Integrand size = 21, antiderivative size = 182 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\frac {x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right )}+\frac {2 a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} \sqrt [3]{b c-a d}}+\frac {a \log \left (c+d x^3\right )}{9 c^{5/3} \sqrt [3]{b c-a d}}-\frac {a \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} \sqrt [3]{b c-a d}} \] Output:
1/3*x*(b*x^3+a)^(2/3)/c/(d*x^3+c)+2/9*a*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^(5/3)/(-a*d+b*c)^(1/3)+1/9*a* ln(d*x^3+c)/c^(5/3)/(-a*d+b*c)^(1/3)-1/3*a*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-( b*x^3+a)^(1/3))/c^(5/3)/(-a*d+b*c)^(1/3)
Result contains complex when optimal does not.
Time = 2.78 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\frac {\frac {6 c^{2/3} x \left (a+b x^3\right )^{2/3}}{c+d x^3}-\frac {2 \sqrt {-6+6 i \sqrt {3}} a \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b c-a d}}+\frac {2 \left (a+i \sqrt {3} a\right ) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}-\frac {i \left (-i+\sqrt {3}\right ) a \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{18 c^{5/3}} \] Input:
Integrate[(a + b*x^3)^(2/3)/(c + d*x^3)^2,x]
Output:
((6*c^(2/3)*x*(a + b*x^3)^(2/3))/(c + d*x^3) - (2*Sqrt[-6 + (6*I)*Sqrt[3]] *a*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sq rt[3])*c^(1/3)*(a + b*x^3)^(1/3))])/(b*c - a*d)^(1/3) + (2*(a + I*Sqrt[3]* a)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)]) /(b*c - a*d)^(1/3) - (I*(-I + Sqrt[3])*a*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)])/(b*c - a*d)^(1/3))/(18*c^(5/3))
Time = 0.41 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {903, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 903 |
| \(\displaystyle \frac {2 a \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c}+\frac {x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right )}\) |
| \(\Big \downarrow \) 901 |
| \(\displaystyle \frac {2 a \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{3 c}+\frac {x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right )}\) |
Input:
Int[(a + b*x^3)^(2/3)/(c + d*x^3)^2,x]
Output:
(x*(a + b*x^3)^(2/3))/(3*c*(c + d*x^3)) + (2*a*(ArcTan[(1 + (2*(b*c - a*d) ^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a* d)^(1/3)) + Log[c + d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d )^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/(3 *c)
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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] - Simp[ c*(q/(a*(p + 1))) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]
Time = 1.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.18
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {a \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) \left (d \,x^{3}+c \right )}{9}+\frac {2 a \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) \left (d \,x^{3}+c \right )}{9}+\frac {x \left (b \,x^{3}+a \right )^{\frac {2}{3}} c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{3}+\frac {2 a \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \left (d \,x^{3}+c \right ) \sqrt {3}}{9}}{c^{2} \left (d \,x^{3}+c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}\) | \(214\) |
Input:
int((b*x^3+a)^(2/3)/(d*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
2/9/((a*d-b*c)/c)^(1/3)*(-1/2*a*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)*(d*x^3+c)+a*ln((((a*d-b*c)/c )^(1/3)*x+(b*x^3+a)^(1/3))/x)*(d*x^3+c)+3/2*x*(b*x^3+a)^(2/3)*c*((a*d-b*c) /c)^(1/3)+a*arctan(1/3*3^(1/2)*(-2/((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/ x)*(d*x^3+c)*3^(1/2))/c^2/(d*x^3+c)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(2/3)/(d*x^3+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{\left (c + d x^{3}\right )^{2}}\, dx \] Input:
integrate((b*x**3+a)**(2/3)/(d*x**3+c)**2,x)
Output:
Integral((a + b*x**3)**(2/3)/(c + d*x**3)**2, x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/(d*x^3+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(2/3)/(d*x^3 + c)^2, x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/(d*x^3+c)^2,x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(2/3)/(d*x^3 + c)^2, x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{2/3}}{{\left (d\,x^3+c\right )}^2} \,d x \] Input:
int((a + b*x^3)^(2/3)/(c + d*x^3)^2,x)
Output:
int((a + b*x^3)^(2/3)/(c + d*x^3)^2, x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx=\frac {-\left (b \,x^{3}+a \right )^{\frac {2}{3}} b x +4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a^{3} c \,d^{2}+4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a^{3} d^{3} x^{3}-10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a^{2} b \,c^{2} d -10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a^{2} b c \,d^{2} x^{3}+6 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a \,b^{2} c^{3}+6 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{2 a b \,d^{3} x^{9}-3 b^{2} c \,d^{2} x^{9}+2 a^{2} d^{3} x^{6}+a b c \,d^{2} x^{6}-6 b^{2} c^{2} d \,x^{6}+4 a^{2} c \,d^{2} x^{3}-4 a b \,c^{2} d \,x^{3}-3 b^{2} c^{3} x^{3}+2 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) a \,b^{2} c^{2} d \,x^{3}}{2 a \,d^{2} x^{3}-3 b c d \,x^{3}+2 a c d -3 b \,c^{2}} \] Input:
int((b*x^3+a)^(2/3)/(d*x^3+c)^2,x)
Output:
( - (a + b*x**3)**(2/3)*b*x + 4*int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 4 *a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + a* b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**3*c*d**2 + 4*int((a + b*x**3)**(2/3)/(2*a**2*c** 2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x* *3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d* x**6 - 3*b**2*c*d**2*x**9),x)*a**3*d**3*x**3 - 10*int((a + b*x**3)**(2/3)/ (2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a* b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b **2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**2*b*c**2*d - 10*int((a + b*x** 3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c **3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3* x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**2*b*c*d**2*x**3 + 6* int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3* x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a*b**2*c* *3 + 6*int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a** 2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d** 3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a* b**2*c**2*d*x**3)/(2*a*c*d + 2*a*d**2*x**3 - 3*b*c**2 - 3*b*c*d*x**3)