Integrand size = 24, antiderivative size = 64 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {1}{3},1,\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
-1/2*(b*x^3+a)^(1/3)*AppellF1(-2/3,-1/3,1,1/3,-b*x^3/a,-d*x^3/c)/c/x^2/(1+ b*x^3/a)^(1/3)
Leaf count is larger than twice the leaf count of optimal. \(327\) vs. \(2(64)=128\).
Time = 10.23 (sec) , antiderivative size = 327, normalized size of antiderivative = 5.11 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\frac {-b d x^6 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {c \left (16 a c \left (b d x^6+a \left (c+3 d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-4 x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}}{8 c^2 x^2 \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[(a + b*x^3)^(1/3)/(x^3*(c + d*x^3)),x]
Output:
(-(b*d*x^6*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]) + (c*(16*a*c*(b*d*x^6 + a*(c + 3*d*x^3))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] - 4*x^3*(a + b*x^3)*(c + d*x^3)*(3*a* d*AppellF1[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)])))/((c + d*x^3)*(-4*a*c*Appe llF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + x^3*(3*a*d*AppellF1[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)]))))/(8*c^2*x^2*(a + b*x^3)^(2/3))
Time = 0.33 (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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt [3]{a+b x^3} \int \frac {\sqrt [3]{\frac {b x^3}{a}+1}}{x^3 \left (d x^3+c\right )}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {1}{3},1,\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac {b x^3}{a}+1}}\) |
Input:
Int[(a + b*x^3)^(1/3)/(x^3*(c + d*x^3)),x]
Output:
-1/2*((a + b*x^3)^(1/3)*AppellF1[-2/3, -1/3, 1, 1/3, -((b*x^3)/a), -((d*x^ 3)/c)])/(c*x^2*(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])
\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{3} \left (d \,x^{3}+c \right )}d x\]
Input:
int((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x)
Output:
int((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\int \frac {\sqrt [3]{a + b x^{3}}}{x^{3} \left (c + d x^{3}\right )}\, dx \] Input:
integrate((b*x**3+a)**(1/3)/x**3/(d*x**3+c),x)
Output:
Integral((a + b*x**3)**(1/3)/(x**3*(c + d*x**3)), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^3), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^3), x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^3\,\left (d\,x^3+c\right )} \,d x \] Input:
int((a + b*x^3)^(1/3)/(x^3*(c + d*x^3)),x)
Output:
int((a + b*x^3)^(1/3)/(x^3*(c + d*x^3)), x)
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^3 \left (c+d x^3\right )} \, dx=\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{6}+c \,x^{3}}d x \] Input:
int((b*x^3+a)^(1/3)/x^3/(d*x^3+c),x)
Output:
int((a + b*x**3)**(1/3)/(c*x**3 + d*x**6),x)