Integrand size = 39, antiderivative size = 73 \[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=2 \text {arctanh}\left (\frac {x \sqrt {x+2 x^4+x^6}}{1+2 x^3+x^5}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x \sqrt {x+2 x^4+x^6}}{1+2 x^3+x^5}\right ) \]
2*arctanh(x*(x^6+2*x^4+x)^(1/2)/(x^5+2*x^3+1))-2*2^(1/2)*arctanh(2^(1/2)*x *(x^6+2*x^4+x)^(1/2)/(x^5+2*x^3+1))
Time = 4.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\frac {2 \sqrt {x+2 x^4+x^6} \left (\text {arctanh}\left (\frac {x^{3/2}}{\sqrt {1+2 x^3+x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x^{3/2}}{\sqrt {1+2 x^3+x^5}}\right )\right )}{\sqrt {x} \sqrt {1+2 x^3+x^5}} \]
(2*Sqrt[x + 2*x^4 + x^6]*(ArcTanh[x^(3/2)/Sqrt[1 + 2*x^3 + x^5]] - Sqrt[2] *ArcTanh[(Sqrt[2]*x^(3/2))/Sqrt[1 + 2*x^3 + x^5]]))/(Sqrt[x]*Sqrt[1 + 2*x^ 3 + x^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.
\(\displaystyle \int \frac {\left (2 x^5-3\right ) \sqrt {x^6+2 x^4+x}}{\left (x^5+1\right ) \left (x^5+x^3+1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x^6+2 x^4+x} \int -\frac {\sqrt {x} \left (3-2 x^5\right ) \sqrt {x^5+2 x^3+1}}{\left (x^5+1\right ) \left (x^5+x^3+1\right )}dx}{\sqrt {x} \sqrt {x^5+2 x^3+1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x^6+2 x^4+x} \int \frac {\sqrt {x} \left (3-2 x^5\right ) \sqrt {x^5+2 x^3+1}}{\left (x^5+1\right ) \left (x^5+x^3+1\right )}dx}{\sqrt {x} \sqrt {x^5+2 x^3+1}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x^6+2 x^4+x} \int \frac {x \left (3-2 x^5\right ) \sqrt {x^5+2 x^3+1}}{\left (x^5+1\right ) \left (x^5+x^3+1\right )}d\sqrt {x}}{\sqrt {x} \sqrt {x^5+2 x^3+1}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x^6+2 x^4+x} \int \left (\frac {x \sqrt {x^5+2 x^3+1} \left (5 x^2+3\right )}{x^5+x^3+1}+\frac {\sqrt {x^5+2 x^3+1}}{x+1}+\frac {\left (-x^3-3 x^2+2 x-1\right ) \sqrt {x^5+2 x^3+1}}{x^4-x^3+x^2-x+1}\right )d\sqrt {x}}{\sqrt {x} \sqrt {x^5+2 x^3+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x^6+2 x^4+x} \left (\frac {1}{2} i \int \frac {\sqrt {x^5+2 x^3+1}}{i-\sqrt {x}}d\sqrt {x}+\frac {1}{2} i \int \frac {\sqrt {x^5+2 x^3+1}}{\sqrt {x}+i}d\sqrt {x}+3 \int \frac {x \sqrt {x^5+2 x^3+1}}{x^5+x^3+1}d\sqrt {x}+5 \int \frac {x^3 \sqrt {x^5+2 x^3+1}}{x^5+x^3+1}d\sqrt {x}+\int \frac {\sqrt {x^5+2 x^3+1}}{-x^4+x^3-x^2+x-1}d\sqrt {x}+2 \int \frac {x \sqrt {x^5+2 x^3+1}}{x^4-x^3+x^2-x+1}d\sqrt {x}-3 \int \frac {x^2 \sqrt {x^5+2 x^3+1}}{x^4-x^3+x^2-x+1}d\sqrt {x}-\int \frac {x^3 \sqrt {x^5+2 x^3+1}}{x^4-x^3+x^2-x+1}d\sqrt {x}\right )}{\sqrt {x} \sqrt {x^5+2 x^3+1}}\) |
3.10.64.3.1 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 = 2.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(-\ln \left (\frac {-x^{2}+\sqrt {x \left (x^{5}+2 x^{3}+1\right )}}{x^{2}}\right )-2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{5}+2 x^{3}+1\right )}\, \sqrt {2}}{2 x^{2}}\right )+\ln \left (\frac {x^{2}+\sqrt {x \left (x^{5}+2 x^{3}+1\right )}}{x^{2}}\right )\) | \(80\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{5}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{3}+4 \sqrt {x^{6}+2 x^{4}+x}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\left (1+x \right ) \left (x^{4}-x^{3}+x^{2}-x +1\right )}\right )+\ln \left (\frac {x^{5}+3 x^{3}+2 \sqrt {x^{6}+2 x^{4}+x}\, x +1}{x^{5}+x^{3}+1}\right )\) | \(117\) |
-ln((-x^2+(x*(x^5+2*x^3+1))^(1/2))/x^2)-2*2^(1/2)*arctanh(1/2*(x*(x^5+2*x^ 3+1))^(1/2)/x^2*2^(1/2))+ln((x^2+(x*(x^5+2*x^3+1))^(1/2))/x^2)
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\frac {1}{2} \, \sqrt {2} \log \left (-\frac {x^{10} + 16 \, x^{8} + 32 \, x^{6} + 2 \, x^{5} + 16 \, x^{3} - 4 \, \sqrt {2} {\left (x^{6} + 4 \, x^{4} + x\right )} \sqrt {x^{6} + 2 \, x^{4} + x} + 1}{x^{10} + 2 \, x^{5} + 1}\right ) + \log \left (-\frac {x^{5} + 3 \, x^{3} + 2 \, \sqrt {x^{6} + 2 \, x^{4} + x} x + 1}{x^{5} + x^{3} + 1}\right ) \]
1/2*sqrt(2)*log(-(x^10 + 16*x^8 + 32*x^6 + 2*x^5 + 16*x^3 - 4*sqrt(2)*(x^6 + 4*x^4 + x)*sqrt(x^6 + 2*x^4 + x) + 1)/(x^10 + 2*x^5 + 1)) + log(-(x^5 + 3*x^3 + 2*sqrt(x^6 + 2*x^4 + x)*x + 1)/(x^5 + x^3 + 1))
Timed out. \[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\int { \frac {\sqrt {x^{6} + 2 \, x^{4} + x} {\left (2 \, x^{5} - 3\right )}}{{\left (x^{5} + x^{3} + 1\right )} {\left (x^{5} + 1\right )}} \,d x } \]
\[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\int { \frac {\sqrt {x^{6} + 2 \, x^{4} + x} {\left (2 \, x^{5} - 3\right )}}{{\left (x^{5} + x^{3} + 1\right )} {\left (x^{5} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {\left (-3+2 x^5\right ) \sqrt {x+2 x^4+x^6}}{\left (1+x^5\right ) \left (1+x^3+x^5\right )} \, dx=\int \frac {\left (2\,x^5-3\right )\,\sqrt {x^6+2\,x^4+x}}{\left (x^5+1\right )\,\left (x^5+x^3+1\right )} \,d x \]