\(\int \frac {1}{\frac {1}{x^2}+x^3} \, dx\) [326]

Optimal result

Integrand size = 9, antiderivative size = 168 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {1-\sqrt {5}-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {1}{2} \sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (1+\sqrt {5}-4 x\right )\right )+\frac {1}{5} \log (1+x)-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1-\sqrt {5}\right ) x+x^2\right )-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1+\sqrt {5}\right ) x+x^2\right ) \] Output:

1/10*(10-2*5^(1/2))^(1/2)*arctan((1-5^(1/2)-4*x)/(10+2*5^(1/2))^(1/2))-1/1 
0*(10+2*5^(1/2))^(1/2)*arctan(1/20*(50+10*5^(1/2))^(1/2)*(1+5^(1/2)-4*x))+ 
1/5*ln(1+x)-1/20*(5^(1/2)+1)*ln(1-1/2*x*(-5^(1/2)+1)+x^2)-1/20*(-5^(1/2)+1 
)*ln(1-1/2*(5^(1/2)+1)*x+x^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\frac {1}{20} \left (-2 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {1+\sqrt {5}-4 x}{\sqrt {10-2 \sqrt {5}}}\right )-2 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {-1+\sqrt {5}+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )+4 \log (1+x)-\left (1+\sqrt {5}\right ) \log \left (1+\frac {1}{2} \left (-1+\sqrt {5}\right ) x+x^2\right )+\left (-1+\sqrt {5}\right ) \log \left (1-\frac {1}{2} \left (1+\sqrt {5}\right ) x+x^2\right )\right ) \] Input:

Integrate[(x^(-2) + x^3)^(-1),x]
 

Output:

(-2*Sqrt[2*(5 + Sqrt[5])]*ArcTan[(1 + Sqrt[5] - 4*x)/Sqrt[10 - 2*Sqrt[5]]] 
 - 2*Sqrt[10 - 2*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
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2027, 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.

\(\displaystyle \int \frac {1}{x^3+\frac {1}{x^2}} \, dx\)

\(\Big \downarrow \) 2027

\(\displaystyle \int \frac {x^2}{x^5+1}dx\)

\(\Big \downarrow \) 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

\(\displaystyle \frac {1}{5} \left (-\sqrt {5} \int \frac {1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx-\frac {1}{4} \left (1+\sqrt {5}\right ) \int -\frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \left (\sqrt {5} \int \frac {1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx-\frac {1}{4} \left (1-\sqrt {5}\right ) \int -\frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \log (x+1)\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx-\sqrt {5} \int \frac {1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \left (\sqrt {5} \int \frac {1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx+\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx\right )+\frac {1}{5} \log (x+1)\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx-2 \sqrt {5} \int \frac {1}{-\left (4 x-\sqrt {5}-1\right )^2-2 \left (5-\sqrt {5}\right )}d\left (4 x-\sqrt {5}-1\right )\right )+\frac {1}{5} \left (\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx+2 \sqrt {5} \int \frac {1}{-\left (4 x+\sqrt {5}-1\right )^2-2 \left (5+\sqrt {5}\right )}d\left (4 x+\sqrt {5}-1\right )\right )+\frac {1}{5} \log (x+1)\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (1+\sqrt {5}\right ) \int \frac {-4 x-\sqrt {5}+1}{2 x^2-\left (1-\sqrt {5}\right ) x+2}dx-\sqrt {\frac {10}{5+\sqrt {5}}} \arctan \left (\frac {4 x+\sqrt {5}-1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )\right )+\frac {1}{5} \left (\frac {1}{4} \left (1-\sqrt {5}\right ) \int \frac {-4 x+\sqrt {5}+1}{2 x^2-\left (1+\sqrt {5}\right ) x+2}dx+\sqrt {\frac {10}{5-\sqrt {5}}} \arctan \left (\frac {4 x-\sqrt {5}-1}{\sqrt {2 \left (5-\sqrt {5}\right )}}\right )\right )+\frac {1}{5} \log (x+1)\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{5} \left (-\sqrt {\frac {10}{5+\sqrt {5}}} \arctan \left (\frac {4 x+\sqrt {5}-1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-\frac {1}{4} \left (1+\sqrt {5}\right ) \log \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )\right )+\frac {1}{5} \left (\sqrt {\frac {10}{5-\sqrt {5}}} \arctan \left (\frac {4 x-\sqrt {5}-1}{\sqrt {2 \left (5-\sqrt {5}\right )}}\right )-\frac {1}{4} \left (1-\sqrt {5}\right ) \log \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )\right )+\frac {1}{5} \log (x+1)\)

