Integrand size = 28, antiderivative size = 157 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2}+\frac {2^{2/3} b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {b^{2/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a d}-\frac {b^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a d} \] Output:
-1/2*(b*x^3+a)^(2/3)/a/d/x^2+1/3*2^(2/3)*b^(2/3)*arctan(1/3*(1+2*2^(1/3)*b ^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/a/d+1/6*b^(2/3)*ln(-b*d*x^3+a*d )*2^(2/3)/a/d-1/2*b^(2/3)*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(2/3)/a/ d
Time = 0.38 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=\frac {-3 \left (a+b x^3\right )^{2/3}+2\ 2^{2/3} \sqrt {3} b^{2/3} x^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-2\ 2^{2/3} b^{2/3} x^2 \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+2^{2/3} b^{2/3} x^2 \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{6 a d x^2} \] Input:
Integrate[(a + b*x^3)^(2/3)/(x^3*(a*d - b*d*x^3)),x]
Output:
(-3*(a + b*x^3)^(2/3) + 2*2^(2/3)*Sqrt[3]*b^(2/3)*x^2*ArcTan[(Sqrt[3]*b^(1 /3)*x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] - 2*2^(2/3)*b^(2/3)*x^2*Lo g[-2*b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3)] + 2^(2/3)*b^(2/3)*x^2*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 )])/(6*a*d*x^2)
Time = 0.42 (sec) , antiderivative size = 154, 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, 901}
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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 975 |
| \(\displaystyle \frac {\int \frac {4 a b}{\left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx}{2 a d}-\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b \int \frac {1}{\left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx}{d}-\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2}\) |
| \(\Big \downarrow \) 901 |
| \(\displaystyle \frac {2 b \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} a \sqrt [3]{b}}+\frac {\log \left (a-b x^3\right )}{6 \sqrt [3]{2} a \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{2} a \sqrt [3]{b}}\right )}{d}-\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2}\) |
Input:
Int[(a + b*x^3)^(2/3)/(x^3*(a*d - b*d*x^3)),x]
Output:
-1/2*(a + b*x^3)^(2/3)/(a*d*x^2) + (2*b*(ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x) /(a + b*x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]*a*b^(1/3)) + Log[a - b*x^3]/ (6*2^(1/3)*a*b^(1/3)) - Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*2^(1 /3)*a*b^(1/3))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
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]
Time = 1.70 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.99
| method | result | size |
| pseudoelliptic | \(\frac {-2 b^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) x^{2}-2 b^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) x^{2}+b^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\frac {b^{\frac {2}{3}} 2^{\frac {2}{3}} x^{2}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {1}{3}} 2^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) x^{2}-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{6 a d \,x^{2}}\) | \(156\) |
Input:
int((b*x^3+a)^(2/3)/x^3/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
Output:
1/6*(-2*b^(2/3)*3^(1/2)*2^(2/3)*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3 )+b^(1/3)*x)/b^(1/3)/x)*x^2-2*b^(2/3)*2^(2/3)*ln((-2^(1/3)*b^(1/3)*x+(b*x^ 3+a)^(1/3))/x)*x^2+b^(2/3)*2^(2/3)*ln((b^(2/3)*2^(2/3)*x^2+(b*x^3+a)^(1/3) *b^(1/3)*2^(1/3)*x+(b*x^3+a)^(2/3))/x^2)*x^2-3*(b*x^3+a)^(2/3))/a/d/x^2
Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (125) = 250\).
