Integrand size = 52, antiderivative size = 481 \[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\frac {\left (1+i \sqrt {3}\right ) (-b+x)^{2/3} \left (\sqrt [3]{d} (a-x)^{2/3}-\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}+(-b+x)^{2/3}\right ) \left (\sqrt [3]{d} (a-x)^{2/3}+\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}+(-b+x)^{2/3}\right ) \left (-a \sqrt [3]{d}+\sqrt [3]{d} x+\sqrt [3]{a-x} (-b+x)^{2/3}\right ) \left (\frac {\left (-3 i+\sqrt {3}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{-b+x}}{-2 \sqrt [6]{d} \sqrt [3]{a-x}+\sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}+\frac {\left (3 i-\sqrt {3}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{-b+x}}{2 \sqrt [6]{d} \sqrt [3]{a-x}+\sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}+\frac {\left (1-i \sqrt {3}\right ) \text {arctanh}\left (\frac {(a-x)^{2/3} \sqrt [3]{-b+x}}{\sqrt [6]{d} (-a+x)}\right )}{2 (a-b) d^{5/6}}+\frac {i \left (i+\sqrt {3}\right ) \text {arctanh}\left (\frac {\sqrt [3]{d} (a-x)^{2/3}+(-b+x)^{2/3}}{\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}\right )}{2 \sqrt [3]{(b-x)^2 (-a+x)} \left (b^2-a^2 d-2 b x+2 a d x-(-1+d) x^2\right )} \]
1/2*(1+I*3^(1/2))*(-b+x)^(2/3)*(d^(1/3)*(a-x)^(2/3)-d^(1/6)*(a-x)^(1/3)*(- b+x)^(1/3)+(-b+x)^(2/3))*(d^(1/3)*(a-x)^(2/3)+d^(1/6)*(a-x)^(1/3)*(-b+x)^( 1/3)+(-b+x)^(2/3))*(-a*d^(1/3)+d^(1/3)*x+(a-x)^(1/3)*(-b+x)^(2/3))*(1/4*(- 3*I+3^(1/2))*arctan(3^(1/2)*(-b+x)^(1/3)/(-2*d^(1/6)*(a-x)^(1/3)+(-b+x)^(1 /3)))/(a-b)/d^(5/6)+1/4*(3*I-3^(1/2))*arctan(3^(1/2)*(-b+x)^(1/3)/(2*d^(1/ 6)*(a-x)^(1/3)+(-b+x)^(1/3)))/(a-b)/d^(5/6)+1/2*(1-I*3^(1/2))*arctanh((a-x )^(2/3)*(-b+x)^(1/3)/d^(1/6)/(-a+x))/(a-b)/d^(5/6)+1/4*I*(3^(1/2)+I)*arcta nh((d^(1/3)*(a-x)^(2/3)+(-b+x)^(2/3))/d^(1/6)/(a-x)^(1/3)/(-b+x)^(1/3))/(a -b)/d^(5/6))/((b-x)^2*(-a+x))^(1/3)/(b^2-a^2*d-2*b*x+2*a*d*x-(-1+d)*x^2)
Time = 1.17 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.49 \[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\frac {(b-x)^{2/3} \sqrt [3]{-a+x} \left (\sqrt {3} \left (\arctan \left (\frac {\sqrt {3} \sqrt [6]{d} \sqrt [3]{-a+x}}{-2 \sqrt [3]{b-x}+\sqrt [6]{d} \sqrt [3]{-a+x}}\right )-\arctan \left (\frac {\sqrt {3} \sqrt [6]{d} \sqrt [3]{-a+x}}{2 \sqrt [3]{b-x}+\sqrt [6]{d} \sqrt [3]{-a+x}}\right )\right )+2 \text {arctanh}\left (\frac {\sqrt [6]{d} \sqrt [3]{-a+x}}{\sqrt [3]{b-x}}\right )+\text {arctanh}\left (\frac {\sqrt [3]{b-x}}{\sqrt [6]{d} \sqrt [3]{-a+x}}+\frac {\sqrt [6]{d} \sqrt [3]{-a+x}}{\sqrt [3]{b-x}}\right )\right )}{2 (a-b) d^{5/6} \sqrt [3]{(b-x)^2 (-a+x)}} \]
Integrate[(-a + x)/(((-a + x)*(-b + x)^2)^(1/3)*(-b^2 + a^2*d + 2*(b - a*d )*x + (-1 + d)*x^2)),x]
((b - x)^(2/3)*(-a + x)^(1/3)*(Sqrt[3]*(ArcTan[(Sqrt[3]*d^(1/6)*(-a + x)^( 1/3))/(-2*(b - x)^(1/3) + d^(1/6)*(-a + x)^(1/3))] - ArcTan[(Sqrt[3]*d^(1/ 6)*(-a + x)^(1/3))/(2*(b - x)^(1/3) + d^(1/6)*(-a + x)^(1/3))]) + 2*ArcTan h[(d^(1/6)*(-a + x)^(1/3))/(b - x)^(1/3)] + ArcTanh[(b - x)^(1/3)/(d^(1/6) *(-a + x)^(1/3)) + (d^(1/6)*(-a + x)^(1/3))/(b - x)^(1/3)]))/(2*(a - b)*d^ (5/6)*((b - x)^2*(-a + x))^(1/3))
Time = 0.99 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.70, 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.
