Integrand size = 95, antiderivative size = 130 \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=-\frac {4 \left (a b x^2+(-a-b) x^3+x^4\right )^{3/4}}{x^2 (-b+x)}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}{a-x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}{a-x}\right ) \]
-4*(a*b*x^2+(-a-b)*x^3+x^4)^(3/4)/x^2/(-b+x)+2*d^(1/4)*arctan(d^(1/4)*(a*b *x^2+(-a-b)*x^3+x^4)^(1/4)/(a-x))-2*d^(1/4)*arctanh(d^(1/4)*(a*b*x^2+(-a-b )*x^3+x^4)^(1/4)/(a-x))
Time = 46.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\frac {4 (a-x)}{\sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\sqrt [4]{d} x \sqrt [4]{\frac {-b+x}{a-x}} \left (\arctan \left (\frac {a-x \left (1+\sqrt {d} \sqrt {\frac {-b+x}{a-x}}\right )}{\sqrt {2} \sqrt [4]{d} \sqrt {\frac {x}{a-x}} (-a+x) \sqrt [4]{\frac {-b+x}{a-x}}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} x \sqrt [4]{\frac {-b+x}{a-x}}}{\sqrt {\frac {x}{2 a-2 x}} \left (a+x \left (-1+\sqrt {d} \sqrt {\frac {-b+x}{a-x}}\right )\right )}\right )\right )}{\sqrt {\frac {x}{2 a-2 x}} \sqrt [4]{x^2 (-a+x) (-b+x)}} \]
Integrate[((-2*a*b + (3*a - b)*x)*(-a^3 + 3*a^2*x - 3*a*x^2 + x^3))/(x*(-b + x)*(x^2*(-a + x)*(-b + x))^(1/4)*(a^3 - 3*a^2*x + (3*a - b*d)*x^2 + (-1 + d)*x^3)),x]
(4*(a - x))/(x^2*(-a + x)*(-b + x))^(1/4) + (d^(1/4)*x*((-b + x)/(a - x))^ (1/4)*(ArcTan[(a - x*(1 + Sqrt[d]*Sqrt[(-b + x)/(a - x)]))/(Sqrt[2]*d^(1/4 )*Sqrt[x/(a - x)]*(-a + x)*((-b + x)/(a - x))^(1/4))] - ArcTanh[(d^(1/4)*x *((-b + x)/(a - x))^(1/4))/(Sqrt[x/(2*a - 2*x)]*(a + x*(-1 + Sqrt[d]*Sqrt[ (-b + x)/(a - x)])))]))/(Sqrt[x/(2*a - 2*x)]*(x^2*(-a + x)*(-b + x))^(1/4) )
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.
| \(\displaystyle \int \frac {\left (-a^3+3 a^2 x-3 a x^2+x^3\right ) (x (3 a-b)-2 a b)}{x (x-b) \sqrt [4]{x^2 (x-a) (x-b)} \left (a^3-3 a^2 x+x^2 (3 a-b d)+(d-1) x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2006 |
| \(\displaystyle \int \frac {(x-a)^3 (x (3 a-b)-2 a b)}{x (x-b) \sqrt [4]{x^2 (x-a) (x-b)} \left (a^3-3 a^2 x+x^2 (3 a-b d)+(d-1) x^3\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt [4]{-x (a+b)+a b+x^2} \int -\frac {(a-x)^3 (2 a b-(3 a-b) x)}{(b-x) x^{3/2} \sqrt [4]{x^2-(a+b) x+a b} \left (a^3-3 x a^2-(1-d) x^3+(3 a-b d) x^2\right )}dx}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{-x (a+b)+a b+x^2} \int \frac {(a-x)^3 (2 a b-(3 a-b) x)}{(b-x) x^{3/2} \sqrt [4]{x^2-(a+b) x+a b} \left (a^3-3 x a^2-(1-d) x^3+(3 a-b d) x^2\right )}dx}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{-x (a+b)+a b+x^2} \int \frac {(a-x)^3 (2 a b-(3 a-b) x)}{(b-x) x \sqrt [4]{x^2-(a+b) x+a b} \left (a^3-3 x a^2-(1-d) x^3+(3 a-b d) x^2\right )}d\sqrt {x}}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a-x} \sqrt [4]{b-x} \int \frac {(a-x)^{11/4} (2 a b-(3 a-b) x)}{(b-x)^{5/4} x \left (a^3-3 x a^2-(1-d) x^3+(3 a-b d) x^2\right )}d\sqrt {x}}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a-x} \sqrt [4]{b-x} \int \left (\frac {2 b (a-x)^{11/4}}{a^2 (b-x)^{5/4} x}+\frac {\left (-\left ((3 a-7 b) a^2\right )+2 b (1-d) x^2-2 b (3 a-b d) x\right ) (a-x)^{11/4}}{a^2 (b-x)^{5/4} \left (a^3-3 x a^2+3 \left (1-\frac {b d}{3 a}\right ) x^2 a-(1-d) x^3\right )}\right )d\sqrt {x}}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a-x} \sqrt [4]{b-x} \left (-(3 a-7 b) \int \frac {(a-x)^{11/4}}{(b-x)^{5/4} \left (a^3-3 x a^2+3 \left (1-\frac {b d}{3 a}\right ) x^2 a-(1-d) x^3\right )}d\sqrt {x}-\frac {2 b (3 a-b d) \int \frac {(a-x)^{11/4} x}{(b-x)^{5/4} \left (a^3-3 x a^2+3 \left (1-\frac {b d}{3 a}\right ) x^2 a-(1-d) x^3\right )}d\sqrt {x}}{a^2}+\frac {2 b (1-d) \int \frac {(a-x)^{11/4} x^2}{(b-x)^{5/4} \left (a^3-3 x a^2+3 \left (1-\frac {b d}{3 a}\right ) x^2 a-(1-d) x^3\right )}d\sqrt {x}}{a^2}-\frac {2 (a-x)^{3/4} \sqrt [4]{1-\frac {x}{b}} \operatorname {AppellF1}\left (-\frac {1}{2},-\frac {11}{4},\frac {5}{4},\frac {1}{2},\frac {x}{a},\frac {x}{b}\right )}{\sqrt {x} \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{b-x}}\right )}{\sqrt [4]{x^2 (a-x) (b-x)}}\) |
