Integrand size = 36, antiderivative size = 157 \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} x^2}{\sqrt [3]{d} x^2+2 \sqrt [3]{-a x^2+x^3}}\right )}{d^{2/3}}+\frac {\log \left (-a \sqrt [3]{d} x^2+a \sqrt [3]{-a x^2+x^3}\right )}{d^{2/3}}-\frac {\log \left (a^2 d^{2/3} x^4+a^2 \sqrt [3]{d} x^2 \sqrt [3]{-a x^2+x^3}+a^2 \left (-a x^2+x^3\right )^{2/3}\right )}{2 d^{2/3}} \]
3^(1/2)*arctan(3^(1/2)*d^(1/3)*x^2/(d^(1/3)*x^2+2*(-a*x^2+x^3)^(1/3)))/d^( 2/3)+ln(-a*d^(1/3)*x^2+a*(-a*x^2+x^3)^(1/3))/d^(2/3)-1/2*ln(a^2*d^(2/3)*x^ 4+a^2*d^(1/3)*x^2*(-a*x^2+x^3)^(1/3)+a^2*(-a*x^2+x^3)^(2/3))/d^(2/3)
Time = 0.72 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=\frac {x^{4/3} (-a+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} x^{4/3}}{\sqrt [3]{d} x^{4/3}+2 \sqrt [3]{-a+x}}\right )+2 \log \left (a \left (-\sqrt [3]{d} x^{4/3}+\sqrt [3]{-a+x}\right )\right )-\log \left (a^2 \left (d^{2/3} x^{8/3}+\sqrt [3]{d} x^{4/3} \sqrt [3]{-a+x}+(-a+x)^{2/3}\right )\right )\right )}{2 d^{2/3} \left (x^2 (-a+x)\right )^{2/3}} \]
(x^(4/3)*(-a + x)^(2/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*x^(4/3))/(d^(1/ 3)*x^(4/3) + 2*(-a + x)^(1/3))] + 2*Log[a*(-(d^(1/3)*x^(4/3)) + (-a + x)^( 1/3))] - Log[a^2*(d^(2/3)*x^(8/3) + d^(1/3)*x^(4/3)*(-a + x)^(1/3) + (-a + x)^(2/3))]))/(2*d^(2/3)*(x^2*(-a + x))^(2/3))
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.
\(\displaystyle \int \frac {x^3 (3 x-4 a)}{\left (x^2 (x-a)\right )^{2/3} \left (a+d x^4-x\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{4/3} (x-a)^{2/3} \int -\frac {(4 a-3 x) x^{5/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}dx}{\left (-\left (x^2 (a-x)\right )\right )^{2/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {x^{4/3} (x-a)^{2/3} \int \frac {(4 a-3 x) x^{5/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}dx}{\left (-\left (x^2 (a-x)\right )\right )^{2/3}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {3 x^{4/3} (x-a)^{2/3} \int \frac {(4 a-3 x) x^{7/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}d\sqrt [3]{x}}{\left (-\left (x^2 (a-x)\right )\right )^{2/3}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3 x^{4/3} (x-a)^{2/3} \int \left (\frac {4 a x^{7/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}-\frac {3 x^{10/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}\right )d\sqrt [3]{x}}{\left (-\left (x^2 (a-x)\right )\right )^{2/3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 x^{4/3} (x-a)^{2/3} \left (4 a \int \frac {x^{7/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}d\sqrt [3]{x}-3 \int \frac {x^{10/3}}{(x-a)^{2/3} \left (d x^4-x+a\right )}d\sqrt [3]{x}\right )}{\left (-\left (x^2 (a-x)\right )\right )^{2/3}}\) |
3.22.50.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.45
method | result | size |
pseudoelliptic | \(a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{12}-3 \textit {\_Z}^{9}+3 \textit {\_Z}^{6}+a^{3} d -\textit {\_Z}^{3}\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (-\left (a -x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R}^{8}-2 \textit {\_R}^{5}+\textit {\_R}^{2}}\right )\) | \(70\) |
a^2*sum(ln((-_R*x+(-(a-x)*x^2)^(1/3))/x)/(_R^8-2*_R^5+_R^2),_R=RootOf(_Z^1 2-3*_Z^9+3*_Z^6+a^3*d-_Z^3))
Time = 0.26 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=-\frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} {\left ({\left (d^{2}\right )}^{\frac {1}{3}} d x^{2} + 2 \, {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} {\left (d^{2}\right )}^{\frac {2}{3}}\right )} {\left (d^{2}\right )}^{\frac {1}{6}}}{3 \, d^{2} x^{2}}\right ) - 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (d^{2}\right )}^{\frac {2}{3}} x^{2} - {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} d}{x^{2}}\right ) + {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (d^{2}\right )}^{\frac {1}{3}} d x^{4} + {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} {\left (d^{2}\right )}^{\frac {2}{3}} x^{2} + {\left (-a x^{2} + x^{3}\right )}^{\frac {2}{3}} d}{x^{4}}\right )}{2 \, d^{2}} \]
-1/2*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(1/3*sqrt(3)*((d^2)^(1/3)*d*x^2 + 2*(- a*x^2 + x^3)^(1/3)*(d^2)^(2/3))*(d^2)^(1/6)/(d^2*x^2)) - 2*(d^2)^(2/3)*log (((d^2)^(2/3)*x^2 - (-a*x^2 + x^3)^(1/3)*d)/x^2) + (d^2)^(2/3)*log(((d^2)^ (1/3)*d*x^4 + (-a*x^2 + x^3)^(1/3)*(d^2)^(2/3)*x^2 + (-a*x^2 + x^3)^(2/3)* d)/x^4))/d^2
Timed out. \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=\int { -\frac {{\left (4 \, a - 3 \, x\right )} x^{3}}{{\left (d x^{4} + a - x\right )} \left (-{\left (a - x\right )} x^{2}\right )^{\frac {2}{3}}} \,d x } \]
Time = 0.35 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=0 \]
Timed out. \[ \int \frac {x^3 (-4 a+3 x)}{\left (x^2 (-a+x)\right )^{2/3} \left (a-x+d x^4\right )} \, dx=\int -\frac {x^3\,\left (4\,a-3\,x\right )}{{\left (-x^2\,\left (a-x\right )\right )}^{2/3}\,\left (d\,x^4-x+a\right )} \,d x \]