Integrand size = 24, antiderivative size = 64 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {5}{3},-\frac {1}{3},1,-\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c x^5 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
-1/5*(b*x^3+a)^(1/3)*AppellF1(-5/3,-1/3,1,-2/3,-b*x^3/a,-d*x^3/c)/c/x^5/(1 +b*x^3/a)^(1/3)
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(64)=128\).
Time = 10.31 (sec) , antiderivative size = 289, normalized size of antiderivative = 4.52 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\frac {-\frac {4 \left (a+b x^3\right ) \left (2 a c+b c x^3-5 a d x^3\right )}{a c^2 x^5}+\frac {b d (-b c+5 a d) x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^3}+\frac {16 \left (b^2 c^2+5 a b c d-10 a^2 d^2\right ) x \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c \left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}}{40 \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[(a + b*x^3)^(1/3)/(x^6*(c + d*x^3)),x]
Output:
((-4*(a + b*x^3)*(2*a*c + b*c*x^3 - 5*a*d*x^3))/(a*c^2*x^5) + (b*d*(-(b*c) + 5*a*d)*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a ), -((d*x^3)/c)])/(a*c^3) + (16*(b^2*c^2 + 5*a*b*c*d - 10*a^2*d^2)*x*Appel lF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(c*(c + d*x^3)*(-4*a*c* AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + x^3*(3*a*d*Appell F1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3, 5/3 , 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))))/(40*(a + b*x^3)^(2/3))
Time = 0.35 (sec) , antiderivative size = 64, 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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt [3]{a+b x^3} \int \frac {\sqrt [3]{\frac {b x^3}{a}+1}}{x^6 \left (d x^3+c\right )}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {5}{3},-\frac {1}{3},1,-\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c x^5 \sqrt [3]{\frac {b x^3}{a}+1}}\) |
Input:
Int[(a + b*x^3)^(1/3)/(x^6*(c + d*x^3)),x]
Output:
-1/5*((a + b*x^3)^(1/3)*AppellF1[-5/3, -1/3, 1, -2/3, -((b*x^3)/a), -((d*x ^3)/c)])/(c*x^5*(1 + (b*x^3)/a)^(1/3))
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])
\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{6} \left (d \,x^{3}+c \right )}d x\]
Input:
int((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x)
Output:
int((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\int \frac {\sqrt [3]{a + b x^{3}}}{x^{6} \left (c + d x^{3}\right )}\, dx \] Input:
integrate((b*x**3+a)**(1/3)/x**6/(d*x**3+c),x)
Output:
Integral((a + b*x**3)**(1/3)/(x**6*(c + d*x**3)), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{6}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^6), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{6}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^6), x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^6\,\left (d\,x^3+c\right )} \,d x \] Input:
int((a + b*x^3)^(1/3)/(x^6*(c + d*x^3)),x)
Output:
int((a + b*x^3)^(1/3)/(x^6*(c + d*x^3)), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (c+d x^3\right )} \, dx=\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{9}+c \,x^{6}}d x \] Input:
int((b*x^3+a)^(1/3)/x^6/(d*x^3+c),x)
Output:
int((a + b*x**3)**(1/3)/(c*x**6 + d*x**9),x)