Integrand size = 21, antiderivative size = 60 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\frac {a x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {4}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
a*x*(b*x^3+a)^(1/3)*AppellF1(1/3,-4/3,2,4/3,-b*x^3/a,-d*x^3/c)/c^2/(1+b*x^ 3/a)^(1/3)
Leaf count is larger than twice the leaf count of optimal. \(341\) vs. \(2(60)=120\).
Time = 10.36 (sec) , antiderivative size = 341, normalized size of antiderivative = 5.68 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\frac {x \left (b (2 b c+a d) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {4 c \left (-4 a c \left (3 a^2 d-b^2 c x^3+a b d x^3\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+(-b c+a d) x^3 \left (a+b x^3\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{12 c^2 d \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[(a + b*x^3)^(4/3)/(c + d*x^3)^2,x]
Output:
(x*(b*(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)] + (4*c*(-4*a*c*(3*a^2*d - b^2*c*x^3 + a*b*d*x^3 )*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + (-(b*c) + a*d)* x^3*(a + b*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*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)])))))/(12*c^2*d*(a + b*x^3)^(2/3))
Time = 0.29 (sec) , antiderivative size = 60, 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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {a \sqrt [3]{a+b x^3} \int \frac {\left (\frac {b x^3}{a}+1\right )^{4/3}}{\left (d x^3+c\right )^2}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {a x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {4}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^2 \sqrt [3]{\frac {b x^3}{a}+1}}\) |
Input:
Int[(a + b*x^3)^(4/3)/(c + d*x^3)^2,x]
Output:
(a*x*(a + b*x^3)^(1/3)*AppellF1[1/3, -4/3, 2, 4/3, -((b*x^3)/a), -((d*x^3) /c)])/(c^2*(1 + (b*x^3)/a)^(1/3))
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])
\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{\left (d \,x^{3}+c \right )^{2}}d x\]
Input:
int((b*x^3+a)^(4/3)/(d*x^3+c)^2,x)
Output:
int((b*x^3+a)^(4/3)/(d*x^3+c)^2,x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(4/3)/(d*x^3+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{\left (c + d x^{3}\right )^{2}}\, dx \] Input:
integrate((b*x**3+a)**(4/3)/(d*x**3+c)**2,x)
Output:
Integral((a + b*x**3)**(4/3)/(c + d*x**3)**2, x)
\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(4/3)/(d*x^3+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(4/3)/(d*x^3 + c)^2, x)
\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^3+a)^(4/3)/(d*x^3+c)^2,x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(4/3)/(d*x^3 + c)^2, x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{4/3}}{{\left (d\,x^3+c\right )}^2} \,d x \] Input:
int((a + b*x^3)^(4/3)/(c + d*x^3)^2,x)
Output:
int((a + b*x^3)^(4/3)/(c + d*x^3)^2, x)
\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((b*x^3+a)^(4/3)/(d*x^3+c)^2,x)
Output:
( - (a + b*x**3)**(1/3)*a*b*x + int((a + b*x**3)**(1/3)/(a**2*c**2*d + 2*a **2*c*d**2*x**3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2 *x**6 + a*b*d**3*x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2* x**9),x)*a**4*c*d**2 + int((a + b*x**3)**(1/3)/(a**2*c**2*d + 2*a**2*c*d** 2*x**3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a *b*d**3*x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2*x**9),x)* a**4*d**3*x**3 - int((a + b*x**3)**(1/3)/(a**2*c**2*d + 2*a**2*c*d**2*x**3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a*b*d** 3*x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2*x**9),x)*a**3*b *c**2*d - int((a + b*x**3)**(1/3)/(a**2*c**2*d + 2*a**2*c*d**2*x**3 + a**2 *d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a*b*d**3*x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2*x**9),x)*a**3*b*c*d**2 *x**3 - int(((a + b*x**3)**(1/3)*x**6)/(a**2*c**2*d + 2*a**2*c*d**2*x**3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a*b*d**3* x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2*x**9),x)*a*b**3*c **2*d - int(((a + b*x**3)**(1/3)*x**6)/(a**2*c**2*d + 2*a**2*c*d**2*x**3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a*b*d**3* x**9 - b**2*c**3*x**3 - 2*b**2*c**2*d*x**6 - b**2*c*d**2*x**9),x)*a*b**3*c *d**2*x**3 + int(((a + b*x**3)**(1/3)*x**6)/(a**2*c**2*d + 2*a**2*c*d**2*x **3 + a**2*d**3*x**6 - a*b*c**3 - a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + a...