Integrand size = 30, antiderivative size = 319 \[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\frac {\sqrt {-3+3 i \sqrt {3}} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} \sqrt [3]{a^2-b} x}{\sqrt [3]{a} \sqrt [3]{a^2-b} x-2 \sqrt [3]{-1} \sqrt [3]{-b^2 x^2+a^3 x^3}}\right )}{\sqrt {2} \sqrt [3]{a} \sqrt [3]{a^2-b} b}-\frac {i \left (-i+\sqrt {3}\right ) \log \left (\sqrt [3]{a} \sqrt [3]{a^2-b} x+\sqrt [3]{-1} \sqrt [3]{-b^2 x^2+a^3 x^3}\right )}{2 \sqrt [3]{a} \sqrt [3]{a^2-b} b}+\frac {\left (1+i \sqrt {3}\right ) \log \left (a^{2/3} \left (a^2-b\right )^{2/3} x^2-\sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{a^2-b} x \sqrt [3]{-b^2 x^2+a^3 x^3}+(-1)^{2/3} \left (-b^2 x^2+a^3 x^3\right )^{2/3}\right )}{4 \sqrt [3]{a} \sqrt [3]{a^2-b} b} \]
1/2*(-3+3*I*3^(1/2))^(1/2)*arctan(3^(1/2)*a^(1/3)*(a^2-b)^(1/3)*x/(a^(1/3) *(a^2-b)^(1/3)*x-2*(-1)^(1/3)*(a^3*x^3-b^2*x^2)^(1/3)))*2^(1/2)/a^(1/3)/(a ^2-b)^(1/3)/b-1/2*I*(-I+3^(1/2))*ln(a^(1/3)*(a^2-b)^(1/3)*x+(-1)^(1/3)*(a^ 3*x^3-b^2*x^2)^(1/3))/a^(1/3)/(a^2-b)^(1/3)/b+1/4*(1+I*3^(1/2))*ln(a^(2/3) *(a^2-b)^(2/3)*x^2-(-1)^(1/3)*a^(1/3)*(a^2-b)^(1/3)*x*(a^3*x^3-b^2*x^2)^(1 /3)+(-1)^(2/3)*(a^3*x^3-b^2*x^2)^(2/3))/a^(1/3)/(a^2-b)^(1/3)/b
Time = 1.45 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\frac {x^{2/3} \sqrt [3]{-b^2+a^3 x} \left (2 \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {3 \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}-\left (3 i+\sqrt {3}\right ) \sqrt [3]{-b^2+a^3 x}}\right )-i \left (-i+\sqrt {3}\right ) \left (2 \log \left (2 \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}+\left (1+i \sqrt {3}\right ) \sqrt [3]{-b^2+a^3 x}\right )-\log \left (-2 i a^{2/3} \left (a^2-b\right )^{2/3} x^{2/3}-\left (-i+\sqrt {3}\right ) \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x} \sqrt [3]{-b^2+a^3 x}+\left (i+\sqrt {3}\right ) \left (-b^2+a^3 x\right )^{2/3}\right )\right )\right )}{4 \sqrt [3]{a} \sqrt [3]{a^2-b} b \sqrt [3]{x^2 \left (-b^2+a^3 x\right )}} \]
(x^(2/3)*(-b^2 + a^3*x)^(1/3)*(2*Sqrt[-6 + (6*I)*Sqrt[3]]*ArcTan[(3*a^(1/3 )*(a^2 - b)^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3)*(a^2 - b)^(1/3)*x^(1/3) - (3*I + Sqrt[3])*(-b^2 + a^3*x)^(1/3))] - I*(-I + Sqrt[3])*(2*Log[2*a^(1/3)*(a^ 2 - b)^(1/3)*x^(1/3) + (1 + I*Sqrt[3])*(-b^2 + a^3*x)^(1/3)] - Log[(-2*I)* a^(2/3)*(a^2 - b)^(2/3)*x^(2/3) - (-I + Sqrt[3])*a^(1/3)*(a^2 - b)^(1/3)*x ^(1/3)*(-b^2 + a^3*x)^(1/3) + (I + Sqrt[3])*(-b^2 + a^3*x)^(2/3)])))/(4*a^ (1/3)*(a^2 - b)^(1/3)*b*(x^2*(-b^2 + a^3*x))^(1/3))
Time = 0.31 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2467, 25, 102}
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.
