \int -\frac {\left (1-\sqrt {5}\right ) a-4 x}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \left (\frac {1}{4} \left (5+\sqrt {5}\right ) a \int \frac {1}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx+\frac {1}{4} \int -\frac {\left (1+\sqrt {5}\right ) a-4 x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\left (1+\sqrt {5}\right ) \left (\frac {1}{4} \left (5-\sqrt {5}\right ) a \int \frac {1}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx-\frac {1}{4} \int \frac {\left (1-\sqrt {5}\right ) a-4 x}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \left (\frac {1}{4} \left (5+\sqrt {5}\right ) a \int \frac {1}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx-\frac {1}{4} \int \frac {\left (1+\sqrt {5}\right ) a-4 x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {-\frac {\left (1+\sqrt {5}\right ) \left (-\frac {1}{4} \int \frac {\left (1-\sqrt {5}\right ) a-4 x}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx-\frac {1}{2} \left (5-\sqrt {5}\right ) a \int \frac {1}{-2 \left (5+\sqrt {5}\right ) a^2-\left (4 x-\left (1-\sqrt {5}\right ) a\right )^2}d\left (4 x-\left (1-\sqrt {5}\right ) a\right )\right )}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \left (-\frac {1}{4} \int \frac {\left (1+\sqrt {5}\right ) a-4 x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx-\frac {1}{2} \left (5+\sqrt {5}\right ) a \int \frac {1}{-2 \left (5-\sqrt {5}\right ) a^2-\left (4 x-\left (1+\sqrt {5}\right ) a\right )^2}d\left (4 x-\left (1+\sqrt {5}\right ) a\right )\right )}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {-\frac {\left (1+\sqrt {5}\right ) \left (\frac {\left (5-\sqrt {5}\right ) \arctan \left (\frac {4 x-\left (1-\sqrt {5}\right ) a}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{2 \sqrt {2 \left (5+\sqrt {5}\right )}}-\frac {1}{4} \int \frac {\left (1-\sqrt {5}\right ) a-4 x}{2 a^2-\left (1-\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \left (\frac {\left (5+\sqrt {5}\right ) \arctan \left (\frac {4 x-\left (1+\sqrt {5}\right ) a}{\sqrt {2 \left (5-\sqrt {5}\right )} a}\right )}{2 \sqrt {2 \left (5-\sqrt {5}\right )}}-\frac {1}{4} \int \frac {\left (1+\sqrt {5}\right ) a-4 x}{2 a^2-\left (1+\sqrt {5}\right ) x a+2 x^2}dx\right )}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}-\frac {1}{2 a^5 x^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {1}{2 a^5 x^2}-\frac {-\frac {\left (1+\sqrt {5}\right ) \left (\frac {1}{4} \log \left (2 a^2-\left (1-\sqrt {5}\right ) a x+2 x^2\right )+\frac {\left (5-\sqrt {5}\right ) \arctan \left (\frac {4 x-\left (1-\sqrt {5}\right ) a}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{2 \sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{5 a^2}-\frac {\left (1-\sqrt {5}\right ) \left (\frac {1}{4} \log \left (2 a^2-\left (1+\sqrt {5}\right ) a x+2 x^2\right )+\frac {\left (5+\sqrt {5}\right ) \arctan \left (\frac {4 x-\left (1+\sqrt {5}\right ) a}{\sqrt {2 \left (5-\sqrt {5}\right )} a}\right )}{2 \sqrt {2 \left (5-\sqrt {5}\right )}}\right )}{5 a^2}+\frac {\log (a+x)}{5 a^2}}{a^5}\) |
