Integrand size = 24, antiderivative size = 62 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {2}{3},1,\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c x \left (1+\frac {b x^3}{a}\right )^{2/3}} \] Output:
-(b*x^3+a)^(2/3)*AppellF1(-1/3,-2/3,1,2/3,-b*x^3/a,-d*x^3/c)/c/x/(1+b*x^3/ a)^(2/3)
Leaf count is larger than twice the leaf count of optimal. \(138\) vs. \(2(62)=124\).
Time = 10.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.23 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\frac {-10 c \left (a+b x^3\right )-5 (-2 b c+a d) x^3 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b d x^6 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{10 c^2 x \sqrt [3]{a+b x^3}} \] Input:
Integrate[(a + b*x^3)^(2/3)/(x^2*(c + d*x^3)),x]
Output:
(-10*c*(a + b*x^3) - 5*(-2*b*c + a*d)*x^3*(1 + (b*x^3)/a)^(1/3)*AppellF1[2 /3, 1/3, 1, 5/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*d*x^6*(1 + (b*x^3)/a)^( 1/3)*AppellF1[5/3, 1/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)])/(10*c^2*x*(a + b*x^3)^(1/3))
Time = 0.35 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1013, 1012}
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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\left (a+b x^3\right )^{2/3} \int \frac {\left (\frac {b x^3}{a}+1\right )^{2/3}}{x^2 \left (d x^3+c\right )}dx}{\left (\frac {b x^3}{a}+1\right )^{2/3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\left (a+b x^3\right )^{2/3} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {2}{3},1,\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c x \left (\frac {b x^3}{a}+1\right )^{2/3}}\) |
Input:
Int[(a + b*x^3)^(2/3)/(x^2*(c + d*x^3)),x]
Output:
-(((a + b*x^3)^(2/3)*AppellF1[-1/3, -2/3, 1, 2/3, -((b*x^3)/a), -((d*x^3)/ c)])/(c*x*(1 + (b*x^3)/a)^(2/3)))
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2} \left (d \,x^{3}+c \right )}d x\]
Input:
int((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x)
Output:
int((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{x^{2} \left (c + d x^{3}\right )}\, dx \] Input:
integrate((b*x**3+a)**(2/3)/x**2/(d*x**3+c),x)
Output:
Integral((a + b*x**3)**(2/3)/(x**2*(c + d*x**3)), x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (d x^{3} + c\right )} x^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^2), x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{{\left (d x^{3} + c\right )} x^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^2), x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{2/3}}{x^2\,\left (d\,x^3+c\right )} \,d x \] Input:
int((a + b*x^3)^(2/3)/(x^2*(c + d*x^3)),x)
Output:
int((a + b*x^3)^(2/3)/(x^2*(c + d*x^3)), x)
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^2 \left (c+d x^3\right )} \, dx=\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{d \,x^{5}+c \,x^{2}}d x \] Input:
int((b*x^3+a)^(2/3)/x^2/(d*x^3+c),x)
Output:
int((a + b*x**3)**(2/3)/(c*x**2 + d*x**5),x)