Integrand size = 39, antiderivative size = 354 \[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}{-2^{2/3} b-2^{2/3} a x+\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}\right )}{2^{2/3} a b^{2/3}}+\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )}{2^{2/3} a b^{2/3}}-\frac {\log \left (b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}}-\frac {\log \left (2^{2/3} b+2^{2/3} a x+2 \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )}{2^{2/3} a b^{2/3}}+\frac {\log \left (\sqrt [3]{2} b^2+2 \sqrt [3]{2} a b x+\sqrt [3]{2} a^2 x^2-2^{2/3} b^{4/3} \sqrt [3]{b^2-a^2 x^2}-2^{2/3} a \sqrt [3]{b} x \sqrt [3]{b^2-a^2 x^2}+2 b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}} \]
-1/2*3^(1/2)*arctan(3^(1/2)*b^(1/3)*(-a^2*x^2+b^2)^(1/3)/(-2^(2/3)*b-2^(2/ 3)*a*x+b^(1/3)*(-a^2*x^2+b^2)^(1/3)))*2^(1/3)/a/b^(2/3)+1/2*ln(b^(1/3)*(-a ^2*x^2+b^2)^(1/3))*2^(1/3)/a/b^(2/3)-1/4*ln(b^(2/3)*(-a^2*x^2+b^2)^(2/3))* 2^(1/3)/a/b^(2/3)-1/2*ln(2^(2/3)*b+2^(2/3)*a*x+2*b^(1/3)*(-a^2*x^2+b^2)^(1 /3))*2^(1/3)/a/b^(2/3)+1/4*ln(2^(1/3)*b^2+2*2^(1/3)*a*b*x+2^(1/3)*a^2*x^2- 2^(2/3)*b^(4/3)*(-a^2*x^2+b^2)^(1/3)-2^(2/3)*a*b^(1/3)*x*(-a^2*x^2+b^2)^(1 /3)+2*b^(2/3)*(-a^2*x^2+b^2)^(2/3))*2^(1/3)/a/b^(2/3)
Time = 0.45 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.85 \[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}{-2^{2/3} b-2^{2/3} a x+\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}}\right )-2 \log \left (\sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )+\log \left (b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )+2 \log \left (2^{2/3} b+2^{2/3} a x+2 \sqrt [3]{b} \sqrt [3]{b^2-a^2 x^2}\right )-\log \left (\sqrt [3]{2} b^2+2 \sqrt [3]{2} a b x+\sqrt [3]{2} a^2 x^2-2^{2/3} b^{4/3} \sqrt [3]{b^2-a^2 x^2}-2^{2/3} a \sqrt [3]{b} x \sqrt [3]{b^2-a^2 x^2}+2 b^{2/3} \left (b^2-a^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} a b^{2/3}} \]
-1/2*(2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*(b^2 - a^2*x^2)^(1/3))/(-(2^(2/3)* b) - 2^(2/3)*a*x + b^(1/3)*(b^2 - a^2*x^2)^(1/3))] - 2*Log[b^(1/3)*(b^2 - a^2*x^2)^(1/3)] + Log[b^(2/3)*(b^2 - a^2*x^2)^(2/3)] + 2*Log[2^(2/3)*b + 2 ^(2/3)*a*x + 2*b^(1/3)*(b^2 - a^2*x^2)^(1/3)] - Log[2^(1/3)*b^2 + 2*2^(1/3 )*a*b*x + 2^(1/3)*a^2*x^2 - 2^(2/3)*b^(4/3)*(b^2 - a^2*x^2)^(1/3) - 2^(2/3 )*a*b^(1/3)*x*(b^2 - a^2*x^2)^(1/3) + 2*b^(2/3)*(b^2 - a^2*x^2)^(2/3)])/(2 ^(2/3)*a*b^(2/3))
Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.47, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1342, 25, 1341}
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.
