Integrand size = 18, antiderivative size = 147 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=-\frac {1}{5} \sqrt {5-2 \sqrt {5}} \arctan \left (\frac {1-\sqrt {5}-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-\frac {1}{5} \sqrt {5+2 \sqrt {5}} \arctan \left (\frac {1}{2} \sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (1+\sqrt {5}-4 x\right )\right )-\frac {\log \left (2-x-\sqrt {5} x+2 x^2\right )}{2 \sqrt {5}}+\frac {\log \left (2-x+\sqrt {5} x+2 x^2\right )}{2 \sqrt {5}} \] Output:
-1/5*(5-2*5^(1/2))^(1/2)*arctan((1-5^(1/2)-4*x)/(10+2*5^(1/2))^(1/2))-1/5* (5+2*5^(1/2))^(1/2)*arctan(1/20*(50+10*5^(1/2))^(1/2)*(1+5^(1/2)-4*x))-1/1 0*ln(2-x-x*5^(1/2)+2*x^2)*5^(1/2)+1/10*ln(2-x+x*5^(1/2)+2*x^2)*5^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.35 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\text {RootSum}\left [1-\text {$\#$1}+\text {$\#$1}^2-\text {$\#$1}^3+\text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{-1+2 \text {$\#$1}-3 \text {$\#$1}^2+4 \text {$\#$1}^3}\&\right ] \] Input:
Integrate[(1 - x + x^2 - x^3 + x^4)^(-1),x]
Output:
RootSum[1 - #1 + #1^2 - #1^3 + #1^4 & , Log[x - #1]/(-1 + 2*#1 - 3*#1^2 + 4*#1^3) & ]
Time = 0.39 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2492, 2009}
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.
\(\displaystyle \int \frac {1}{x^4-x^3+x^2-x+1} \, dx\) |
\(\Big \downarrow \) 2492 |
\(\displaystyle \int \left (\frac {-2 x+\sqrt {5}+1}{\sqrt {5} \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )}-\frac {-2 x-\sqrt {5}+1}{\sqrt {5} \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{5} \sqrt {5-2 \sqrt {5}} \arctan \left (\frac {-4 x-\sqrt {5}+1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )-\frac {1}{5} \sqrt {5+2 \sqrt {5}} \arctan \left (\frac {-4 x+\sqrt {5}+1}{\sqrt {2 \left (5-\sqrt {5}\right )}}\right )+\frac {\log \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )}{2 \sqrt {5}}-\frac {\log \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )}{2 \sqrt {5}}\) |
Input:
Int[(1 - x + x^2 - x^3 + x^4)^(-1),x]
Output:
-1/5*(Sqrt[5 - 2*Sqrt[5]]*ArcTan[(1 - Sqrt[5] - 4*x)/Sqrt[2*(5 + Sqrt[5])] ]) - (Sqrt[5 + 2*Sqrt[5]]*ArcTan[(1 + Sqrt[5] - 4*x)/Sqrt[2*(5 - Sqrt[5])] ])/5 + Log[2 - (1 - Sqrt[5])*x + 2*x^2]/(2*Sqrt[5]) - Log[2 - (1 + Sqrt[5] )*x + 2*x^2]/(2*Sqrt[5])
Int[(Px_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2 + (d_.)*(x_)^3 + (e_.)*(x_)^4) ^(p_), x_Symbol] :> Simp[e^p Int[ExpandIntegrand[Px*(b/d + ((d + Sqrt[e*( (b^2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p*(b/d + ((d - Sqrt[e*((b^ 2 - 4*a*c)/a) + 8*a*d*(e/b)])/(2*e))*x + x^2)^p, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x] && ILtQ[p, 0] && EqQ[a*d^2 - b^2*e, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}-3 \textit {\_R}^{2}+2 \textit {\_R} -1}\) | \(45\) |
