Integrand size = 29, antiderivative size = 23 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\text {arctanh}\left (\frac {2 \sqrt {-x+x^4}}{1+2 x^2}\right ) \] Output:
arctanh(2*(x^4-x)^(1/2)/(2*x^2+1))
Time = 7.75 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\frac {2 \sqrt {x} \sqrt {-1+x^3} \text {arctanh}\left (\frac {(-1+x) \sqrt {x}}{\sqrt {-1+x^3}}\right )}{\sqrt {x \left (-1+x^3\right )}} \] Input:
Integrate[(-1 + 2*x + 2*x^2)/((1 + 2*x)*Sqrt[-x + x^4]),x]
Output:
(2*Sqrt[x]*Sqrt[-1 + x^3]*ArcTanh[((-1 + x)*Sqrt[x])/Sqrt[-1 + x^3]])/Sqrt [x*(-1 + x^3)]
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.
\(\displaystyle \int \frac {2 x^2+2 x-1}{(2 x+1) \sqrt {x^4-x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^3-1} \int -\frac {-2 x^2-2 x+1}{\sqrt {x} (2 x+1) \sqrt {x^3-1}}dx}{\sqrt {x^4-x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {x^3-1} \int \frac {-2 x^2-2 x+1}{\sqrt {x} (2 x+1) \sqrt {x^3-1}}dx}{\sqrt {x^4-x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \int \frac {-2 x^2-2 x+1}{(2 x+1) \sqrt {x^3-1}}d\sqrt {x}}{\sqrt {x^4-x}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \int \left (-\frac {x}{\sqrt {x^3-1}}+\frac {3}{2 (2 x+1) \sqrt {x^3-1}}-\frac {1}{2 \sqrt {x^3-1}}\right )d\sqrt {x}}{\sqrt {x^4-x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \left (\frac {3}{4} i \int \frac {1}{\left (i-\sqrt {2} \sqrt {x}\right ) \sqrt {x^3-1}}d\sqrt {x}+\frac {3}{4} i \int \frac {1}{\left (\sqrt {2} \sqrt {x}+i\right ) \sqrt {x^3-1}}d\sqrt {x}-\frac {(1-x) \sqrt {x} \sqrt {\frac {x^2+x+1}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x}{1-\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {-\frac {(1-x) x}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x^3-1}}-\frac {1}{3} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x^3-1}}\right )\right )}{\sqrt {x^4-x}}\) |
Input:
Int[(-1 + 2*x + 2*x^2)/((1 + 2*x)*Sqrt[-x + x^4]),x]
Output:
$Aborted
Time = 0.65 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26
method | result | size |
trager | \(\ln \left (-\frac {2 x^{2}+2 \sqrt {x^{4}-x}+1}{2 x +1}\right )\) | \(29\) |
default | \(-\frac {\ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{3}+\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \left (-1+x \right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )}{\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (-1+x \right ) \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \left (-1+x \right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}\, \left (\operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )+2 \operatorname {EllipticPi}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \frac {-\frac {3}{2}+\frac {3 i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (-1+x \right ) \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\) | \(509\) |
elliptic | \(\text {Expression too large to display}\) | \(774\) |
Input:
int((2*x^2+2*x-1)/(2*x+1)/(x^4-x)^(1/2),x,method=_RETURNVERBOSE)
Output:
ln(-(2*x^2+2*(x^4-x)^(1/2)+1)/(2*x+1))
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\log \left (-\frac {2 \, x^{2} + 2 \, \sqrt {x^{4} - x} + 1}{2 \, x + 1}\right ) \] Input:
integrate((2*x^2+2*x-1)/(1+2*x)/(x^4-x)^(1/2),x, algorithm="fricas")
Output:
log(-(2*x^2 + 2*sqrt(x^4 - x) + 1)/(2*x + 1))
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int \frac {2 x^{2} + 2 x - 1}{\sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (2 x + 1\right )}\, dx \] Input:
integrate((2*x**2+2*x-1)/(1+2*x)/(x**4-x)**(1/2),x)
Output:
Integral((2*x**2 + 2*x - 1)/(sqrt(x*(x - 1)*(x**2 + x + 1))*(2*x + 1)), x)
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (2 \, x + 1\right )}} \,d x } \] Input:
integrate((2*x^2+2*x-1)/(1+2*x)/(x^4-x)^(1/2),x, algorithm="maxima")
Output:
integrate((2*x^2 + 2*x - 1)/(sqrt(x^4 - x)*(2*x + 1)), x)
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (2 \, x + 1\right )}} \,d x } \] Input:
integrate((2*x^2+2*x-1)/(1+2*x)/(x^4-x)^(1/2),x, algorithm="giac")
Output:
integrate((2*x^2 + 2*x - 1)/(sqrt(x^4 - x)*(2*x + 1)), x)
Timed out. \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int \frac {2\,x^2+2\,x-1}{\sqrt {x^4-x}\,\left (2\,x+1\right )} \,d x \] Input:
int((2*x + 2*x^2 - 1)/((x^4 - x)^(1/2)*(2*x + 1)),x)
Output:
int((2*x + 2*x^2 - 1)/((x^4 - x)^(1/2)*(2*x + 1)), x)
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=2 \left (\int \frac {\sqrt {x}\, \sqrt {x^{3}-1}\, x}{2 x^{4}+x^{3}-2 x -1}d x \right )-\left (\int \frac {\sqrt {x}\, \sqrt {x^{3}-1}}{2 x^{5}+x^{4}-2 x^{2}-x}d x \right )+2 \left (\int \frac {\sqrt {x}\, \sqrt {x^{3}-1}}{2 x^{4}+x^{3}-2 x -1}d x \right ) \] Input:
int((2*x^2+2*x-1)/(1+2*x)/(x^4-x)^(1/2),x)
Output:
2*int((sqrt(x)*sqrt(x**3 - 1)*x)/(2*x**4 + x**3 - 2*x - 1),x) - int((sqrt( x)*sqrt(x**3 - 1))/(2*x**5 + x**4 - 2*x**2 - x),x) + 2*int((sqrt(x)*sqrt(x **3 - 1))/(2*x**4 + x**3 - 2*x - 1),x)