Integrand size = 37, antiderivative size = 52 \[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt {-x+x^5}}\right )}{a^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt {-x+x^5}}\right )}{a^{3/4}} \]
\[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx \]
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.
\(\displaystyle \int \frac {3 x^5+x}{\sqrt {x^5-x} \left (-a x^2+x^8-2 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {x \left (3 x^4+1\right )}{\sqrt {x^5-x} \left (-a x^2+x^8-2 x^4+1\right )}dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^4-1} \int \frac {\sqrt {x} \left (3 x^4+1\right )}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}dx}{\sqrt {x^5-x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4-1} \int \frac {x \left (3 x^4+1\right )}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}d\sqrt {x}}{\sqrt {x^5-x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4-1} \int \left (\frac {3 x^5}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}+\frac {x}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}\right )d\sqrt {x}}{\sqrt {x^5-x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4-1} \left (\int \frac {x}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}d\sqrt {x}+3 \int \frac {x^5}{\sqrt {x^4-1} \left (x^8-2 x^4-a x^2+1\right )}d\sqrt {x}\right )}{\sqrt {x^5-x}}\) |
3.7.60.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
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]
Time = 2.50 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (\frac {-a^{\frac {1}{4}} x -\sqrt {x^{5}-x}}{a^{\frac {1}{4}} x -\sqrt {x^{5}-x}}\right )+2 \arctan \left (\frac {\sqrt {x^{5}-x}}{a^{\frac {1}{4}} x}\right )}{2 a^{\frac {3}{4}}}\) | \(65\) |
-1/2*(ln((-a^(1/4)*x-(x^5-x)^(1/2))/(a^(1/4)*x-(x^5-x)^(1/2)))+2*arctan(1/ a^(1/4)/x*(x^5-x)^(1/2)))/a^(3/4)
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 405, normalized size of antiderivative = 7.79 \[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=-\frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} - 2 \, x^{4} + a x^{2} + 2 \, \sqrt {x^{5} - x} {\left (a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (a x^{4} - a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} + 2 \, {\left (a^{2} x^{5} - a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} - 2 \, x^{4} - a x^{2} + 1}\right ) + \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} - 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} - x} {\left (a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (a x^{4} - a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} + 2 \, {\left (a^{2} x^{5} - a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} - 2 \, x^{4} - a x^{2} + 1}\right ) - \frac {1}{4} i \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} - 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} - x} {\left (i \, a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (-i \, a x^{4} + i \, a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} - 2 \, {\left (a^{2} x^{5} - a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} - 2 \, x^{4} - a x^{2} + 1}\right ) + \frac {1}{4} i \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} - 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} - x} {\left (-i \, a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (i \, a x^{4} - i \, a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} - 2 \, {\left (a^{2} x^{5} - a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} - 2 \, x^{4} - a x^{2} + 1}\right ) \]
-1/4*(a^(-3))^(1/4)*log((x^8 - 2*x^4 + a*x^2 + 2*sqrt(x^5 - x)*(a^3*(a^(-3 ))^(3/4)*x + (a*x^4 - a)*(a^(-3))^(1/4)) + 2*(a^2*x^5 - a^2*x)*sqrt(a^(-3) ) + 1)/(x^8 - 2*x^4 - a*x^2 + 1)) + 1/4*(a^(-3))^(1/4)*log((x^8 - 2*x^4 + a*x^2 - 2*sqrt(x^5 - x)*(a^3*(a^(-3))^(3/4)*x + (a*x^4 - a)*(a^(-3))^(1/4) ) + 2*(a^2*x^5 - a^2*x)*sqrt(a^(-3)) + 1)/(x^8 - 2*x^4 - a*x^2 + 1)) - 1/4 *I*(a^(-3))^(1/4)*log((x^8 - 2*x^4 + a*x^2 - 2*sqrt(x^5 - x)*(I*a^3*(a^(-3 ))^(3/4)*x + (-I*a*x^4 + I*a)*(a^(-3))^(1/4)) - 2*(a^2*x^5 - a^2*x)*sqrt(a ^(-3)) + 1)/(x^8 - 2*x^4 - a*x^2 + 1)) + 1/4*I*(a^(-3))^(1/4)*log((x^8 - 2 *x^4 + a*x^2 - 2*sqrt(x^5 - x)*(-I*a^3*(a^(-3))^(3/4)*x + (I*a*x^4 - I*a)* (a^(-3))^(1/4)) - 2*(a^2*x^5 - a^2*x)*sqrt(a^(-3)) + 1)/(x^8 - 2*x^4 - a*x ^2 + 1))
Timed out. \[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\int { \frac {3 \, x^{5} + x}{{\left (x^{8} - 2 \, x^{4} - a x^{2} + 1\right )} \sqrt {x^{5} - x}} \,d x } \]
\[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\int { \frac {3 \, x^{5} + x}{{\left (x^{8} - 2 \, x^{4} - a x^{2} + 1\right )} \sqrt {x^{5} - x}} \,d x } \]
Time = 9.34 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.58 \[ \int \frac {x+3 x^5}{\sqrt {-x+x^5} \left (1-a x^2-2 x^4+x^8\right )} \, dx=\frac {\ln \left (\frac {a^2+2\,\sqrt {x^5-x}\,{\left (a^3\right )}^{3/4}-a^2\,x^4-a\,x\,\sqrt {a^3}}{a-a\,x^4+x\,\sqrt {a^3}}\right )}{2\,{\left (a^3\right )}^{1/4}}+\frac {\ln \left (\frac {a^2\,1{}\mathrm {i}-2\,\sqrt {x^5-x}\,{\left (a^3\right )}^{3/4}-a^2\,x^4\,1{}\mathrm {i}+a\,x\,\sqrt {a^3}\,1{}\mathrm {i}}{a\,x^4-a+x\,\sqrt {a^3}}\right )\,1{}\mathrm {i}}{2\,{\left (a^3\right )}^{1/4}} \]