| \(\displaystyle \int \frac {1}{(a x-b) \sqrt [3]{a^3 x^3-b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \sqrt [3]{a^3 x-b^2} \int -\frac {1}{x^{2/3} (b-a x) \sqrt [3]{a^3 x-b^2}}dx}{\sqrt [3]{a^3 x^3-b^2 x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \sqrt [3]{a^3 x-b^2} \int \frac {1}{x^{2/3} (b-a x) \sqrt [3]{a^3 x-b^2}}dx}{\sqrt [3]{a^3 x^3-b^2 x^2}}\) |
| \(\Big \downarrow \) 102 |
| \(\displaystyle -\frac {x^{2/3} \sqrt [3]{a^3 x-b^2} \left (\frac {\log (b-a x)}{2 \sqrt [3]{a} b \sqrt [3]{a^2-b}}-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{a^3 x-b^2}}{\sqrt {3} \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{a^2-b}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{a} b \sqrt [3]{a^2-b}}-\frac {3 \log \left (\frac {\sqrt [3]{a^3 x-b^2}}{\sqrt [3]{a} \sqrt [3]{a^2-b}}-\sqrt [3]{x}\right )}{2 \sqrt [3]{a} b \sqrt [3]{a^2-b}}\right )}{\sqrt [3]{a^3 x^3-b^2 x^2}}\) |
-((x^(2/3)*(-b^2 + a^3*x)^(1/3)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(-b^2 + a^3*x)^(1/3))/(Sqrt[3]*a^(1/3)*(a^2 - b)^(1/3)*x^(1/3))])/(a^(1/3)*(a^2 - b)^(1/3)*b)) + Log[b - a*x]/(2*a^(1/3)*(a^2 - b)^(1/3)*b) - (3*Log[-x^(1/3 ) + (-b^2 + a^3*x)^(1/3)/(a^(1/3)*(a^2 - b)^(1/3))])/(2*a^(1/3)*(a^2 - b)^ (1/3)*b)))/(-(b^2*x^2) + a^3*x^3)^(1/3))
3.29.97.3.1 Defintions of rubi rules used
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.) *(x_))), x_] :> With[{q = Rt[(d*e - c*f)/(b*e - a*f), 3]}, Simp[(-Sqrt[3])* q*(ArcTan[1/Sqrt[3] + 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))]/(d*e - c*f)), x] + (Simp[q*(Log[e + f*x]/(2*(d*e - c*f))), x] - Simp[3*q*(Log[q *(a + b*x)^(1/3) - (c + d*x)^(1/3)]/(2*(d*e - c*f))), x])] /; FreeQ[{a, b, c, d, e, f}, 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]
Time = 0.75 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.58
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left ({\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x +2 \left (x^{2} \left (a^{3} x -b^{2}\right )\right )^{\frac {1}{3}}\right )}{3 {\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-{\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x +\left (x^{2} \left (a^{3} x -b^{2}\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {{\left (a \left (a^{2}-b \right )\right )}^{\frac {2}{3}} x^{2}+{\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} \left (x^{2} \left (a^{3} x -b^{2}\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (a^{3} x -b^{2}\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 {\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} b}\) | \(185\) |
1/2*(2*3^(1/2)*arctan(1/3*3^(1/2)*((a*(a^2-b))^(1/3)*x+2*(x^2*(a^3*x-b^2)) ^(1/3))/(a*(a^2-b))^(1/3)/x)+2*ln((-(a*(a^2-b))^(1/3)*x+(x^2*(a^3*x-b^2))^ (1/3))/x)-ln(((a*(a^2-b))^(2/3)*x^2+(a*(a^2-b))^(1/3)*(x^2*(a^3*x-b^2))^(1 /3)*x+(x^2*(a^3*x-b^2))^(2/3))/x^2))/(a*(a^2-b))^(1/3)/b
Time = 0.26 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.72 \[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\left [\frac {\sqrt {3} {\left (a^{3} - a b\right )} \sqrt {-\frac {1}{{\left (a^{3} - a b\right )}^{\frac {2}{3}}}} \log \left (-\frac {2 \, b^{2} x - {\left (3 \, a^{3} - a b\right )} x^{2} + 3 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {2}{3}} x + \sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {4}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )} x - 2 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}} {\left (a^{3} - a b\right )}^{\frac {2}{3}}\right )} \sqrt {-\frac {1}{{\left (a^{3} - a b\right )}^{\frac {2}{3}}}}}{a x^{2} - b x}\right ) + 2 \, {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} x - {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (\frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {1}{3}} x + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, {\left (a^{3} b - a b^{2}\right )}}, \frac {2 \, \sqrt {3} {\left (a^{3} - a b\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {1}{3}} x + 2 \, {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - a b\right )}^{\frac {1}{3}} x}\right ) + 2 \, {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} x - {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left (\frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} x^{2} + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a^{3} - a b\right )}^{\frac {1}{3}} x + {\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, {\left (a^{3} b - a b^{2}\right )}}\right ] \]
[1/2*(sqrt(3)*(a^3 - a*b)*sqrt(-1/(a^3 - a*b)^(2/3))*log(-(2*b^2*x - (3*a^ 3 - a*b)*x^2 + 3*(a^3*x^3 - b^2*x^2)^(1/3)*(a^3 - a*b)^(2/3)*x + sqrt(3)*( (a^3 - a*b)^(4/3)*x^2 + (a^3*x^3 - b^2*x^2)^(1/3)*(a^3 - a*b)*x - 2*(a^3*x ^3 - b^2*x^2)^(2/3)*(a^3 - a*b)^(2/3))*sqrt(-1/(a^3 - a*b)^(2/3)))/(a*x^2 - b*x)) + 2*(a^3 - a*b)^(2/3)*log(-((a^3 - a*b)^(1/3)*x - (a^3*x^3 - b^2*x ^2)^(1/3))/x) - (a^3 - a*b)^(2/3)*log(((a^3 - a*b)^(2/3)*x^2 + (a^3*x^3 - b^2*x^2)^(1/3)*(a^3 - a*b)^(1/3)*x + (a^3*x^3 - b^2*x^2)^(2/3))/x^2))/(a^3 *b - a*b^2), 1/2*(2*sqrt(3)*(a^3 - a*b)^(2/3)*arctan(1/3*sqrt(3)*((a^3 - a *b)^(1/3)*x + 2*(a^3*x^3 - b^2*x^2)^(1/3))/((a^3 - a*b)^(1/3)*x)) + 2*(a^3 - a*b)^(2/3)*log(-((a^3 - a*b)^(1/3)*x - (a^3*x^3 - b^2*x^2)^(1/3))/x) - (a^3 - a*b)^(2/3)*log(((a^3 - a*b)^(2/3)*x^2 + (a^3*x^3 - b^2*x^2)^(1/3)*( a^3 - a*b)^(1/3)*x + (a^3*x^3 - b^2*x^2)^(2/3))/x^2))/(a^3*b - a*b^2)]
\[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\int \frac {1}{\sqrt [3]{x^{2} \left (a^{3} x - b^{2}\right )} \left (a x - b\right )}\, dx \]
\[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\int { \frac {1}{{\left (a^{3} x^{3} - b^{2} x^{2}\right )}^{\frac {1}{3}} {\left (a x - b\right )}} \,d x } \]
Time = 19.94 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=\frac {3 \, {\left (a^{3} - a b\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {1}{3}} + 2 \, {\left (a^{3} - \frac {b^{2}}{x}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - a b\right )}^{\frac {1}{3}}}\right )}{\sqrt {3} a^{3} b - \sqrt {3} a b^{2}} - \frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left ({\left (a^{3} - a b\right )}^{\frac {2}{3}} + {\left (a^{3} - a b\right )}^{\frac {1}{3}} {\left (a^{3} - \frac {b^{2}}{x}\right )}^{\frac {1}{3}} + {\left (a^{3} - \frac {b^{2}}{x}\right )}^{\frac {2}{3}}\right )}{2 \, {\left (a^{3} b - a b^{2}\right )}} + \frac {{\left (a^{3} - a b\right )}^{\frac {2}{3}} \log \left ({\left | -{\left (a^{3} - a b\right )}^{\frac {1}{3}} + {\left (a^{3} - \frac {b^{2}}{x}\right )}^{\frac {1}{3}} \right |}\right )}{a^{3} b - a b^{2}} \]
3*(a^3 - a*b)^(2/3)*arctan(1/3*sqrt(3)*((a^3 - a*b)^(1/3) + 2*(a^3 - b^2/x )^(1/3))/(a^3 - a*b)^(1/3))/(sqrt(3)*a^3*b - sqrt(3)*a*b^2) - 1/2*(a^3 - a *b)^(2/3)*log((a^3 - a*b)^(2/3) + (a^3 - a*b)^(1/3)*(a^3 - b^2/x)^(1/3) + (a^3 - b^2/x)^(2/3))/(a^3*b - a*b^2) + (a^3 - a*b)^(2/3)*log(abs(-(a^3 - a *b)^(1/3) + (a^3 - b^2/x)^(1/3)))/(a^3*b - a*b^2)
Timed out. \[ \int \frac {1}{(-b+a x) \sqrt [3]{-b^2 x^2+a^3 x^3}} \, dx=-\int \frac {1}{{\left (a^3\,x^3-b^2\,x^2\right )}^{1/3}\,\left (b-a\,x\right )} \,d x \]