Integrand size = 24, antiderivative size = 173 \[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3}}{a c x}+\frac {d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{4/3} (b c-a d)^{2/3}}-\frac {d \log \left (c+d x^3\right )}{6 c^{4/3} (b c-a d)^{2/3}}+\frac {d \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3} (b c-a d)^{2/3}} \] Output:
-(b*x^3+a)^(1/3)/a/c/x+1/3*d*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b *x^3+a)^(1/3))*3^(1/2))*3^(1/2)/c^(4/3)/(-a*d+b*c)^(2/3)-1/6*d*ln(d*x^3+c) /c^(4/3)/(-a*d+b*c)^(2/3)+1/2*d*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1 /3))/c^(4/3)/(-a*d+b*c)^(2/3)
Result contains complex when optimal does not.
Time = 1.54 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.78 \[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {-12 \sqrt [3]{c} (b c-a d)^{2/3} \sqrt [3]{a+b x^3}-2 \sqrt {-6-6 i \sqrt {3}} a d x \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+2 i \left (i+\sqrt {3}\right ) a d x \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+a \left (d-i \sqrt {3} d\right ) x \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 a c^{4/3} (b c-a d)^{2/3} x} \] Input:
Integrate[1/(x^2*(a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
(-12*c^(1/3)*(b*c - a*d)^(2/3)*(a + b*x^3)^(1/3) - 2*Sqrt[-6 - (6*I)*Sqrt[ 3]]*a*d*x*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3 *I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))] + (2*I)*(I + Sqrt[3])*a*d*x*Log[ 2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)] + a*(d - I*Sqrt[3]*d)*x*Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b *c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[3])*c^(2/3)*(a + b*x^3)^ (2/3)])/(12*a*c^(4/3)*(b*c - a*d)^(2/3)*x)
Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {980, 25, 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.
| \(\displaystyle \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 980 |
| \(\displaystyle \frac {\int -\frac {a d x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{a c}-\frac {\sqrt [3]{a+b x^3}}{a c x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {a d x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{a c}-\frac {\sqrt [3]{a+b x^3}}{a c x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d \int \frac {x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{c}-\frac {\sqrt [3]{a+b x^3}}{a c x}\) |
| \(\Big \downarrow \) 992 |
| \(\displaystyle -\frac {d \left (-\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\log \left (c+d x^3\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} (b c-a d)^{2/3}}\right )}{c}-\frac {\sqrt [3]{a+b x^3}}{a c x}\) |
Input:
Int[1/(x^2*(a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
-((a + b*x^3)^(1/3)/(a*c*x)) - (d*(-(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/( c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(1/3)*(b*c - a*d)^(2/3))) + Log[c + d*x^3]/(6*c^(1/3)*(b*c - a*d)^(2/3)) - Log[((b*c - a*d)^(1/3)*x) /c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(1/3)*(b*c - a*d)^(2/3))))/c
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 + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1)) Int[(e*x)^(m + n)*( a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[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 = 1.73 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.13
| method | result | size |
| pseudoelliptic | \(-\frac {-a \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) x d +3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} c \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}+a d \left (\arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) x}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} a \,c^{2} x}\) | \(195\) |
Input:
int(1/x^2/(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
-1/3/((a*d-b*c)/c)^(2/3)*(-a*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x) *x*d+3*(b*x^3+a)^(1/3)*c*((a*d-b*c)/c)^(2/3)+a*d*(arctan(1/3*3^(1/2)*(-2/( (a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/x)*3^(1/2)+1/2*ln((((a*d-b*c)/c)^(2/ 3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*x)/a/c ^2/x
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^{2} \left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] Input:
integrate(1/x**2/(b*x**3+a)**(2/3)/(d*x**3+c),x)
Output:
Integral(1/(x**2*(a + b*x**3)**(2/3)*(c + d*x**3)), x)
\[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)*x^2), x)
\[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^2\,{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \] Input:
int(1/(x^2*(a + b*x^3)^(2/3)*(c + d*x^3)),x)
Output:
int(1/(x^2*(a + b*x^3)^(2/3)*(c + d*x^3)), x)
\[ \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c \,x^{2}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{5}}d x \] Input:
int(1/x^2/(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
int(1/((a + b*x**3)**(2/3)*c*x**2 + (a + b*x**3)**(2/3)*d*x**5),x)