Time = 76.82 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.76 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=-\frac {2 \cdot 4^{\frac {1}{3}} \sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \arctan \left (\frac {3 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (5 \, b^{2} x^{7} - 4 \, a b x^{4} - a^{2} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b^{2}\right )^{\frac {2}{3}} + 6 \cdot 4^{\frac {1}{3}} \sqrt {3} {\left (19 \, b^{3} x^{8} + 16 \, a b^{2} x^{5} + a^{2} b x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} - \sqrt {3} {\left (71 \, b^{4} x^{9} + 111 \, a b^{3} x^{6} + 33 \, a^{2} b^{2} x^{3} + a^{3} b\right )}}{3 \, {\left (109 \, b^{4} x^{9} + 105 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{3} - a^{3} b\right )}}\right ) - 2 \cdot 4^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (\frac {3 \cdot 4^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x^{2} - 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 4^{\frac {1}{3}} {\left (b x^{3} - a\right )} \left (-b^{2}\right )^{\frac {1}{3}}}{b x^{3} - a}\right ) + 4^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (-\frac {6 \cdot 4^{\frac {1}{3}} {\left (5 \, b^{2} x^{4} + a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b^{2}\right )^{\frac {1}{3}} - 4^{\frac {2}{3}} {\left (19 \, b^{2} x^{6} + 16 \, a b x^{3} + a^{2}\right )} \left (-b^{2}\right )^{\frac {2}{3}} - 24 \, {\left (2 \, b^{3} x^{5} + a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{2} x^{6} - 2 \, a b x^{3} + a^{2}}\right ) + 9 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{18 \, a d x^{2}} \] Input:
integrate((b*x^3+a)^(2/3)/x^3/(-b*d*x^3+a*d),x, algorithm="fricas")
Output:
-1/18*(2*4^(1/3)*sqrt(3)*(-b^2)^(1/3)*x^2*arctan(1/3*(3*4^(2/3)*sqrt(3)*(5 *b^2*x^7 - 4*a*b*x^4 - a^2*x)*(b*x^3 + a)^(2/3)*(-b^2)^(2/3) + 6*4^(1/3)*s qrt(3)*(19*b^3*x^8 + 16*a*b^2*x^5 + a^2*b*x^2)*(b*x^3 + a)^(1/3)*(-b^2)^(1 /3) - sqrt(3)*(71*b^4*x^9 + 111*a*b^3*x^6 + 33*a^2*b^2*x^3 + a^3*b))/(109* b^4*x^9 + 105*a*b^3*x^6 + 3*a^2*b^2*x^3 - a^3*b)) - 2*4^(1/3)*(-b^2)^(1/3) *x^2*log((3*4^(2/3)*(b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x^2 - 6*(b*x^3 + a)^(2/ 3)*b*x + 4^(1/3)*(b*x^3 - a)*(-b^2)^(1/3))/(b*x^3 - a)) + 4^(1/3)*(-b^2)^( 1/3)*x^2*log(-(6*4^(1/3)*(5*b^2*x^4 + a*b*x)*(b*x^3 + a)^(2/3)*(-b^2)^(1/3 ) - 4^(2/3)*(19*b^2*x^6 + 16*a*b*x^3 + a^2)*(-b^2)^(2/3) - 24*(2*b^3*x^5 + a*b^2*x^2)*(b*x^3 + a)^(1/3))/(b^2*x^6 - 2*a*b*x^3 + a^2)) + 9*(b*x^3 + a )^(2/3))/(a*d*x^2)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=- \frac {\int \frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{- a x^{3} + b x^{6}}\, dx}{d} \] Input:
integrate((b*x**3+a)**(2/3)/x**3/(-b*d*x**3+a*d),x)
Output:
-Integral((a + b*x**3)**(2/3)/(-a*x**3 + b*x**6), x)/d
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (b d x^{3} - a d\right )} x^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/x^3/(-b*d*x^3+a*d),x, algorithm="maxima")
Output:
-integrate((b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^3), x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (b d x^{3} - a d\right )} x^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/x^3/(-b*d*x^3+a*d),x, algorithm="giac")
Output:
integrate(-(b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^3), x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{2/3}}{x^3\,\left (a\,d-b\,d\,x^3\right )} \,d x \] Input:
int((a + b*x^3)^(2/3)/(x^3*(a*d - b*d*x^3)),x)
Output:
int((a + b*x^3)^(2/3)/(x^3*(a*d - b*d*x^3)), x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^3 \left (a d-b d x^3\right )} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}+4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{-b^{2} x^{9}+a^{2} x^{3}}d x \right ) a^{2} x^{2}}{2 a d \,x^{2}} \] Input:
int((b*x^3+a)^(2/3)/x^3/(-b*d*x^3+a*d),x)
Output:
((a + b*x**3)**(2/3) + 4*int((a + b*x**3)**(2/3)/(a**2*x**3 - b**2*x**9),x )*a**2*x**2)/(2*a*d*x**2)