| \(\displaystyle \int \frac {x-a}{\sqrt [3]{(x-a) (x-b)^2} \left (a^2 d+2 x (b-a d)-b^2+(d-1) x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt [3]{x-a} (x-b)^{2/3} \int -\frac {(x-a)^{2/3}}{(x-b)^{2/3} \left (-d a^2+b^2+(1-d) x^2-2 (b-a d) x\right )}dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \int \frac {(x-a)^{2/3}}{(x-b)^{2/3} \left (-d a^2+b^2+(1-d) x^2-2 (b-a d) x\right )}dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 1205 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \int \left (\frac {(x-a)^{2/3} (1-d)}{(a-b) \sqrt {d} (x-b)^{2/3} \left (-2 \sqrt {d} (a-b)+2 b-2 a d-2 (1-d) x\right )}+\frac {(x-a)^{2/3} (1-d)}{(a-b) \sqrt {d} (x-b)^{2/3} \left (-2 \sqrt {d} (a-b)-2 b+2 a d+2 (1-d) x\right )}\right )dx}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [3]{x-a} (x-b)^{2/3} \left (\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [6]{d} \sqrt [3]{x-a}}{\sqrt {3} \sqrt [3]{x-b}}\right )}{2 d^{5/6} (a-b)}-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [6]{d} \sqrt [3]{x-a}}{\sqrt {3} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{2 d^{5/6} (a-b)}+\frac {\log \left (2 \left (\sqrt {d}+1\right ) \left (b-a \sqrt {d}\right )-2 (1-d) x\right )}{4 d^{5/6} (a-b)}-\frac {\log \left (2 (1-d) x-2 \left (1-\sqrt {d}\right ) \left (a \sqrt {d}+b\right )\right )}{4 d^{5/6} (a-b)}+\frac {3 \log \left (-\sqrt [6]{d} \sqrt [3]{x-a}-\sqrt [3]{x-b}\right )}{4 d^{5/6} (a-b)}-\frac {3 \log \left (\sqrt [6]{d} \sqrt [3]{x-a}-\sqrt [3]{x-b}\right )}{4 d^{5/6} (a-b)}\right )}{\sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\) |
-(((-a + x)^(1/3)*(-b + x)^(2/3)*((Sqrt[3]*ArcTan[1/Sqrt[3] - (2*d^(1/6)*( -a + x)^(1/3))/(Sqrt[3]*(-b + x)^(1/3))])/(2*(a - b)*d^(5/6)) - (Sqrt[3]*A rcTan[1/Sqrt[3] + (2*d^(1/6)*(-a + x)^(1/3))/(Sqrt[3]*(-b + x)^(1/3))])/(2 *(a - b)*d^(5/6)) + Log[2*(1 + Sqrt[d])*(b - a*Sqrt[d]) - 2*(1 - d)*x]/(4* (a - b)*d^(5/6)) - Log[-2*(1 - Sqrt[d])*(b + a*Sqrt[d]) + 2*(1 - d)*x]/(4* (a - b)*d^(5/6)) + (3*Log[-(d^(1/6)*(-a + x)^(1/3)) - (-b + x)^(1/3)])/(4* (a - b)*d^(5/6)) - (3*Log[d^(1/6)*(-a + x)^(1/3) - (-b + x)^(1/3)])/(4*(a - b)*d^(5/6))))/(-((a - x)*(b - x)^2))^(1/3))
3.31.67.3.1 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]
\[\int \frac {-a +x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (-b^{2}+a^{2} d +2 \left (-a d +b \right ) x +\left (-1+d \right ) x^{2}\right )}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1826 vs. \(2 (364) = 728\).
Time = 0.27 (sec) , antiderivative size = 1826, normalized size of antiderivative = 3.80 \[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\text {Too large to display} \]
integrate((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x ^2),x, algorithm="fricas")
1/4*(sqrt(-3) - 1)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b ^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log(-1/2*(((a^5 - 5*a^4*b + 10*a^3*b^2 - 1 0*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^ 2*b^4 + 5*a*b^5 - b^6)*d^4 + sqrt(-3)*((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^ 2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^ 4 + 5*a*b^5 - b^6)*d^4))*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15 *a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) + 2*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3))/(b - x)) - 1/4*(sqrt(-3) - 1)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log(1/2 *(((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5* b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4 + sqrt(-3)*(( a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4))*(1/((a^6 - 6*a^ 5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) - 2*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3))/(b - x)) - 1/4*( sqrt(-3) + 1)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log(-1/2*(((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2 *b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4 - sqrt(-3)*((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 ...
Timed out. \[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {1}{3}} {\left (a^{2} d + {\left (d - 1\right )} x^{2} - b^{2} - 2 \, {\left (a d - b\right )} x\right )}} \,d x } \]
integrate((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x ^2),x, algorithm="maxima")
-integrate((a - x)/((-(a - x)*(b - x)^2)^(1/3)*(a^2*d + (d - 1)*x^2 - b^2 - 2*(a*d - b)*x)), x)
\[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {1}{3}} {\left (a^{2} d + {\left (d - 1\right )} x^{2} - b^{2} - 2 \, {\left (a d - b\right )} x\right )}} \,d x } \]
integrate((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x ^2),x, algorithm="giac")
integrate(-(a - x)/((-(a - x)*(b - x)^2)^(1/3)*(a^2*d + (d - 1)*x^2 - b^2 - 2*(a*d - b)*x)), x)
Timed out. \[ \int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx=\int -\frac {a-x}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\left (a^2\,d+2\,x\,\left (b-a\,d\right )-b^2+x^2\,\left (d-1\right )\right )} \,d x \]