Input:

Int[(x^(-2) + x^3)^(-1),x]
 

Output:

Log[1 + x]/5 + (-(Sqrt[10/(5 + Sqrt[5])]*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[10/(5 - Sqrt[5])]*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
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 822

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]
 

rule 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.20

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{2}+x \right )\right )}{5}+\frac {\ln \left (x +1\right )}{5}\)	\(34\)
default	\(\frac {\ln \left (x +1\right )}{5}-\frac {\left (-\sqrt {5}+1\right ) \ln \left (2-x -\sqrt {5}\, x +2 x^{2}\right )}{20}-\frac {2 \left (-\sqrt {5}+1-\frac {\left (-\sqrt {5}+1\right ) \left (-\sqrt {5}-1\right )}{4}\right ) \arctan \left (\frac {-1-\sqrt {5}+4 x}{\sqrt {10-2 \sqrt {5}}}\right )}{5 \sqrt {10-2 \sqrt {5}}}+\frac {\left (-\sqrt {5}-1\right ) \ln \left (2-x +\sqrt {5}\, x +2 x^{2}\right )}{20}+\frac {2 \left (-\sqrt {5}-1-\frac {\left (-\sqrt {5}-1\right ) \left (\sqrt {5}-1\right )}{4}\right ) \arctan \left (\frac {-1+\sqrt {5}+4 x}{\sqrt {10+2 \sqrt {5}}}\right )}{5 \sqrt {10+2 \sqrt {5}}}\)	\(165\)

Input:

int(1/(1/x^2+x^3),x,method=_RETURNVERBOSE)
 

Output:

1/5*sum(_R*ln(_R^2+x),_R=RootOf(_Z^4+_Z^3+_Z^2+_Z+1))+1/5*ln(x+1)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=-\frac {1}{20} \, {\left (\sqrt {5} + 1\right )} \log \left (2 \, x^{2} + \sqrt {5} x - x + 2\right ) + \frac {1}{20} \, {\left (\sqrt {5} - 1\right )} \log \left (2 \, x^{2} - \sqrt {5} x - x + 2\right ) + \frac {1}{5} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (\sqrt {5} {\left (4 \, x - 1\right )} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) - \frac {1}{5} \, \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (\sqrt {5} {\left (4 \, x - 1\right )} + 5\right )} \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) + \frac {1}{5} \, \log \left (x + 1\right ) \] Input:

integrate(1/(1/x^2+x^3),x, algorithm="fricas")
 

Output:

-1/20*(sqrt(5) + 1)*log(2*x^2 + sqrt(5)*x - x + 2) + 1/20*(sqrt(5) - 1)*lo 
g(2*x^2 - sqrt(5)*x - x + 2) + 1/5*sqrt(1/2*sqrt(5) + 5/2)*arctan(1/10*(sq 
rt(5)*(4*x - 1) - 5)*sqrt(1/2*sqrt(5) + 5/2)) - 1/5*sqrt(-1/2*sqrt(5) + 5/ 
2)*arctan(1/10*(sqrt(5)*(4*x - 1) + 5)*sqrt(-1/2*sqrt(5) + 5/2)) + 1/5*log 
(x + 1)
 

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\frac {\log {\left (x + 1 \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} + 125 t^{3} + 25 t^{2} + 5 t + 1, \left ( t \mapsto t \log {\left (25 t^{2} + x \right )} \right )\right )} \] Input:

integrate(1/(1/x**2+x**3),x)
 

