Integrand size = 46, antiderivative size = 173 \[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} x}{\sqrt [3]{d} x+2 \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{\sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d} x+\sqrt [3]{a b x+(-a-b) x^2+x^3}\right )}{\sqrt [3]{d}}+\frac {\log \left (d^{2/3} x^2+\sqrt [3]{d} x \sqrt [3]{a b x+(-a-b) x^2+x^3}+\left (a b x+(-a-b) x^2+x^3\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
3^(1/2)*arctan(3^(1/2)*d^(1/3)*x/(d^(1/3)*x+2*(a*b*x+(-a-b)*x^2+x^3)^(1/3) ))/d^(1/3)-ln(-d^(1/3)*x+(a*b*x+(-a-b)*x^2+x^3)^(1/3))/d^(1/3)+1/2*ln(d^(2 /3)*x^2+d^(1/3)*x*(a*b*x+(-a-b)*x^2+x^3)^(1/3)+(a*b*x+(-a-b)*x^2+x^3)^(2/3 ))/d^(1/3)
Time = 15.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.79 \[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} x}{\sqrt [3]{d} x+2 \sqrt [3]{x (-a+x) (-b+x)}}\right )-2 \log \left (-\sqrt [3]{d} x+\sqrt [3]{x (-a+x) (-b+x)}\right )+\log \left (d^{2/3} x^2+\sqrt [3]{d} x \sqrt [3]{x (-a+x) (-b+x)}+(x (-a+x) (-b+x))^{2/3}\right )}{2 \sqrt [3]{d}} \]
Integrate[(-2*a*b + (a + b)*x)/((x*(-a + x)*(-b + x))^(1/3)*(-(a*b) + (a + b)*x + (-1 + d)*x^2)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*x)/(d^(1/3)*x + 2*(x*(-a + x)*(-b + x)) ^(1/3))] - 2*Log[-(d^(1/3)*x) + (x*(-a + x)*(-b + x))^(1/3)] + Log[d^(2/3) *x^2 + d^(1/3)*x*(x*(-a + x)*(-b + x))^(1/3) + (x*(-a + x)*(-b + x))^(2/3) ])/(2*d^(1/3))
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.
\(\displaystyle \int \frac {x (a+b)-2 a b}{\sqrt [3]{x (x-a) (x-b)} \left (x (a+b)-a b+(d-1) x^2\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {2 a b-(a+b) x}{\sqrt [3]{x} \sqrt [3]{x^2-(a+b) x+a b} \left ((1-d) x^2-(a+b) x+a b\right )}dx}{\sqrt [3]{x (a-x) (b-x)}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {\sqrt [3]{x} (2 a b-(a+b) x)}{\sqrt [3]{x^2-(a+b) x+a b} \left ((1-d) x^2-(a+b) x+a b\right )}d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \left (\frac {(-a-b) x^{4/3}}{\sqrt [3]{x^2-(a+b) x+a b} \left ((1-d) x^2-(a+b) x+a b\right )}+\frac {2 a b \sqrt [3]{x}}{\sqrt [3]{x^2-(a+b) x+a b} \left ((1-d) x^2-(a+b) x+a b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \left (\frac {4 a b (1-d) \int \frac {\sqrt [3]{x}}{\left (a+b-2 (1-d) x-\sqrt {a^2-2 b a+4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {a^2-2 a b (1-2 d)+b^2}}-\frac {(a+b) \left (-\sqrt {a^2+4 a b d-2 a b+b^2}+a+b\right ) \int \frac {\sqrt [3]{x}}{\left (a+b-2 (1-d) x-\sqrt {a^2-2 b a+4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {a^2-2 a b (1-2 d)+b^2}}+\frac {(a+b) \left (\sqrt {a^2+4 a b d-2 a b+b^2}+a+b\right ) \int \frac {\sqrt [3]{x}}{\left (a+b-2 (1-d) x+\sqrt {a^2-2 b a+4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {a^2-2 a b (1-2 d)+b^2}}+\frac {4 a b (1-d) \int \frac {\sqrt [3]{x}}{\left (-a-b+2 (1-d) x-\sqrt {a^2-2 b a+4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {a^2-2 a b (1-2 d)+b^2}}\right )}{\sqrt [3]{x (a-x) (b-x)}}\) |
3.23.71.3.1 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_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 1.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (d^{\frac {1}{3}} x +2 \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}\right )}{3 d^{\frac {1}{3}} x}\right )+\ln \left (\frac {-d^{\frac {1}{3}} x +\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {d^{\frac {2}{3}} x^{2}+d^{\frac {1}{3}} \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}} x +\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}}{d^{\frac {1}{3}}}\) | \(119\) |
int((-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(-a*b+(a+b)*x+(-1+d)*x^2),x,m ethod=_RETURNVERBOSE)
-(3^(1/2)*arctan(1/3*3^(1/2)*(d^(1/3)*x+2*(x*(a-x)*(b-x))^(1/3))/d^(1/3)/x )+ln((-d^(1/3)*x+(x*(a-x)*(b-x))^(1/3))/x)-1/2*ln((d^(2/3)*x^2+d^(1/3)*(x* (a-x)*(b-x))^(1/3)*x+(x*(a-x)*(b-x))^(2/3))/x^2))/d^(1/3)
Timed out. \[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(-a*b+(a+b)*x+(-1+d)*x^ 2),x, algorithm="fricas")
Timed out. \[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\int { -\frac {2 \, a b - {\left (a + b\right )} x}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}} {\left ({\left (d - 1\right )} x^{2} - a b + {\left (a + b\right )} x\right )}} \,d x } \]
integrate((-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(-a*b+(a+b)*x+(-1+d)*x^ 2),x, algorithm="maxima")
-integrate((2*a*b - (a + b)*x)/(((a - x)*(b - x)*x)^(1/3)*((d - 1)*x^2 - a *b + (a + b)*x)), x)
\[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=\int { -\frac {2 \, a b - {\left (a + b\right )} x}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}} {\left ({\left (d - 1\right )} x^{2} - a b + {\left (a + b\right )} x\right )}} \,d x } \]
integrate((-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(-a*b+(a+b)*x+(-1+d)*x^ 2),x, algorithm="giac")
integrate(-(2*a*b - (a + b)*x)/(((a - x)*(b - x)*x)^(1/3)*((d - 1)*x^2 - a *b + (a + b)*x)), x)
Timed out. \[ \int \frac {-2 a b+(a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a b+(a+b) x+(-1+d) x^2\right )} \, dx=-\int \frac {2\,a\,b-x\,\left (a+b\right )}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (\left (d-1\right )\,x^2+\left (a+b\right )\,x-a\,b\right )} \,d x \]