| \(\displaystyle \int \frac {a x-3 b}{\sqrt [3]{b^2-a^2 x^2} \left (a^2 x^2+3 b^2\right )} \, dx\) |
| \(\Big \downarrow \) 1342 |
| \(\displaystyle \frac {\sqrt [3]{1-\frac {a^2 x^2}{b^2}} \int -\frac {3 b-a x}{\left (3 b^2+a^2 x^2\right ) \sqrt [3]{1-\frac {a^2 x^2}{b^2}}}dx}{\sqrt [3]{b^2-a^2 x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{1-\frac {a^2 x^2}{b^2}} \int \frac {3 b-a x}{\left (3 b^2+a^2 x^2\right ) \sqrt [3]{1-\frac {a^2 x^2}{b^2}}}dx}{\sqrt [3]{b^2-a^2 x^2}}\) |
| \(\Big \downarrow \) 1341 |
| \(\displaystyle -\frac {\sqrt [3]{1-\frac {a^2 x^2}{b^2}} \left (-\frac {\log \left (a^2 x^2+3 b^2\right )}{2\ 2^{2/3} a}-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (\frac {a x}{b}+1\right )^{2/3}}{\sqrt {3} \sqrt [3]{1-\frac {a x}{b}}}\right )}{2^{2/3} a}+\frac {3 \log \left (\left (\frac {a x}{b}+1\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{1-\frac {a x}{b}}\right )}{2\ 2^{2/3} a}\right )}{\sqrt [3]{b^2-a^2 x^2}}\) |
-(((1 - (a^2*x^2)/b^2)^(1/3)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] - (2^(2/3)*(1 + (a*x)/b)^(2/3))/(Sqrt[3]*(1 - (a*x)/b)^(1/3))])/(2^(2/3)*a)) - Log[3*b^2 + a^2*x^2]/(2*2^(2/3)*a) + (3*Log[2^(1/3)*(1 - (a*x)/b)^(1/3) + (1 + (a*x)/ b)^(2/3)])/(2*2^(2/3)*a)))/(b^2 - a^2*x^2)^(1/3))
3.30.50.3.1 Defintions of rubi rules used
Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)^(1/3)*((d_) + (f_.)*(x_)^2)) , x_Symbol] :> Simp[Sqrt[3]*h*(ArcTan[1/Sqrt[3] - 2^(2/3)*((1 - 3*h*(x/g))^ (2/3)/(Sqrt[3]*(1 + 3*h*(x/g))^(1/3)))]/(2^(2/3)*a^(1/3)*f)), x] + (-Simp[3 *h*(Log[(1 - 3*h*(x/g))^(2/3) + 2^(1/3)*(1 + 3*h*(x/g))^(1/3)]/(2^(5/3)*a^( 1/3)*f)), x] + Simp[h*(Log[d + f*x^2]/(2^(5/3)*a^(1/3)*f)), x]) /; FreeQ[{a , c, d, f, g, h}, x] && EqQ[c*d + 3*a*f, 0] && EqQ[c*g^2 + 9*a*h^2, 0] && G tQ[a, 0]
Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)^(1/3)*((d_) + (f_.)*(x_)^2)) , x_Symbol] :> Simp[(1 + c*(x^2/a))^(1/3)/(a + c*x^2)^(1/3) Int[(g + h*x) /((1 + c*(x^2/a))^(1/3)*(d + f*x^2)), x], x] /; FreeQ[{a, c, d, f, g, h}, x ] && EqQ[c*d + 3*a*f, 0] && EqQ[c*g^2 + 9*a*h^2, 0] && !GtQ[a, 0]
\[\int \frac {a x -3 b}{\left (-a^{2} x^{2}+b^{2}\right )^{\frac {1}{3}} \left (a^{2} x^{2}+3 b^{2}\right )}d x\]
Timed out. \[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=\int \frac {a x - 3 b}{\sqrt [3]{- \left (a x - b\right ) \left (a x + b\right )} \left (a^{2} x^{2} + 3 b^{2}\right )}\, dx \]
\[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=\int { \frac {a x - 3 \, b}{{\left (a^{2} x^{2} + 3 \, b^{2}\right )} {\left (-a^{2} x^{2} + b^{2}\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=\int { \frac {a x - 3 \, b}{{\left (a^{2} x^{2} + 3 \, b^{2}\right )} {\left (-a^{2} x^{2} + b^{2}\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {-3 b+a x}{\sqrt [3]{b^2-a^2 x^2} \left (3 b^2+a^2 x^2\right )} \, dx=\int -\frac {3\,b-a\,x}{{\left (b^2-a^2\,x^2\right )}^{1/3}\,\left (a^2\,x^2+3\,b^2\right )} \,d x \]