Integrand size = 36, antiderivative size = 89 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\frac {1}{2} \text {arctanh}\left (\frac {x}{1-2 x+x^2-\sqrt {1-x+x^2-x^3+x^4}}\right )+\frac {3 \text {arctanh}\left (\frac {\sqrt {5} x}{1+2 x+x^2-\sqrt {1-x+x^2-x^3+x^4}}\right )}{2 \sqrt {5}} \] Output:
1/2*arctanh(x/(1-2*x+x^2-(x^4-x^3+x^2-x+1)^(1/2)))+3/10*arctanh(5^(1/2)*x/ (1+2*x+x^2-(x^4-x^3+x^2-x+1)^(1/2)))*5^(1/2)
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\frac {1}{2} \text {arctanh}\left (\frac {x}{1-2 x+x^2-\sqrt {1-x+x^2-x^3+x^4}}\right )+\frac {3 \text {arctanh}\left (\frac {\sqrt {5} x}{1+2 x+x^2-\sqrt {1-x+x^2-x^3+x^4}}\right )}{2 \sqrt {5}} \] Input:
Integrate[(1 - x + x^2)/((-1 + x^2)*Sqrt[1 - x + x^2 - x^3 + x^4]),x]
Output:
ArcTanh[x/(1 - 2*x + x^2 - Sqrt[1 - x + x^2 - x^3 + x^4])]/2 + (3*ArcTanh[ (Sqrt[5]*x)/(1 + 2*x + x^2 - Sqrt[1 - x + x^2 - x^3 + x^4])])/(2*Sqrt[5])
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.
\(\displaystyle \int \frac {x^2-x+1}{\left (x^2-1\right ) \sqrt {x^4-x^3+x^2-x+1}} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {2-x}{\left (x^2-1\right ) \sqrt {x^4-x^3+x^2-x+1}}+\frac {1}{\sqrt {x^4-x^3+x^2-x+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\sqrt {x^4-x^3+x^2-x+1}}dx-\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^4-x^3+x^2-x+1}}dx-\frac {3}{2} \int \frac {1}{(x+1) \sqrt {x^4-x^3+x^2-x+1}}dx\) |
Input:
Int[(1 - x + x^2)/((-1 + x^2)*Sqrt[1 - x + x^2 - x^3 + x^4]),x]
Output:
$Aborted
Time = 1.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\frac {3 x^{2}-4 x +3}{2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}}\right )}{4}+\frac {3 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (x^{2}+1\right ) \sqrt {5}}{2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}}\right )}{20}\) | \(69\) |
pseudoelliptic | \(-\frac {\operatorname {arctanh}\left (\frac {3 x^{2}-4 x +3}{2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}}\right )}{4}+\frac {3 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (x^{2}+1\right ) \sqrt {5}}{2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}}\right )}{20}\) | \(69\) |
trager | \(-\frac {\ln \left (-\frac {3 x^{2}+2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}-4 x +3}{\left (-1+x \right )^{2}}\right )}{4}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )-2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}}{\left (1+x \right )^{2}}\right )}{20}\) | \(94\) |
elliptic | \(\text {Expression too large to display}\) | \(101423\) |
Input:
int((x^2-x+1)/(x^2-1)/(x^4-x^3+x^2-x+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4*arctanh(1/2*(3*x^2-4*x+3)/(x^4-x^3+x^2-x+1)^(1/2))+3/20*5^(1/2)*arcta nh(1/2*(x^2+1)*5^(1/2)/(x^4-x^3+x^2-x+1)^(1/2))
Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.36 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\frac {3}{40} \, \sqrt {5} \log \left (-\frac {9 \, x^{4} - 4 \, x^{3} + 4 \, \sqrt {5} \sqrt {x^{4} - x^{3} + x^{2} - x + 1} {\left (x^{2} + 1\right )} + 14 \, x^{2} - 4 \, x + 9}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1}\right ) + \frac {1}{4} \, \log \left (\frac {3 \, x^{2} - 4 \, x - 2 \, \sqrt {x^{4} - x^{3} + x^{2} - x + 1} + 3}{x^{2} - 2 \, x + 1}\right ) \] Input:
integrate((x^2-x+1)/(x^2-1)/(x^4-x^3+x^2-x+1)^(1/2),x, algorithm="fricas")
Output:
3/40*sqrt(5)*log(-(9*x^4 - 4*x^3 + 4*sqrt(5)*sqrt(x^4 - x^3 + x^2 - x + 1) *(x^2 + 1) + 14*x^2 - 4*x + 9)/(x^4 + 4*x^3 + 6*x^2 + 4*x + 1)) + 1/4*log( (3*x^2 - 4*x - 2*sqrt(x^4 - x^3 + x^2 - x + 1) + 3)/(x^2 - 2*x + 1))
\[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\int \frac {x^{2} - x + 1}{\left (x - 1\right ) \left (x + 1\right ) \sqrt {x^{4} - x^{3} + x^{2} - x + 1}}\, dx \] Input:
integrate((x**2-x+1)/(x**2-1)/(x**4-x**3+x**2-x+1)**(1/2),x)
Output:
Integral((x**2 - x + 1)/((x - 1)*(x + 1)*sqrt(x**4 - x**3 + x**2 - x + 1)) , x)
\[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\int { \frac {x^{2} - x + 1}{\sqrt {x^{4} - x^{3} + x^{2} - x + 1} {\left (x^{2} - 1\right )}} \,d x } \] Input:
integrate((x^2-x+1)/(x^2-1)/(x^4-x^3+x^2-x+1)^(1/2),x, algorithm="maxima")
Output:
integrate((x^2 - x + 1)/(sqrt(x^4 - x^3 + x^2 - x + 1)*(x^2 - 1)), x)
\[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\int { \frac {x^{2} - x + 1}{\sqrt {x^{4} - x^{3} + x^{2} - x + 1} {\left (x^{2} - 1\right )}} \,d x } \] Input:
integrate((x^2-x+1)/(x^2-1)/(x^4-x^3+x^2-x+1)^(1/2),x, algorithm="giac")
Output:
integrate((x^2 - x + 1)/(sqrt(x^4 - x^3 + x^2 - x + 1)*(x^2 - 1)), x)
Timed out. \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=\int \frac {x^2-x+1}{\left (x^2-1\right )\,\sqrt {x^4-x^3+x^2-x+1}} \,d x \] Input:
int((x^2 - x + 1)/((x^2 - 1)*(x^2 - x - x^3 + x^4 + 1)^(1/2)),x)
Output:
int((x^2 - x + 1)/((x^2 - 1)*(x^2 - x - x^3 + x^4 + 1)^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt {1-x+x^2-x^3+x^4}} \, dx=-\frac {3 \sqrt {5}\, \mathrm {log}\left (x^{2}+2 x +1\right )}{20}+\frac {3 \sqrt {5}\, \mathrm {log}\left (2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}\, \sqrt {5}+5 x^{2}+5\right )}{20}-\frac {\mathrm {log}\left (x^{2}-2 x +1\right )}{4}+\frac {\mathrm {log}\left (-2 \sqrt {x^{4}-x^{3}+x^{2}-x +1}+3 x^{2}-4 x +3\right )}{4} \] Input:
int((x^2-x+1)/(x^2-1)/(x^4-x^3+x^2-x+1)^(1/2),x)
Output:
( - 3*sqrt(5)*log(x**2 + 2*x + 1) + 3*sqrt(5)*log(2*sqrt(x**4 - x**3 + x** 2 - x + 1)*sqrt(5) + 5*x**2 + 5) - 5*log(x**2 - 2*x + 1) + 5*log( - 2*sqrt (x**4 - x**3 + x**2 - x + 1) + 3*x**2 - 4*x + 3))/20