Integrand size = 35, antiderivative size = 60 \[ \int \frac {x-3 x^5}{\sqrt {x+x^5} \left (1-a x^2+2 x^4+x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x+x^5}}{1+x^4}\right )}{a^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x+x^5}}{1+x^4}\right )}{a^{3/4}} \]
-arctan(a^(1/4)*(x^5+x)^(1/2)/(x^4+1))/a^(3/4)+arctanh(a^(1/4)*(x^5+x)^(1/ 2)/(x^4+1))/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.
| \(\displaystyle \int \frac {x-3 x^5}{\sqrt {x^5+x} \left (-a x^2+x^8+2 x^4+1\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x \left (1-3 x^4\right )}{\sqrt {x^5+x} \left (-a x^2+x^8+2 x^4+1\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt {x^4+1} \int \frac {\sqrt {x} \left (1-3 x^4\right )}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}dx}{\sqrt {x^5+x}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4+1} \int \frac {x \left (1-3 x^4\right )}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}d\sqrt {x}}{\sqrt {x^5+x}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4+1} \int \left (\frac {x}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}-\frac {3 x^5}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}\right )d\sqrt {x}}{\sqrt {x^5+x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {x^4+1} \left (\int \frac {x}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}d\sqrt {x}-3 \int \frac {x^5}{\sqrt {x^4+1} \left (x^8+2 x^4-a x^2+1\right )}d\sqrt {x}\right )}{\sqrt {x^5+x}}\) |
3.8.88.3.1 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 = 2.76 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08
| method | result | size |
| pseudoelliptic | \(\frac {\ln \left (\frac {-a^{\frac {1}{4}} x -\sqrt {x \left (x^{4}+1\right )}}{a^{\frac {1}{4}} x -\sqrt {x \left (x^{4}+1\right )}}\right )+2 \arctan \left (\frac {\sqrt {x \left (x^{4}+1\right )}}{x \,a^{\frac {1}{4}}}\right )}{2 a^{\frac {3}{4}}}\) | \(65\) |
1/2/a^(3/4)*(ln((-a^(1/4)*x-(x*(x^4+1))^(1/2))/(a^(1/4)*x-(x*(x^4+1))^(1/2 )))+2*arctan((x*(x^4+1))^(1/2)/x/a^(1/4)))
Result contains complex when optimal does not.
Time = 0.31 (sec) , antiderivative size = 389, normalized size of antiderivative = 6.48 \[ \int \frac {x-3 x^5}{\sqrt {x+x^5} \left (1-a x^2+2 x^4+x^8\right )} \, dx=\frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} + 2 \, x^{4} + a x^{2} + 2 \, \sqrt {x^{5} + x} {\left (a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (a x^{4} + a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} + 2 \, {\left (a^{2} x^{5} + a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} + 2 \, x^{4} - a x^{2} + 1}\right ) - \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} + 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} + x} {\left (a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (a x^{4} + a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} + 2 \, {\left (a^{2} x^{5} + a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} + 2 \, x^{4} - a x^{2} + 1}\right ) + \frac {1}{4} i \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} + 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} + x} {\left (i \, a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (-i \, a x^{4} - i \, a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} - 2 \, {\left (a^{2} x^{5} + a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} + 2 \, x^{4} - a x^{2} + 1}\right ) - \frac {1}{4} i \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {x^{8} + 2 \, x^{4} + a x^{2} - 2 \, \sqrt {x^{5} + x} {\left (-i \, a^{3} \frac {1}{a^{3}}^{\frac {3}{4}} x + {\left (i \, a x^{4} + i \, a\right )} \frac {1}{a^{3}}^{\frac {1}{4}}\right )} - 2 \, {\left (a^{2} x^{5} + a^{2} x\right )} \sqrt {\frac {1}{a^{3}}} + 1}{x^{8} + 2 \, x^{4} - a x^{2} + 1}\right ) \]
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 = 10.73 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.62 \[ \int \frac {x-3 x^5}{\sqrt {x+x^5} \left (1-a x^2+2 x^4+x^8\right )} \, dx=\frac {\ln \left (\frac {512\,\sqrt {x^5+x}\,{\left (a^3\right )}^{7/4}+256\,a^5-27\,a^7+256\,a^5\,x^4-27\,x\,{\left (a^3\right )}^{5/2}-27\,a^7\,x^4-54\,a^5\,\sqrt {x^5+x}\,{\left (a^3\right )}^{3/4}+256\,a^4\,x\,\sqrt {a^3}}{a+a\,x^4-x\,\sqrt {a^3}}\right )}{2\,{\left (a^3\right )}^{1/4}}+\frac {\ln \left (\frac {54\,a^6\,\sqrt {x^5+x}\,{\left (a^3\right )}^{3/4}-512\,a\,\sqrt {x^5+x}\,{\left (a^3\right )}^{7/4}+a^6\,256{}\mathrm {i}-a^8\,27{}\mathrm {i}+a^6\,x^4\,256{}\mathrm {i}-a^8\,x^4\,27{}\mathrm {i}-a^5\,x\,\sqrt {a^3}\,256{}\mathrm {i}+a^7\,x\,\sqrt {a^3}\,27{}\mathrm {i}}{a+a\,x^4+x\,\sqrt {a^3}}\right )\,1{}\mathrm {i}}{2\,{\left (a^3\right )}^{1/4}} \]
log((512*(x + x^5)^(1/2)*(a^3)^(7/4) + 256*a^5 - 27*a^7 + 256*a^5*x^4 - 27 *x*(a^3)^(5/2) - 27*a^7*x^4 - 54*a^5*(x + x^5)^(1/2)*(a^3)^(3/4) + 256*a^4 *x*(a^3)^(1/2))/(a + a*x^4 - x*(a^3)^(1/2)))/(2*(a^3)^(1/4)) + (log((a^6*2 56i - a^8*27i + a^6*x^4*256i - a^8*x^4*27i + 54*a^6*(x + x^5)^(1/2)*(a^3)^ (3/4) - a^5*x*(a^3)^(1/2)*256i + a^7*x*(a^3)^(1/2)*27i - 512*a*(x + x^5)^( 1/2)*(a^3)^(7/4))/(a + a*x^4 + x*(a^3)^(1/2)))*1i)/(2*(a^3)^(1/4))