-1/2*1/(a^5*x^2) - (Log[a + x]/(5*a^2) - ((1 + Sqrt[5])*(((5 - Sqrt[5])*Ar cTan[(-((1 - Sqrt[5])*a) + 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)])/(2*Sqrt[2*(5 + Sqrt[5])]) + Log[2*a^2 - (1 - Sqrt[5])*a*x + 2*x^2]/4))/(5*a^2) - ((1 - S qrt[5])*(((5 + Sqrt[5])*ArcTan[(-((1 + Sqrt[5])*a) + 4*x)/(Sqrt[2*(5 - Sqr t[5])]*a)])/(2*Sqrt[2*(5 - Sqrt[5])]) + Log[2*a^2 - (1 + Sqrt[5])*a*x + 2* x^2]/4))/(5*a^2))/a^5
3.2.43.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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 = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.37
| method | result | size |
| risch | \(-\frac {1}{2 a^{5} x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{28} \textit {\_Z}^{4}-a^{21} \textit {\_Z}^{3}+a^{14} \textit {\_Z}^{2}-a^{7} \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\left (-6 \textit {\_R}^{5} a^{35}-5\right ) x +a^{15} \textit {\_R}^{2}\right )\right )}{5}-\frac {\ln \left (a +x \right )}{5 a^{7}}\) | \(78\) |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-a \,\textit {\_Z}^{3}+\textit {\_Z}^{2} a^{2}-a^{3} \textit {\_Z} +a^{4}\right )}{\sum }\frac {\left (\textit {\_R}^{3}-2 \textit {\_R}^{2} a -2 a^{2} \textit {\_R} +a^{3}\right ) \ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}-3 \textit {\_R}^{2} a +2 a^{2} \textit {\_R} -a^{3}}}{5 a^{7}}-\frac {1}{2 a^{5} x^{2}}-\frac {\ln \left (a +x \right )}{5 a^{7}}\) | \(105\) |
-1/2/a^5/x^2+1/5*sum(_R*ln((-6*_R^5*a^35-5)*x+a^15*_R^2),_R=RootOf(_Z^4*a^ 28-_Z^3*a^21+_Z^2*a^14-_Z*a^7+1))-1/5*ln(a+x)/a^7
Result contains complex when optimal does not.
Time = 1.02 (sec) , antiderivative size = 15499, normalized size of antiderivative = 73.45 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=\text {Too large to display} \]
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.24 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=- \frac {1}{2 a^{5} x^{2}} + \frac {- \frac {\log {\left (a + x \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} - 125 t^{3} + 25 t^{2} - 5 t + 1, \left ( t \mapsto t \log {\left (25 t^{2} a + x \right )} \right )\right )}}{a^{7}} \]
-1/(2*a**5*x**2) + (-log(a + x)/5 + RootSum(625*_t**4 - 125*_t**3 + 25*_t* *2 - 5*_t + 1, Lambda(_t, _t*log(25*_t**2*a + x))))/a**7
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=\frac {\frac {2 \, \sqrt {5} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{a^{2} \sqrt {2 \, \sqrt {5} + 10}} - \frac {2 \, \sqrt {5} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{a^{2} \sqrt {-2 \, \sqrt {5} + 10}} - \frac {\log \left (a + x\right )}{a^{2}} - \frac {\log \left (-a x {\left (\sqrt {5} + 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a^{2} {\left (\sqrt {5} + 1\right )}} + \frac {\log \left (a x {\left (\sqrt {5} - 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a^{2} {\left (\sqrt {5} - 1\right )}}}{5 \, a^{5}} - \frac {1}{2 \, a^{5} x^{2}} \]
