Integrand size = 52, antiderivative size = 541 \[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} (a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}{-2 a (-a+b)^{2/3}+2 (-a+b)^{2/3} x+(a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}\right )}{2 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}}-\frac {\log \left (a (-a+b)^{2/3}-(-a+b)^{2/3} x+(a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}\right )}{2 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}}+\frac {\log \left (a^3 \sqrt [3]{-a+b}-a^2 b \sqrt [3]{-a+b}-2 a^2 \sqrt [3]{-a+b} x+2 a b \sqrt [3]{-a+b} x+a \sqrt [3]{-a+b} x^2-b \sqrt [3]{-a+b} x^2+\left (a (a-b)^{2/3} (-a+b)^{2/3} \sqrt [3]{d}-(a-b)^{2/3} (-a+b)^{2/3} \sqrt [3]{d} x\right ) \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}+\left (-a \sqrt [3]{a-b} d^{2/3}+\sqrt [3]{a-b} b d^{2/3}\right ) \left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{2/3}\right )}{4 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}} \]
-1/2*3^(1/2)*arctan(3^(1/2)*(a-b)^(2/3)*d^(1/3)*(-a*b^2+(2*a*b+b^2)*x+(-a- 2*b)*x^2+x^3)^(1/3)/(-2*a*(-a+b)^(2/3)+2*(-a+b)^(2/3)*x+(a-b)^(2/3)*d^(1/3 )*(-a*b^2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(1/3)))/(a-b)^(1/3)/(-a+b)^(2/3) /d^(2/3)-1/2*ln(a*(-a+b)^(2/3)-(-a+b)^(2/3)*x+(a-b)^(2/3)*d^(1/3)*(-a*b^2+ (2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(1/3))/(a-b)^(1/3)/(-a+b)^(2/3)/d^(2/3)+1/ 4*ln(a^3*(-a+b)^(1/3)-a^2*b*(-a+b)^(1/3)-2*a^2*(-a+b)^(1/3)*x+2*a*b*(-a+b) ^(1/3)*x+a*(-a+b)^(1/3)*x^2-b*(-a+b)^(1/3)*x^2+(a*(a-b)^(2/3)*(-a+b)^(2/3) *d^(1/3)-(a-b)^(2/3)*(-a+b)^(2/3)*d^(1/3)*x)*(-a*b^2+(2*a*b+b^2)*x+(-a-2*b )*x^2+x^3)^(1/3)+(-a*(a-b)^(1/3)*d^(2/3)+(a-b)^(1/3)*b*d^(2/3))*(-a*b^2+(2 *a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(2/3))/(a-b)^(1/3)/(-a+b)^(2/3)/d^(2/3)
Time = 0.01 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.47 \[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\frac {(b-x)^{2/3} \sqrt [3]{-a+x} \left (-2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}}{\sqrt {3}}\right )+\log \left (1+\frac {\sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}-\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )-2 \log \left (-1+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )-2 \log \left (1+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )+\log \left (1+\frac {\sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )\right )}{4 (a-b) d^{2/3} \sqrt [3]{(b-x)^2 (-a+x)}} \]
Integrate[(-b + x)/(((-a + x)*(-b + x)^2)^(1/3)*(-a^2 + b^2*d + 2*(a - b*d )*x + (-1 + d)*x^2)),x]
((b - x)^(2/3)*(-a + x)^(1/3)*(-2*Sqrt[3]*ArcTan[(1 + (2*d^(1/3)*(b - x)^( 2/3))/(-a + x)^(2/3))/Sqrt[3]] + Log[1 + (d^(1/3)*(b - x)^(2/3))/(-a + x)^ (2/3) - (d^(1/6)*(b - x)^(1/3))/(-a + x)^(1/3)] - 2*Log[-1 + (d^(1/6)*(b - x)^(1/3))/(-a + x)^(1/3)] - 2*Log[1 + (d^(1/6)*(b - x)^(1/3))/(-a + x)^(1 /3)] + Log[1 + (d^(1/3)*(b - x)^(2/3))/(-a + x)^(2/3) + (d^(1/6)*(b - x)^( 1/3))/(-a + x)^(1/3)]))/(4*(a - b)*d^(2/3)*((b - x)^2*(-a + x))^(1/3))
Time = 0.91 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7270, 25, 1205, 2009}
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.
