Integrand size = 47, antiderivative size = 291 \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{-2 b+2 x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{2/3}}+\frac {\log \left (b-x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}\right )}{d^{2/3}}-\frac {\log \left (b^2-2 b x+x^2+\left (-b \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}+d^{2/3} \left (-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4\right )^{2/3}\right )}{2 d^{2/3}} \]
3^(1/2)*arctan(3^(1/2)*d^(1/3)*(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)*x^3+x^4) ^(1/3)/(-2*b+2*x+d^(1/3)*(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)*x^3+x^4)^(1/3) ))/d^(2/3)+ln(b-x+d^(1/3)*(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)*x^3+x^4)^(1/3 ))/d^(2/3)-1/2*ln(b^2-2*b*x+x^2+(-b*d^(1/3)+d^(1/3)*x)*(-a*b^2*x+(2*a*b+b^ 2)*x^2+(-a-2*b)*x^3+x^4)^(1/3)+d^(2/3)*(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)* x^3+x^4)^(2/3))/d^(2/3)
Time = 15.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.69 \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{(b-x)^2 x (-a+x)}}{-2 b+2 x+\sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}}\right )+2 \log \left (-b+x-\sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}\right )-\log \left (b^2-2 b x+x^2-b \sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}+\sqrt [3]{d} x \sqrt [3]{x (-a+x) (-b+x)^2}+d^{2/3} \left (x (-a+x) (-b+x)^2\right )^{2/3}\right )}{2 d^{2/3}} \]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*((b - x)^2*x*(-a + x))^(1/3))/(-2*b + 2 *x + d^(1/3)*(x*(-a + x)*(-b + x)^2)^(1/3))] + 2*Log[-b + x - d^(1/3)*(x*( -a + x)*(-b + x)^2)^(1/3)] - Log[b^2 - 2*b*x + x^2 - b*d^(1/3)*(x*(-a + x) *(-b + x)^2)^(1/3) + d^(1/3)*x*(x*(-a + x)*(-b + x)^2)^(1/3) + d^(2/3)*(x* (-a + x)*(-b + x)^2)^(2/3)])/(2*d^(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.
| \(\displaystyle \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (x-a) (x-b)^2} \left (-x (a d+1)+b+d x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \frac {x^2-2 b x+a b}{\sqrt [3]{x} \left (d x^2-(a d+1) x+b\right ) \sqrt [3]{x^3-(a+2 b) x^2+b (2 a+b) x-a b^2}}dx}{\sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \frac {\sqrt [3]{x} \left (x^2-2 b x+a b\right )}{\left (d x^2-(a d+1) x+b\right ) \sqrt [3]{x^3-(a+2 b) x^2+b (2 a+b) x-a b^2}}d\sqrt [3]{x}}{\sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \left (\frac {\sqrt [3]{x} ((a d-2 b d+1) x-b (1-a d))}{d \left (d x^2+(-a d-1) x+b\right ) \sqrt [3]{x^3-(a+2 b) x^2+b (2 a+b) x-a b^2}}+\frac {\sqrt [3]{x}}{d \sqrt [3]{x^3-(a+2 b) x^2+b (2 a+b) x-a b^2}}\right )d\sqrt [3]{x}}{\sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \frac {\sqrt [3]{x} \left (x^2-2 b x+a b\right )}{\sqrt [3]{(b-x)^2 (x-a)} \left (d x^2-(a d+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 2058 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x-a} (b-x)^{2/3} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \frac {\sqrt [3]{x} \left (x^2-2 b x+a b\right )}{(b-x)^{2/3} \sqrt [3]{x-a} \left (d x^2-(a d+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )} \sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x-a} (b-x)^{2/3} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \int \left (\frac {\sqrt [3]{x} ((a d-2 b d+1) x-b (1-a d))}{d (b-x)^{2/3} \sqrt [3]{x-a} \left (d x^2+(-a d-1) x+b\right )}+\frac {\sqrt [3]{x}}{d (b-x)^{2/3} \sqrt [3]{x-a}}\right )d\sqrt [3]{x}}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )} \sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x-a} (b-x)^{2/3} \sqrt [3]{-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3} \left (\frac {(a d-2 b d+1) \left (-\sqrt {a^2 d^2+2 a d-4 b d+1}+a d+1\right ) \int \frac {\sqrt [3]{x}}{(b-x)^{2/3} \sqrt [3]{x-a} \left (a d-2 x d-\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [3]{x}}{d \sqrt {a^2 d^2+2 a d-4 b d+1}}-\frac {(a d-2 b d+1) \left (\sqrt {a^2 d^2+2 a d-4 b d+1}+a d+1\right ) \int \frac {\sqrt [3]{x}}{(b-x)^{2/3} \sqrt [3]{x-a} \left (a d-2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [3]{x}}{d \sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {2 b (1-a d) \int \frac {\sqrt [3]{x}}{(b-x)^{2/3} \sqrt [3]{x-a} \left (a d-2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [3]{x}}{\sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {2 b (1-a d) \int \frac {\sqrt [3]{x}}{(b-x)^{2/3} \sqrt [3]{x-a} \left (-a d+2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}-1\right )}d\sqrt [3]{x}}{\sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {x^{2/3} \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {x}{a},\frac {x}{b}\right )}{2 d \sqrt [3]{x-a} (b-x)^{2/3}}\right )}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )} \sqrt [3]{-\left (x (a-x) (b-x)^2\right )}}\) |
3.29.42.3.1 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[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ (r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))] Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, 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]
\[\int \frac {a b -2 b x +x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (b -\left (a d +1\right ) x +d \,x^{2}\right )}d x\]
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
integrate((a*b-2*b*x+x^2)/(x*(-a+x)*(-b+x)^2)^(1/3)/(b-(a*d+1)*x+d*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{3}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
integrate((a*b-2*b*x+x^2)/(x*(-a+x)*(-b+x)^2)^(1/3)/(b-(a*d+1)*x+d*x^2),x, algorithm="maxima")
\[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{3}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int \frac {x^2-2\,b\,x+a\,b}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\left (d\,x^2+\left (-a\,d-1\right )\,x+b\right )} \,d x \]