1/5*(2*sqrt(5)*arctan((a*(sqrt(5) - 1) + 4*x)/(a*sqrt(2*sqrt(5) + 10)))/(a ^2*sqrt(2*sqrt(5) + 10)) - 2*sqrt(5)*arctan(-(a*(sqrt(5) + 1) - 4*x)/(a*sq rt(-2*sqrt(5) + 10)))/(a^2*sqrt(-2*sqrt(5) + 10)) - log(a + x)/a^2 - log(- a*x*(sqrt(5) + 1) + 2*a^2 + 2*x^2)/(a^2*(sqrt(5) + 1)) + log(a*x*(sqrt(5) - 1) + 2*a^2 + 2*x^2)/(a^2*(sqrt(5) - 1)))/a^5 - 1/2/(a^5*x^2)
Time = 0.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=\frac {\sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{10 \, a^{7}} - \frac {\sqrt {2 \, \sqrt {5} + 10} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{10 \, a^{7}} - \frac {\sqrt {5} \log \left (a^{2} - \frac {1}{2} \, {\left (\sqrt {5} a + a\right )} x + x^{2}\right )}{20 \, a^{7}} + \frac {\sqrt {5} \log \left (a^{2} + \frac {1}{2} \, {\left (\sqrt {5} a - a\right )} x + x^{2}\right )}{20 \, a^{7}} + \frac {\log \left ({\left | a^{4} - a^{3} x + a^{2} x^{2} - a x^{3} + x^{4} \right |}\right )}{20 \, a^{7}} - \frac {\log \left ({\left | a + x \right |}\right )}{5 \, a^{7}} - \frac {1}{2 \, a^{5} x^{2}} \]
1/10*sqrt(-2*sqrt(5) + 10)*arctan((a*(sqrt(5) - 1) + 4*x)/(a*sqrt(2*sqrt(5 ) + 10)))/a^7 - 1/10*sqrt(2*sqrt(5) + 10)*arctan(-(a*(sqrt(5) + 1) - 4*x)/ (a*sqrt(-2*sqrt(5) + 10)))/a^7 - 1/20*sqrt(5)*log(a^2 - 1/2*(sqrt(5)*a + a )*x + x^2)/a^7 + 1/20*sqrt(5)*log(a^2 + 1/2*(sqrt(5)*a - a)*x + x^2)/a^7 + 1/20*log(abs(a^4 - a^3*x + a^2*x^2 - a*x^3 + x^4))/a^7 - 1/5*log(abs(a + x))/a^7 - 1/2/(a^5*x^2)
Time = 0.82 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=\frac {\ln \left (a^{20}-\frac {a^{19}\,x\,{\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}^3}{64}\right )\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^7}-\frac {1}{2\,a^5\,x^2}-\frac {\ln \left (a^{20}+\frac {x\,{\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}^3\,a^{19}}{64}\right )\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}{20\,a^7}-\frac {\ln \left (a+x\right )}{5\,a^7}+\frac {\ln \left (a^{20}-\frac {a^{19}\,x\,{\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}^3}{64}\right )\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^7}+\frac {\ln \left (a^{20}-\frac {a^{19}\,x\,{\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}^3}{64}\right )\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{20\,a^7} \]
(log(a^20 - (a^19*x*(5^(1/2) + (2*5^(1/2) - 10)^(1/2) + 1)^3)/64)*(5^(1/2) + (2*5^(1/2) - 10)^(1/2) + 1))/(20*a^7) - 1/(2*a^5*x^2) - (log(a^20 + (a^ 19*x*(5^(1/2) + (- 2*5^(1/2) - 10)^(1/2) - 1)^3)/64)*(5^(1/2) + (- 2*5^(1/ 2) - 10)^(1/2) - 1))/(20*a^7) - log(a + x)/(5*a^7) + (log(a^20 - (a^19*x*( 5^(1/2) - (2*5^(1/2) - 10)^(1/2) + 1)^3)/64)*(5^(1/2) - (2*5^(1/2) - 10)^( 1/2) + 1))/(20*a^7) + (log(a^20 - (a^19*x*((- 2*5^(1/2) - 10)^(1/2) - 5^(1 /2) + 1)^3)/64)*((- 2*5^(1/2) - 10)^(1/2) - 5^(1/2) + 1))/(20*a^7)
\[ \int \frac {1}{x^3 \left (a^5+x^5\right )} \, dx=\frac {-2 \left (\int \frac {x^{2}}{a^{5}+x^{5}}d x \right ) x^{2}-1}{2 a^{5} x^{2}} \]