Integrand size = 28, antiderivative size = 156 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3}}{a d x}-\frac {\sqrt [3]{2} \sqrt [3]{b} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {\sqrt [3]{b} \log \left (a d-b d x^3\right )}{3\ 2^{2/3} a d}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a d} \]
-(b*x^3+a)^(1/3)/a/d/x+1/6*b^(1/3)*ln(-b*d*x^3+a*d)*2^(1/3)/a/d-1/2*b^(1/3 )*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(1/3)/a/d-1/3*2^(1/3)*b^(1/3)*ar ctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/a/d*3^(1/2)
Time = 0.52 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=-\frac {6 \sqrt [3]{a+b x^3}+2 \sqrt [3]{2} \sqrt {3} \sqrt [3]{b} x \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )+2 \sqrt [3]{2} \sqrt [3]{b} x \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )-\sqrt [3]{2} \sqrt [3]{b} x \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{6 a d x} \]
-1/6*(6*(a + b*x^3)^(1/3) + 2*2^(1/3)*Sqrt[3]*b^(1/3)*x*ArcTan[(Sqrt[3]*b^ (1/3)*x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] + 2*2^(1/3)*b^(1/3)*x*Lo g[-2*b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3)] - 2^(1/3)*b^(1/3)*x*Log[2*b^(2 /3)*x^2 + 2^(2/3)*b^(1/3)*x*(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)] )/(a*d*x)
Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {975, 27, 992}
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.
| \(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 975 |
| \(\displaystyle \frac {\int \frac {2 a b x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{a d}-\frac {\sqrt [3]{a+b x^3}}{a d x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b \int \frac {x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{d}-\frac {\sqrt [3]{a+b x^3}}{a d x}\) |
| \(\Big \downarrow \) 992 |
| \(\displaystyle \frac {2 b \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a b^{2/3}}+\frac {\log \left (a-b x^3\right )}{6\ 2^{2/3} a b^{2/3}}-\frac {\log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2\ 2^{2/3} a b^{2/3}}\right )}{d}-\frac {\sqrt [3]{a+b x^3}}{a d x}\) |
-((a + b*x^3)^(1/3)/(a*d*x)) + (2*b*(-(ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/( a + b*x^3)^(1/3))/Sqrt[3]]/(2^(2/3)*Sqrt[3]*a*b^(2/3))) + Log[a - b*x^3]/( 6*2^(2/3)*a*b^(2/3)) - Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*2^(2/ 3)*a*b^(2/3))))/d
3.6.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/ (a*e*(m + 1))), x] - Simp[1/(a*e^n*(m + 1)) Int[(e*x)^(m + n)*(a + b*x^n) ^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomi alQ[a, b, c, d, e, m, n, p, q, x]
Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3 ))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c* q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Time = 4.80 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96
| method | result | size |
| pseudoelliptic | \(\frac {2 b^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) x -2 b^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) x +b^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\frac {2^{\frac {2}{3}} b^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) x -6 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{6 a d x}\) | \(150\) |
1/6*(2*b^(1/3)*2^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3) +b^(1/3)*x)/b^(1/3)/x)*x-2*b^(1/3)*2^(1/3)*ln((-2^(1/3)*b^(1/3)*x+(b*x^3+a )^(1/3))/x)*x+b^(1/3)*2^(1/3)*ln((2^(2/3)*b^(2/3)*x^2+2^(1/3)*b^(1/3)*(b*x ^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)*x-6*(b*x^3+a)^(1/3))/a/d/x
Leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (125) = 250\).
Time = 84.46 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.53 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=-\frac {2 \, \sqrt {3} 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \arctan \left (\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} {\left (19 \, b^{2} x^{8} + 16 \, a b x^{5} + a^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} + 6 \, \sqrt {3} 2^{\frac {1}{3}} {\left (5 \, b^{2} x^{7} - 4 \, a b x^{4} - a^{2} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {1}{3}} + \sqrt {3} {\left (71 \, b^{3} x^{9} + 111 \, a b^{2} x^{6} + 33 \, a^{2} b x^{3} + a^{3}\right )}}{3 \, {\left (109 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} - a^{3}\right )}}\right ) - 2 \cdot 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \log \left (-\frac {6 \cdot 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} b x^{2} + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2^{\frac {2}{3}} {\left (b x^{3} - a\right )} \left (-b\right )^{\frac {2}{3}}}{b x^{3} - a}\right ) + 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (5 \, b x^{4} + a x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (19 \, b^{2} x^{6} + 16 \, a b x^{3} + a^{2}\right )} \left (-b\right )^{\frac {1}{3}} + 12 \, {\left (2 \, b^{2} x^{5} + a b x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{2} x^{6} - 2 \, a b x^{3} + a^{2}}\right ) + 18 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{18 \, a d x} \]
-1/18*(2*sqrt(3)*2^(1/3)*(-b)^(1/3)*x*arctan(1/3*(6*sqrt(3)*2^(2/3)*(19*b^ 2*x^8 + 16*a*b*x^5 + a^2*x^2)*(b*x^3 + a)^(1/3)*(-b)^(2/3) + 6*sqrt(3)*2^( 1/3)*(5*b^2*x^7 - 4*a*b*x^4 - a^2*x)*(b*x^3 + a)^(2/3)*(-b)^(1/3) + sqrt(3 )*(71*b^3*x^9 + 111*a*b^2*x^6 + 33*a^2*b*x^3 + a^3))/(109*b^3*x^9 + 105*a* b^2*x^6 + 3*a^2*b*x^3 - a^3)) - 2*2^(1/3)*(-b)^(1/3)*x*log(-(6*2^(1/3)*(b* x^3 + a)^(1/3)*(-b)^(1/3)*b*x^2 + 6*(b*x^3 + a)^(2/3)*b*x + 2^(2/3)*(b*x^3 - a)*(-b)^(2/3))/(b*x^3 - a)) + 2^(1/3)*(-b)^(1/3)*x*log((3*2^(2/3)*(5*b* x^4 + a*x)*(b*x^3 + a)^(2/3)*(-b)^(2/3) - 2^(1/3)*(19*b^2*x^6 + 16*a*b*x^3 + a^2)*(-b)^(1/3) + 12*(2*b^2*x^5 + a*b*x^2)*(b*x^3 + a)^(1/3))/(b^2*x^6 - 2*a*b*x^3 + a^2)) + 18*(b*x^3 + a)^(1/3))/(a*d*x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=- \frac {\int \frac {\sqrt [3]{a + b x^{3}}}{- a x^{2} + b x^{5}}\, dx}{d} \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{2}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^2 \left (a d-b d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^2\,\left (a\,d-b\,d\,x^3\right )} \,d x \]