Integrand size = 39, antiderivative size = 87 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt {x+2 x^2-2 x^4}}{-1-2 x+2 x^3}\right )-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x \sqrt {x+2 x^2-2 x^4}}{-1-2 x+2 x^3}\right ) \]
2*2^(1/2)*arctan(2^(1/2)*x*(-2*x^4+2*x^2+x)^(1/2)/(2*x^3-2*x-1))-2*3^(1/2) *arctan(3^(1/2)*x*(-2*x^4+2*x^2+x)^(1/2)/(2*x^3-2*x-1))
Time = 2.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\frac {2 \sqrt {x+2 x^2-2 x^4} \left (\sqrt {2} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-\frac {1}{2}-x+x^3}}\right )-\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x^{3/2}}{\sqrt {-1-2 x+2 x^3}}\right )\right )}{\sqrt {x} \sqrt {-1-2 x+2 x^3}} \]
(2*Sqrt[x + 2*x^2 - 2*x^4]*(Sqrt[2]*ArcTanh[x^(3/2)/Sqrt[-1/2 - x + x^3]] - Sqrt[3]*ArcTanh[(Sqrt[3]*x^(3/2))/Sqrt[-1 - 2*x + 2*x^3]]))/(Sqrt[x]*Sqr t[-1 - 2*x + 2*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.
\(\displaystyle \int \frac {(4 x+3) \sqrt {-2 x^4+2 x^2+x}}{(2 x+1) \left (x^3+2 x+1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {-2 x^4+2 x^2+x} \int \frac {\sqrt {x} (4 x+3) \sqrt {-2 x^3+2 x+1}}{(2 x+1) \left (x^3+2 x+1\right )}dx}{\sqrt {x} \sqrt {-2 x^3+2 x+1}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \sqrt {-2 x^4+2 x^2+x} \int \frac {x (4 x+3) \sqrt {-2 x^3+2 x+1}}{(2 x+1) \left (x^3+2 x+1\right )}d\sqrt {x}}{\sqrt {x} \sqrt {-2 x^3+2 x+1}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {2 \sqrt {-2 x^4+2 x^2+x} \int \left (\frac {\sqrt {-2 x^3+2 x+1} \left (-2 x^2+3 x-4\right )}{x^3+2 x+1}+\frac {4 \sqrt {-2 x^3+2 x+1}}{2 x+1}\right )d\sqrt {x}}{\sqrt {x} \sqrt {-2 x^3+2 x+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {-2 x^4+2 x^2+x} \left (2 i \int \frac {\sqrt {-2 x^3+2 x+1}}{i-\sqrt {2} \sqrt {x}}d\sqrt {x}+2 i \int \frac {\sqrt {-2 x^3+2 x+1}}{\sqrt {2} \sqrt {x}+i}d\sqrt {x}-4 \int \frac {\sqrt {-2 x^3+2 x+1}}{x^3+2 x+1}d\sqrt {x}+3 \int \frac {x \sqrt {-2 x^3+2 x+1}}{x^3+2 x+1}d\sqrt {x}-2 \int \frac {x^2 \sqrt {-2 x^3+2 x+1}}{x^3+2 x+1}d\sqrt {x}\right )}{\sqrt {x} \sqrt {-2 x^3+2 x+1}}\) |
3.12.80.3.1 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 = 19.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67
method | result | size |
default | \(2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {2}}{2 x^{2}}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {3}}{3 x^{2}}\right )\) | \(58\) |
pseudoelliptic | \(2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {2}}{2 x^{2}}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {3}}{3 x^{2}}\right )\) | \(58\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{3}+4 \sqrt {-2 x^{4}+2 x^{2}+x}\, x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{2 x +1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{3}+6 \sqrt {-2 x^{4}+2 x^{2}+x}\, x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{x^{3}+2 x +1}\right )\) | \(129\) |
elliptic | \(\text {Expression too large to display}\) | \(413672\) |
2*2^(1/2)*arctan(1/2*(-2*x^4+2*x^2+x)^(1/2)/x^2*2^(1/2))-2*3^(1/2)*arctan( 1/3*(-2*x^4+2*x^2+x)^(1/2)/x^2*3^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (75) = 150\).
Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.93 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\frac {2}{5} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x^{3} - 4 \, x^{2} - x + 1\right )}}{16 \, x^{5} - 16 \, x^{4} - 12 \, x^{3} + 8 \, x^{2} + 4 \, x - 1}\right ) - \frac {1}{5} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x^{2} + 5 \, x + 2\right )}}{32 \, x^{5} + 80 \, x^{4} + 84 \, x^{3} + 40 \, x^{2} + 6 \, x - 1}\right ) - \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} x}{5 \, x^{3} - 2 \, x - 1}\right ) \]
2/5*sqrt(2)*arctan(2*sqrt(2)*sqrt(-2*x^4 + 2*x^2 + x)*(4*x^3 - 4*x^2 - x + 1)/(16*x^5 - 16*x^4 - 12*x^3 + 8*x^2 + 4*x - 1)) - 1/5*sqrt(2)*arctan(2*s qrt(2)*sqrt(-2*x^4 + 2*x^2 + x)*(4*x^2 + 5*x + 2)/(32*x^5 + 80*x^4 + 84*x^ 3 + 40*x^2 + 6*x - 1)) - sqrt(3)*arctan(2*sqrt(3)*sqrt(-2*x^4 + 2*x^2 + x) *x/(5*x^3 - 2*x - 1))
Timed out. \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int { \frac {\sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x + 3\right )}}{{\left (x^{3} + 2 \, x + 1\right )} {\left (2 \, x + 1\right )}} \,d x } \]
\[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int { \frac {\sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x + 3\right )}}{{\left (x^{3} + 2 \, x + 1\right )} {\left (2 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int \frac {\left (4\,x+3\right )\,\sqrt {-2\,x^4+2\,x^2+x}}{\left (2\,x+1\right )\,\left (x^3+2\,x+1\right )} \,d x \]