Integrand size = 39, antiderivative size = 62 \[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\frac {2 \left (3 a+x+3 a x^4\right ) \sqrt {x+x^5}}{3 \left (1+x^4\right )^2}-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {x+x^5}}{\sqrt {a} \left (1+x^4\right )}\right ) \]
2/3*(3*a*x^4+3*a+x)*(x^5+x)^(1/2)/(x^4+1)^2-2*a^(3/2)*arctanh((x^5+x)^(1/2 )/a^(1/2)/(x^4+1))
Time = 22.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\frac {2 \sqrt {x+x^5} \left (x+3 a \left (1+x^4\right )\right )}{3 \left (1+x^4\right )^2}-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {x+x^5}}{\sqrt {a} \left (1+x^4\right )}\right ) \]
(2*Sqrt[x + x^5]*(x + 3*a*(1 + x^4)))/(3*(1 + x^4)^2) - 2*a^(3/2)*ArcTanh[ Sqrt[x + x^5]/(Sqrt[a]*(1 + x^4))]
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.
\(\displaystyle \int \frac {x^2 \left (3 x^4-1\right )}{\left (x^4+1\right )^2 \sqrt {x^5+x} \left (a x^4+a-x\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^4+1} \int -\frac {x^{3/2} \left (1-3 x^4\right )}{\left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}dx}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {x^4+1} \int \frac {x^{3/2} \left (1-3 x^4\right )}{\left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}dx}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \int \frac {x^2 \left (1-3 x^4\right )}{\left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}d\sqrt {x}}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \int \left (\frac {(4 a-3 x) x^2}{a \left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}-\frac {3 x^2}{a \left (x^4+1\right )^{5/2}}\right )d\sqrt {x}}{\sqrt {x^5+x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^4+1} \left (-\frac {3 \int \frac {x^3}{\left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}d\sqrt {x}}{a}+4 \int \frac {x^2}{\left (x^4+1\right )^{5/2} \left (a x^4-x+a\right )}d\sqrt {x}+\frac {7 x^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-x^4\right )}{80 a}-\frac {7 x^{5/2}}{16 a \sqrt {x^4+1}}-\frac {x^{5/2}}{4 a \left (x^4+1\right )^{3/2}}\right )}{\sqrt {x^5+x}}\) |
3.9.19.3.1 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]
Time = 1.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {-6 a^{\frac {3}{2}} \sqrt {x \left (x^{4}+1\right )}\, \left (x^{4}+1\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{4}+1\right )}\, \sqrt {a}}{x}\right )+6 x \left (a \,x^{4}+a +\frac {1}{3} x \right )}{\sqrt {x \left (x^{4}+1\right )}\, \left (3 x^{4}+3\right )}\) | \(70\) |
6/(x*(x^4+1))^(1/2)*(-a^(3/2)*(x*(x^4+1))^(1/2)*(x^4+1)*arctanh((x*(x^4+1) )^(1/2)/x*a^(1/2))+x*(a*x^4+a+1/3*x))/(3*x^4+3)
Time = 0.31 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.73 \[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\left [\frac {3 \, {\left (a x^{8} + 2 \, a x^{4} + a\right )} \sqrt {a} \log \left (\frac {a^{2} x^{8} + 2 \, a^{2} x^{4} + 6 \, a x^{5} - 4 \, {\left (a x^{4} + a + x\right )} \sqrt {x^{5} + x} \sqrt {a} + a^{2} + 6 \, a x + x^{2}}{a^{2} x^{8} + 2 \, a^{2} x^{4} - 2 \, a x^{5} + a^{2} - 2 \, a x + x^{2}}\right ) + 4 \, {\left (3 \, a x^{4} + 3 \, a + x\right )} \sqrt {x^{5} + x}}{6 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}}, \frac {3 \, {\left (a x^{8} + 2 \, a x^{4} + a\right )} \sqrt {-a} \arctan \left (\frac {{\left (a x^{4} + a + x\right )} \sqrt {x^{5} + x} \sqrt {-a}}{2 \, {\left (a x^{5} + a x\right )}}\right ) + 2 \, {\left (3 \, a x^{4} + 3 \, a + x\right )} \sqrt {x^{5} + x}}{3 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}}\right ] \]
[1/6*(3*(a*x^8 + 2*a*x^4 + a)*sqrt(a)*log((a^2*x^8 + 2*a^2*x^4 + 6*a*x^5 - 4*(a*x^4 + a + x)*sqrt(x^5 + x)*sqrt(a) + a^2 + 6*a*x + x^2)/(a^2*x^8 + 2 *a^2*x^4 - 2*a*x^5 + a^2 - 2*a*x + x^2)) + 4*(3*a*x^4 + 3*a + x)*sqrt(x^5 + x))/(x^8 + 2*x^4 + 1), 1/3*(3*(a*x^8 + 2*a*x^4 + a)*sqrt(-a)*arctan(1/2* (a*x^4 + a + x)*sqrt(x^5 + x)*sqrt(-a)/(a*x^5 + a*x)) + 2*(3*a*x^4 + 3*a + x)*sqrt(x^5 + x))/(x^8 + 2*x^4 + 1)]
Timed out. \[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\text {Timed out} \]
\[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\int { \frac {{\left (3 \, x^{4} - 1\right )} x^{2}}{{\left (a x^{4} + a - x\right )} \sqrt {x^{5} + x} {\left (x^{4} + 1\right )}^{2}} \,d x } \]
\[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=\int { \frac {{\left (3 \, x^{4} - 1\right )} x^{2}}{{\left (a x^{4} + a - x\right )} \sqrt {x^{5} + x} {\left (x^{4} + 1\right )}^{2}} \,d x } \]
Time = 6.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int \frac {x^2 \left (-1+3 x^4\right )}{\left (1+x^4\right )^2 \left (a-x+a x^4\right ) \sqrt {x+x^5}} \, dx=a^{3/2}\,\ln \left (\frac {a+x-2\,\sqrt {a}\,\sqrt {x^5+x}+a\,x^4}{a\,x^4-x+a}\right )+\frac {2\,a\,\sqrt {x^5+x}}{x^4+1}+\frac {2\,x\,\sqrt {x^5+x}}{3\,{\left (x^4+1\right )}^2} \]