Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c \left (a+b x^3\right )^{2/3}} \] Output:
x*(1+b*x^3/a)^(2/3)*AppellF1(1/3,5/3,1,4/3,-b*x^3/a,-d*x^3/c)/a/c/(b*x^3+a )^(2/3)
Leaf count is larger than twice the leaf count of optimal. \(332\) vs. \(2(62)=124\).
Time = 10.29 (sec) , antiderivative size = 332, normalized size of antiderivative = 5.35 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\frac {x \left (-\frac {b d x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c}+\frac {4 \left (4 a c \left (2 a d-b \left (2 c+d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b x^3 \left (c+d x^3\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{8 a (-b c+a d) \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[1/((a + b*x^3)^(5/3)*(c + d*x^3)),x]
Output:
(x*(-((b*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)])/c) + (4*(4*a*c*(2*a*d - b*(2*c + d*x^3))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + b*x^3*(c + d*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)])))/((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*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*a*(-(b*c) + a*d)*(a + b*x^3)^(2/3))
Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {937, 936}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{5/3} \left (d x^3+c\right )}dx}{a \left (a+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c \left (a+b x^3\right )^{2/3}}\) |
Input:
Int[1/((a + b*x^3)^(5/3)*(c + d*x^3)),x]
Output:
(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 5/3, 1, 4/3, -((b*x^3)/a), -((d*x^3 )/c)])/(a*c*(a + b*x^3)^(2/3))
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (d \,x^{3}+c \right )}d x\]
Input:
int(1/(b*x^3+a)^(5/3)/(d*x^3+c),x)
Output:
int(1/(b*x^3+a)^(5/3)/(d*x^3+c),x)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {5}{3}} \left (c + d x^{3}\right )}\, dx \] Input:
integrate(1/(b*x**3+a)**(5/3)/(d*x**3+c),x)
Output:
Integral(1/((a + b*x**3)**(5/3)*(c + d*x**3)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{5/3}\,\left (d\,x^3+c\right )} \,d x \] Input:
int(1/((a + b*x^3)^(5/3)*(c + d*x^3)),x)
Output:
int(1/((a + b*x^3)^(5/3)*(c + d*x^3)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a c +\left (b \,x^{3}+a \right )^{\frac {2}{3}} a d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b c \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b d \,x^{6}}d x \] Input:
int(1/(b*x^3+a)^(5/3)/(d*x^3+c),x)
Output:
int(1/((a + b*x**3)**(2/3)*a*c + (a + b*x**3)**(2/3)*a*d*x**3 + (a + b*x** 3)**(2/3)*b*c*x**3 + (a + b*x**3)**(2/3)*b*d*x**6),x)