default | \(\frac {\ln \left (2-x +\sqrt {5}\, x +2 x^{2}\right ) \sqrt {5}}{10}+\frac {2 \left (-\frac {\sqrt {5}\, \left (\sqrt {5}-1\right )}{2}+5-\sqrt {5}\right ) \arctan \left (\frac {-1+\sqrt {5}+4 x}{\sqrt {10+2 \sqrt {5}}}\right )}{5 \sqrt {10+2 \sqrt {5}}}-\frac {\ln \left (2-x -\sqrt {5}\, x +2 x^{2}\right ) \sqrt {5}}{10}-\frac {2 \left (-\frac {\sqrt {5}\, \left (-\sqrt {5}-1\right )}{2}-5-\sqrt {5}\right ) \arctan \left (\frac {-1-\sqrt {5}+4 x}{\sqrt {10-2 \sqrt {5}}}\right )}{5 \sqrt {10-2 \sqrt {5}}}\) | \(143\) |
Input:
int(1/(x^4-x^3+x^2-x+1),x,method=_RETURNVERBOSE)
Output:
sum(1/(4*_R^3-3*_R^2+2*_R-1)*ln(x-_R),_R=RootOf(_Z^4-_Z^3+_Z^2-_Z+1))
Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=-\frac {1}{5} \, \sqrt {2 \, \sqrt {5} + 5} \arctan \left (\frac {1}{5} \, {\left (\sqrt {5} {\left (x + 1\right )} - 5 \, x\right )} \sqrt {2 \, \sqrt {5} + 5}\right ) + \frac {1}{5} \, \sqrt {-2 \, \sqrt {5} + 5} \arctan \left (\frac {1}{5} \, {\left (\sqrt {5} {\left (x + 1\right )} + 5 \, x\right )} \sqrt {-2 \, \sqrt {5} + 5}\right ) + \frac {1}{10} \, \sqrt {5} \log \left (2 \, x^{2} + \sqrt {5} x - x + 2\right ) - \frac {1}{10} \, \sqrt {5} \log \left (2 \, x^{2} - \sqrt {5} x - x + 2\right ) \] Input:
integrate(1/(x^4-x^3+x^2-x+1),x, algorithm="fricas")
Output:
-1/5*sqrt(2*sqrt(5) + 5)*arctan(1/5*(sqrt(5)*(x + 1) - 5*x)*sqrt(2*sqrt(5) + 5)) + 1/5*sqrt(-2*sqrt(5) + 5)*arctan(1/5*(sqrt(5)*(x + 1) + 5*x)*sqrt( -2*sqrt(5) + 5)) + 1/10*sqrt(5)*log(2*x^2 + sqrt(5)*x - x + 2) - 1/10*sqrt (5)*log(2*x^2 - sqrt(5)*x - x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 1287 vs. \(2 (122) = 244\).
Time = 0.60 (sec) , antiderivative size = 1287, normalized size of antiderivative = 8.76 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\text {Too large to display} \] Input:
integrate(1/(x**4-x**3+x**2-x+1),x)
Output:
sqrt(5)*log(x**2 + x*(-48/11 - 21*sqrt(5)/11 + 4*sqrt(10)*sqrt(sqrt(5) + 3 )/11 + 45*sqrt(2)*sqrt(sqrt(5) + 3)/22) - 1381*sqrt(10)*sqrt(sqrt(5) + 3)/ 484 - 3045*sqrt(2)*sqrt(sqrt(5) + 3)/484 + 2213*sqrt(5)/242 + 5217/242)/10 - sqrt(5)*log(x**2 + x*(-48/11 - 45*sqrt(2)*sqrt(3 - sqrt(5))/22 + 4*sqrt (10)*sqrt(3 - sqrt(5))/11 + 21*sqrt(5)/11) - 2213*sqrt(5)/242 - 1381*sqrt( 10)*sqrt(3 - sqrt(5))/484 + 3045*sqrt(2)*sqrt(3 - sqrt(5))/484 + 5217/242) /10 + 2*sqrt(-sqrt(10)*sqrt(3 - sqrt(5))/50 + 3/20)*atan(44*x/(-8*sqrt(5)* sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 3*sqrt(10)*sqrt(3 - sqrt(5))*sq rt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 18*sqrt(-2*sqrt(10)*sqrt(3 - sqrt (5)) + 15)) - 96/(-8*sqrt(5)*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 3* sqrt(10)*sqrt(3 - sqrt(5))*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 18*s qrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15)) - 45*sqrt(2)*sqrt(3 - sqrt(5))/(- 8*sqrt(5)*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 3*sqrt(10)*sqrt(3 - s qrt(5))*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 18*sqrt(-2*sqrt(10)*sqr t(3 - sqrt(5)) + 15)) + 8*sqrt(10)*sqrt(3 - sqrt(5))/(-8*sqrt(5)*sqrt(-2*s qrt(10)*sqrt(3 - sqrt(5)) + 15) + 3*sqrt(10)*sqrt(3 - sqrt(5))*sqrt(-2*sqr t(10)*sqrt(3 - sqrt(5)) + 15) + 18*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15 )) + 42*sqrt(5)/(-8*sqrt(5)*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 3*s qrt(10)*sqrt(3 - sqrt(5))*sqrt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15) + 18*sq rt(-2*sqrt(10)*sqrt(3 - sqrt(5)) + 15))) + 2*sqrt(-sqrt(10)*sqrt(sqrt(5...
\[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\int { \frac {1}{x^{4} - x^{3} + x^{2} - x + 1} \,d x } \] Input:
integrate(1/(x^4-x^3+x^2-x+1),x, algorithm="maxima")
Output:
integrate(1/(x^4 - x^3 + x^2 - x + 1), x)
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.69 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\frac {1}{5} \, \sqrt {-2 \, \sqrt {5} + 5} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) + \frac {1}{5} \, \sqrt {2 \, \sqrt {5} + 5} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) - \frac {1}{10} \, \sqrt {5} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {5} + 1\right )} + 1\right ) + \frac {1}{10} \, \sqrt {5} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {5} - 1\right )} + 1\right ) \] Input:
integrate(1/(x^4-x^3+x^2-x+1),x, algorithm="giac")
Output:
1/5*sqrt(-2*sqrt(5) + 5)*arctan((4*x + sqrt(5) - 1)/sqrt(2*sqrt(5) + 10)) + 1/5*sqrt(2*sqrt(5) + 5)*arctan((4*x - sqrt(5) - 1)/sqrt(-2*sqrt(5) + 10) ) - 1/10*sqrt(5)*log(x^2 - 1/2*x*(sqrt(5) + 1) + 1) + 1/10*sqrt(5)*log(x^2 + 1/2*x*(sqrt(5) - 1) + 1)
Time = 21.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44 \[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\sum _{k=1}^4\ln \left (\mathrm {root}\left (z^4-\frac {z}{25}+\frac {1}{125},z,k\right )\,\left (-4\,x+\mathrm {root}\left (z^4-\frac {z}{25}+\frac {1}{125},z,k\right )\,\left (25\,\mathrm {root}\left (z^4-\frac {z}{25}+\frac {1}{125},z,k\right )+15\,x-15\right )+1\right )\right )\,\mathrm {root}\left (z^4-\frac {z}{25}+\frac {1}{125},z,k\right ) \] Input:
int(1/(x^2 - x - x^3 + x^4 + 1),x)
Output:
symsum(log(root(z^4 - z/25 + 1/125, z, k)*(root(z^4 - z/25 + 1/125, z, k)* (25*root(z^4 - z/25 + 1/125, z, k) + 15*x - 15) - 4*x + 1))*root(z^4 - z/2 5 + 1/125, z, k), k, 1, 4)
\[ \int \frac {1}{1-x+x^2-x^3+x^4} \, dx=\int \frac {1}{x^{4}-x^{3}+x^{2}-x +1}d x \] Input:
int(1/(x^4-x^3+x^2-x+1),x)
Output:
int(1/(x**4 - x**3 + x**2 - x + 1),x)