Int[((-2*a*b + (3*a - b)*x)*(-a^3 + 3*a^2*x - 3*a*x^2 + x^3))/(x*(-b + x)* (x^2*(-a + x)*(-b + x))^(1/4)*(a^3 - 3*a^2*x + (3*a - b*d)*x^2 + (-1 + d)* x^3)),x]
3.19.80.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Int[(u_.)*(Px_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^Expon[Px , x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; PolyQ[Px, x] && GtQ[Expon[P x, x], 1] && NeQ[Coeff[Px, x, 0], 0] && !MatchQ[Px, (a_.)*(v_)^Expon[Px, x ] /; FreeQ[a, x] && LinearQ[v, x]]
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]
\[\int \frac {\left (-2 a b +\left (3 a -b \right ) x \right ) \left (-a^{3}+3 a^{2} x -3 a \,x^{2}+x^{3}\right )}{x \left (-b +x \right ) \left (x^{2} \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{4}} \left (a^{3}-3 a^{2} x +\left (-b d +3 a \right ) x^{2}+\left (-1+d \right ) x^{3}\right )}d x\]
int((-2*a*b+(3*a-b)*x)*(-a^3+3*a^2*x-3*a*x^2+x^3)/x/(-b+x)/(x^2*(-a+x)*(-b +x))^(1/4)/(a^3-3*a^2*x+(-b*d+3*a)*x^2+(-1+d)*x^3),x)
int((-2*a*b+(3*a-b)*x)*(-a^3+3*a^2*x-3*a*x^2+x^3)/x/(-b+x)/(x^2*(-a+x)*(-b +x))^(1/4)/(a^3-3*a^2*x+(-b*d+3*a)*x^2+(-1+d)*x^3),x)
Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
integrate((-2*a*b+(3*a-b)*x)*(-a^3+3*a^2*x-3*a*x^2+x^3)/x/(-b+x)/(x^2*(-a+ x)*(-b+x))^(1/4)/(a^3-3*a^2*x+(-b*d+3*a)*x^2+(-1+d)*x^3),x, algorithm="fri cas")
Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
integrate((-2*a*b+(3*a-b)*x)*(-a**3+3*a**2*x-3*a*x**2+x**3)/x/(-b+x)/(x**2 *(-a+x)*(-b+x))**(1/4)/(a**3-3*a**2*x+(-b*d+3*a)*x**2+(-1+d)*x**3),x)
\[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\int { -\frac {{\left (a^{3} - 3 \, a^{2} x + 3 \, a x^{2} - x^{3}\right )} {\left (2 \, a b - {\left (3 \, a - b\right )} x\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - 3 \, a^{2} x - {\left (b d - 3 \, a\right )} x^{2}\right )} {\left (b - x\right )} x} \,d x } \]
integrate((-2*a*b+(3*a-b)*x)*(-a^3+3*a^2*x-3*a*x^2+x^3)/x/(-b+x)/(x^2*(-a+ x)*(-b+x))^(1/4)/(a^3-3*a^2*x+(-b*d+3*a)*x^2+(-1+d)*x^3),x, algorithm="max ima")
-integrate((a^3 - 3*a^2*x + 3*a*x^2 - x^3)*(2*a*b - (3*a - b)*x)/(((a - x) *(b - x)*x^2)^(1/4)*((d - 1)*x^3 + a^3 - 3*a^2*x - (b*d - 3*a)*x^2)*(b - x )*x), x)
\[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\int { -\frac {{\left (a^{3} - 3 \, a^{2} x + 3 \, a x^{2} - x^{3}\right )} {\left (2 \, a b - {\left (3 \, a - b\right )} x\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - 3 \, a^{2} x - {\left (b d - 3 \, a\right )} x^{2}\right )} {\left (b - x\right )} x} \,d x } \]
integrate((-2*a*b+(3*a-b)*x)*(-a^3+3*a^2*x-3*a*x^2+x^3)/x/(-b+x)/(x^2*(-a+ x)*(-b+x))^(1/4)/(a^3-3*a^2*x+(-b*d+3*a)*x^2+(-1+d)*x^3),x, algorithm="gia c")
integrate(-(a^3 - 3*a^2*x + 3*a*x^2 - x^3)*(2*a*b - (3*a - b)*x)/(((a - x) *(b - x)*x^2)^(1/4)*((d - 1)*x^3 + a^3 - 3*a^2*x - (b*d - 3*a)*x^2)*(b - x )*x), x)
Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=-\int \frac {\left (2\,a\,b-x\,\left (3\,a-b\right )\right )\,\left (a^3-3\,a^2\,x+3\,a\,x^2-x^3\right )}{x\,\left (b-x\right )\,{\left (x^2\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/4}\,\left (x^2\,\left (3\,a-b\,d\right )-3\,a^2\,x+a^3+x^3\,\left (d-1\right )\right )} \,d x \]
int(-((2*a*b - x*(3*a - b))*(3*a*x^2 - 3*a^2*x + a^3 - x^3))/(x*(b - x)*(x ^2*(a - x)*(b - x))^(1/4)*(x^2*(3*a - b*d) - 3*a^2*x + a^3 + x^3*(d - 1))) ,x)