| \(\displaystyle \int \frac {x-b}{\sqrt [3]{(x-a) (x-b)^2} \left (-a^2+2 x (a-b d)+b^2 d+(d-1) x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt [3]{x-a} (x-b)^{2/3} \int -\frac {\sqrt [3]{x-b}}{\sqrt [3]{x-a} \left (a^2+(1-d) x^2-b^2 d-2 (a-b d) x\right )}dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \int \frac {\sqrt [3]{x-b}}{\sqrt [3]{x-a} \left (a^2+(1-d) x^2-b^2 d-2 (a-b d) x\right )}dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 1205 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \int \left (\frac {\sqrt [3]{x-b} (1-d)}{(a-b) \sqrt {d} \sqrt [3]{x-a} \left (2 a-2 b d-2 (1-d) x-2 (a-b) \sqrt {d}\right )}+\frac {\sqrt [3]{x-b} (1-d)}{(a-b) \sqrt {d} \sqrt [3]{x-a} \left (-2 a+2 b d+2 (1-d) x-2 (a-b) \sqrt {d}\right )}\right )dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \left (\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x-a}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-b}}\right )}{2 d^{2/3} (a-b)}+\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x-a}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{2 d^{2/3} (a-b)}-\frac {\log \left (2 \left (1-\sqrt {d}\right ) \left (a+b \sqrt {d}\right )-2 (1-d) x\right )}{4 d^{2/3} (a-b)}-\frac {\log \left (2 (1-d) x-2 \left (\sqrt {d}+1\right ) \left (a-b \sqrt {d}\right )\right )}{4 d^{2/3} (a-b)}+\frac {3 \log \left (-\frac {\sqrt [3]{x-a}}{\sqrt [6]{d}}-\sqrt [3]{x-b}\right )}{4 d^{2/3} (a-b)}+\frac {3 \log \left (\frac {\sqrt [3]{x-a}}{\sqrt [6]{d}}-\sqrt [3]{x-b}\right )}{4 d^{2/3} (a-b)}\right )}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
-(((-a + x)^(1/3)*(-b + x)^(2/3)*((Sqrt[3]*ArcTan[1/Sqrt[3] - (2*(-a + x)^ (1/3))/(Sqrt[3]*d^(1/6)*(-b + x)^(1/3))])/(2*(a - b)*d^(2/3)) + (Sqrt[3]*A rcTan[1/Sqrt[3] + (2*(-a + x)^(1/3))/(Sqrt[3]*d^(1/6)*(-b + x)^(1/3))])/(2 *(a - b)*d^(2/3)) - Log[2*(1 - Sqrt[d])*(a + b*Sqrt[d]) - 2*(1 - d)*x]/(4* (a - b)*d^(2/3)) - Log[-2*(1 + Sqrt[d])*(a - b*Sqrt[d]) + 2*(1 - d)*x]/(4* (a - b)*d^(2/3)) + (3*Log[-((-a + x)^(1/3)/d^(1/6)) - (-b + x)^(1/3)])/(4* (a - b)*d^(2/3)) + (3*Log[(-a + x)^(1/3)/d^(1/6) - (-b + x)^(1/3)])/(4*(a - b)*d^(2/3))))/(-((a - x)*(b - x)^2))^(1/3))
3.31.94.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x _) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^ n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
\[\int \frac {-b +x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (-a^{2}+b^{2} d +2 \left (-b d +a \right ) x +\left (-1+d \right ) x^{2}\right )}d x\]
Time = 0.26 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.60 \[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} {\left ({\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} + 2 \, {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} {\left (d^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, {\left (b^{2} d^{2} - 2 \, b d^{2} x + d^{2} x^{2}\right )}}\right ) - 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2} - 2 \, b x + x^{2}\right )} {\left (d^{2}\right )}^{\frac {2}{3}} - {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} d}{b^{2} - 2 \, b x + x^{2}}\right ) + {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {1}{3}} {\left (a d - d x\right )} - {\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} - {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} {\left (d^{2}\right )}^{\frac {2}{3}}}{b^{2} - 2 \, b x + x^{2}}\right )}{4 \, {\left (a - b\right )} d^{2}} \]
integrate((-b+x)/((-a+x)*(-b+x)^2)^(1/3)/(-a^2+b^2*d+2*(-b*d+a)*x+(-1+d)*x ^2),x, algorithm="fricas")
1/4*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(1/3*sqrt(3)*(d^2)^(1/6)*((b^2*d - 2*b* d*x + d*x^2)*(d^2)^(1/3) + 2*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2) *x)^(2/3)*(d^2)^(2/3))/(b^2*d^2 - 2*b*d^2*x + d^2*x^2)) - 2*(d^2)^(2/3)*lo g(-((b^2 - 2*b*x + x^2)*(d^2)^(2/3) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a *b + b^2)*x)^(2/3)*d)/(b^2 - 2*b*x + x^2)) + (d^2)^(2/3)*log(-((-a*b^2 - ( a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(a*d - d*x) - (b^2*d - 2*b*d*x + d*x^2)*(d^2)^(1/3) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^( 2/3)*(d^2)^(2/3))/(b^2 - 2*b*x + x^2)))/((a - b)*d^2)
Timed out. \[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {b - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {1}{3}} {\left (b^{2} d + {\left (d - 1\right )} x^{2} - a^{2} - 2 \, {\left (b d - a\right )} x\right )}} \,d x } \]
integrate((-b+x)/((-a+x)*(-b+x)^2)^(1/3)/(-a^2+b^2*d+2*(-b*d+a)*x+(-1+d)*x ^2),x, algorithm="maxima")
-integrate((b - x)/((-(a - x)*(b - x)^2)^(1/3)*(b^2*d + (d - 1)*x^2 - a^2 - 2*(b*d - a)*x)), x)
\[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {b - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {1}{3}} {\left (b^{2} d + {\left (d - 1\right )} x^{2} - a^{2} - 2 \, {\left (b d - a\right )} x\right )}} \,d x } \]
integrate((-b+x)/((-a+x)*(-b+x)^2)^(1/3)/(-a^2+b^2*d+2*(-b*d+a)*x+(-1+d)*x ^2),x, algorithm="giac")
integrate(-(b - x)/((-(a - x)*(b - x)^2)^(1/3)*(b^2*d + (d - 1)*x^2 - a^2 - 2*(b*d - a)*x)), x)
Timed out. \[ \int \frac {-b+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-a^2+b^2 d+2 (a-b d) x+(-1+d) x^2\right )} \, dx=\int -\frac {b-x}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\left (b^2\,d+2\,x\,\left (a-b\,d\right )-a^2+x^2\,\left (d-1\right )\right )} \,d x \]