Integrand size = 29, antiderivative size = 23 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\text {arctanh}\left (\frac {2 \sqrt {-x+x^4}}{1+2 x^2}\right ) \]
Time = 7.68 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\frac {2 \sqrt {x} \sqrt {-1+x^3} \text {arctanh}\left (\frac {(-1+x) \sqrt {x}}{\sqrt {-1+x^3}}\right )}{\sqrt {x \left (-1+x^3\right )}} \]
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 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \left (\frac {3}{4} i \int \frac {1}{\left (i-\sqrt {2} \sqrt {x}\right ) \sqrt {x^3-1}}d\sqrt {x}+\frac {3}{4} i \int \frac {1}{\left (\sqrt {2} \sqrt {x}+i\right ) \sqrt {x^3-1}}d\sqrt {x}-\frac {(1-x) \sqrt {x} \sqrt {\frac {x^2+x+1}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x}{1-\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {-\frac {(1-x) x}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x^3-1}}-\frac {1}{3} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x^3-1}}\right )\right )}{\sqrt {x^4-x}}\) |
3.3.19.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 3.64 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
method | result | size |
trager | \(-\ln \left (-\frac {-2 x^{2}+2 \sqrt {x^{4}-x}-1}{2 x +1}\right )\) | \(31\) |
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 (x -1\right )}}\, \left (x -1\right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\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 (x -1\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 (x -1\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 (x -1\right )}}\, \left (x -1\right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\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 (x -1\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 (x -1\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 (x -1\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 | \(\text {Expression too large to display}\) | \(774\) |
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\log \left (-\frac {2 \, x^{2} + 2 \, \sqrt {x^{4} - x} + 1}{2 \, x + 1}\right ) \]
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int \frac {2 x^{2} + 2 x - 1}{\sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (2 x + 1\right )}\, dx \]
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (2 \, x + 1\right )}} \,d x } \]
\[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (2 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {-1+2 x+2 x^2}{(1+2 x) \sqrt {-x+x^4}} \, dx=\int \frac {2\,x^2+2\,x-1}{\sqrt {x^4-x}\,\left (2\,x+1\right )} \,d x \]