Integrand size = 24, antiderivative size = 64 \[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {x^7 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {7}{3},\frac {2}{3},1,\frac {10}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \left (a+b x^3\right )^{2/3}} \] Output:
1/7*x^7*(1+b*x^3/a)^(2/3)*AppellF1(7/3,2/3,1,10/3,-b*x^3/a,-d*x^3/c)/c/(b* x^3+a)^(2/3)
Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(64)=128\).
Time = 8.40 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.89 \[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {x \left (-\frac {(2 b c+a d) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{b c}+4 \left (\frac {a}{b}+x^3+\frac {4 a^2 c^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{b \left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )\right )}{8 d \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[x^6/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
(x*(-(((2*b*c + a*d)*x^3*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(b*c)) + 4*(a/b + x^3 + (4*a^2*c^2*AppellF1[1 /3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(b*(c + d*x^3)*(-4*a*c*Appel lF1[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*d*(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.
\(\displaystyle \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {x^6}{\left (\frac {b x^3}{a}+1\right )^{2/3} \left (d x^3+c\right )}dx}{\left (a+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^7 \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {7}{3},\frac {2}{3},1,\frac {10}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \left (a+b x^3\right )^{2/3}}\) |
Input:
Int[x^6/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
(x^7*(1 + (b*x^3)/a)^(2/3)*AppellF1[7/3, 2/3, 1, 10/3, -((b*x^3)/a), -((d* x^3)/c)])/(7*c*(a + b*x^3)^(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])
\[\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )}d x\]
Input:
int(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
int(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x)
Timed out. \[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{6}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] Input:
integrate(x**6/(b*x**3+a)**(2/3)/(d*x**3+c),x)
Output:
Integral(x**6/((a + b*x**3)**(2/3)*(c + d*x**3)), x)
\[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{6}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:
integrate(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
Output:
integrate(x^6/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)
\[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{6}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:
integrate(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
Output:
integrate(x^6/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)
Timed out. \[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^6}{{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \] Input:
int(x^6/((a + b*x^3)^(2/3)*(c + d*x^3)),x)
Output:
int(x^6/((a + b*x^3)^(2/3)*(c + d*x^3)), x)
\[ \int \frac {x^6}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c +\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{3}}d x \] Input:
int(x^6/(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
int(x**6/((a + b*x**3)**(2/3)*c + (a + b*x**3)**(2/3)*d*x**3),x)