Output:

log(x + 1)/5 + RootSum(625*_t**4 + 125*_t**3 + 25*_t**2 + 5*_t + 1, Lambda 
(_t, _t*log(25*_t**2 + x)))
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=-\frac {2 \, \sqrt {5} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {2 \, \sqrt {5} + 10}} + \frac {2 \, \sqrt {5} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {-2 \, \sqrt {5} + 10}} + \frac {\log \left (2 \, x^{2} - x {\left (\sqrt {5} + 1\right )} + 2\right )}{5 \, {\left (\sqrt {5} + 1\right )}} - \frac {\log \left (2 \, x^{2} + x {\left (\sqrt {5} - 1\right )} + 2\right )}{5 \, {\left (\sqrt {5} - 1\right )}} + \frac {1}{5} \, \log \left (x + 1\right ) \] Input:

integrate(1/(1/x^2+x^3),x, algorithm="maxima")
 

Output:

-2/5*sqrt(5)*arctan((4*x + sqrt(5) - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt( 
5) + 10) + 2/5*sqrt(5)*arctan((4*x - sqrt(5) - 1)/sqrt(-2*sqrt(5) + 10))/s 
qrt(-2*sqrt(5) + 10) + 1/5*log(2*x^2 - x*(sqrt(5) + 1) + 2)/(sqrt(5) + 1) 
- 1/5*log(2*x^2 + x*(sqrt(5) - 1) + 2)/(sqrt(5) - 1) + 1/5*log(x + 1)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\frac {1}{20} \, {\left (\sqrt {5} - 1\right )} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {5} + 1\right )} + 1\right ) - \frac {1}{20} \, {\left (\sqrt {5} + 1\right )} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {5} - 1\right )} + 1\right ) - \frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) + \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) + \frac {1}{5} \, \log \left ({\left | x + 1 \right |}\right ) \] Input:

integrate(1/(1/x^2+x^3),x, algorithm="giac")
 

Output:

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))
 

Mupad [B] (verification not implemented)

Time = 9.65 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\frac {\ln \left (x+1\right )}{5}-\ln \left (1-\frac {x\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}-\sqrt {5}+1\right )}^3}{64}\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}-\frac {\sqrt {5}}{20}+\frac {1}{20}\right )+\ln \left (\frac {x\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}+\sqrt {5}-1\right )}^3}{64}+1\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}+\frac {\sqrt {5}}{20}-\frac {1}{20}\right )-\ln \left (1-\frac {x\,{\left (\sqrt {5}+\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{64}\right )\,\left (\frac {\sqrt {5}}{20}+\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )-\ln \left (1-\frac {x\,{\left (\sqrt {5}-\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{64}\right )\,\left (\frac {\sqrt {5}}{20}-\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right ) \] Input:

int(1/(1/x^2 + x^3),x)
 

Output:

log(x + 1)/5 - log(1 - (x*(2^(1/2)*(- 5^(1/2) - 5)^(1/2) - 5^(1/2) + 1)^3) 
/64)*((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 - 5^(1/2)/20 + 1/20) + log((x*(2^ 
(1/2)*(- 5^(1/2) - 5)^(1/2) + 5^(1/2) - 1)^3)/64 + 1)*((2^(1/2)*(- 5^(1/2) 
 - 5)^(1/2))/20 + 5^(1/2)/20 - 1/20) - log(1 - (x*(5^(1/2) + 2^(1/2)*(5^(1 
/2) - 5)^(1/2) + 1)^3)/64)*(5^(1/2)/20 + (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 
+ 1/20) - log(1 - (x*(5^(1/2) - 2^(1/2)*(5^(1/2) - 5)^(1/2) + 1)^3)/64)*(5 
^(1/2)/20 - (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20)
 

Reduce [F]

\[ \int \frac {1}{\frac {1}{x^2}+x^3} \, dx=\int \frac {x^{2}}{x^{5}+1}d x \] Input:

int(1/(1/x^2+x^3),x)
 

Output:

int(x**2/(x**5 + 1),x)