\(\int \frac {(a+b x^3)^{4/3}}{(c+d x^3)^3} \, dx\) [169]

Optimal result

Integrand size = 21, antiderivative size = 60 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\frac {a x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {4}{3},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

a*x*(b*x^3+a)^(1/3)*AppellF1(1/3,-4/3,3,4/3,-b*x^3/a,-d*x^3/c)/c^3/(1+b*x^ 
3/a)^(1/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(285\) vs. \(2(60)=120\).

Time = 10.51 (sec) , antiderivative size = 285, normalized size of antiderivative = 4.75 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\frac {x \left (b (2 b c+5 a d) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {4 c \left (\left (a+b x^3\right ) \left (-b c \left (c-2 d x^3\right )+a d \left (8 c+5 d x^3\right )\right )+\frac {4 a^2 c (b c+10 a d) \left (c+d x^3\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}\right )}{\left (c+d x^3\right )^2}\right )}{72 c^3 d \left (a+b x^3\right )^{2/3}} \] Input:

Integrate[(a + b*x^3)^(4/3)/(c + d*x^3)^3,x]
 

Output:

(x*(b*(2*b*c + 5*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*((a + b*x^3)*(-(b*c*(c - 2*d*x^3)) + a 
*d*(8*c + 5*d*x^3)) + (4*a^2*c*(b*c + 10*a*d)*(c + d*x^3)*AppellF1[1/3, 2/ 
3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(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)]))))/(c + d*x^3)^2))/(72*c^3*d*(a + b*x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.30 (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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {a \sqrt [3]{a+b x^3} \int \frac {\left (\frac {b x^3}{a}+1\right )^{4/3}}{\left (d x^3+c\right )^3}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {a x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {4}{3},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \sqrt [3]{\frac {b x^3}{a}+1}}\)

Input:

Int[(a + b*x^3)^(4/3)/(c + d*x^3)^3,x]
 

Output:

(a*x*(a + b*x^3)^(1/3)*AppellF1[1/3, -4/3, 3, 4/3, -((b*x^3)/a), -((d*x^3) 
/c)])/(c^3*(1 + (b*x^3)/a)^(1/3))
 

Defintions of rubi rules used

rule 936

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])
 

rule 937

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])
 

Maple [F]

\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{\left (d \,x^{3}+c \right )^{3}}d x\]

Input:

int((b*x^3+a)^(4/3)/(d*x^3+c)^3,x)
 

Output:

int((b*x^3+a)^(4/3)/(d*x^3+c)^3,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(4/3)/(d*x^3+c)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**3+a)**(4/3)/(d*x**3+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/(d*x^3+c)^3,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)/(d*x^3 + c)^3, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/(d*x^3+c)^3,x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)/(d*x^3 + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{4/3}}{{\left (d\,x^3+c\right )}^3} \,d x \] Input:

int((a + b*x^3)^(4/3)/(c + d*x^3)^3,x)
 

Output:

int((a + b*x^3)^(4/3)/(c + d*x^3)^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{\left (c+d x^3\right )^3} \, dx=\text {too large to display} \] Input:

int((b*x^3+a)^(4/3)/(d*x^3+c)^3,x)
 

Output:

( - 2*(a + b*x**3)**(1/3)*a*b*x + 25*int((a + b*x**3)**(1/3)/(5*a**2*c**3* 
d + 15*a**2*c**2*d**2*x**3 + 15*a**2*c*d**3*x**6 + 5*a**2*d**4*x**9 - 2*a* 
b*c**4 - a*b*c**3*d*x**3 + 9*a*b*c**2*d**2*x**6 + 13*a*b*c*d**3*x**9 + 5*a 
*b*d**4*x**12 - 2*b**2*c**4*x**3 - 6*b**2*c**3*d*x**6 - 6*b**2*c**2*d**2*x 
**9 - 2*b**2*c*d**3*x**12),x)*a**4*c**2*d**2 + 50*int((a + b*x**3)**(1/3)/ 
(5*a**2*c**3*d + 15*a**2*c**2*d**2*x**3 + 15*a**2*c*d**3*x**6 + 5*a**2*d** 
4*x**9 - 2*a*b*c**4 - a*b*c**3*d*x**3 + 9*a*b*c**2*d**2*x**6 + 13*a*b*c*d* 
*3*x**9 + 5*a*b*d**4*x**12 - 2*b**2*c**4*x**3 - 6*b**2*c**3*d*x**6 - 6*b** 
2*c**2*d**2*x**9 - 2*b**2*c*d**3*x**12),x)*a**4*c*d**3*x**3 + 25*int((a + 
b*x**3)**(1/3)/(5*a**2*c**3*d + 15*a**2*c**2*d**2*x**3 + 15*a**2*c*d**3*x* 
*6 + 5*a**2*d**4*x**9 - 2*a*b*c**4 - a*b*c**3*d*x**3 + 9*a*b*c**2*d**2*x** 
6 + 13*a*b*c*d**3*x**9 + 5*a*b*d**4*x**12 - 2*b**2*c**4*x**3 - 6*b**2*c**3 
*d*x**6 - 6*b**2*c**2*d**2*x**9 - 2*b**2*c*d**3*x**12),x)*a**4*d**4*x**6 - 
 10*int((a + b*x**3)**(1/3)/(5*a**2*c**3*d + 15*a**2*c**2*d**2*x**3 + 15*a 
**2*c*d**3*x**6 + 5*a**2*d**4*x**9 - 2*a*b*c**4 - a*b*c**3*d*x**3 + 9*a*b* 
c**2*d**2*x**6 + 13*a*b*c*d**3*x**9 + 5*a*b*d**4*x**12 - 2*b**2*c**4*x**3 
- 6*b**2*c**3*d*x**6 - 6*b**2*c**2*d**2*x**9 - 2*b**2*c*d**3*x**12),x)*a** 
3*b*c**3*d - 20*int((a + b*x**3)**(1/3)/(5*a**2*c**3*d + 15*a**2*c**2*d**2 
*x**3 + 15*a**2*c*d**3*x**6 + 5*a**2*d**4*x**9 - 2*a*b*c**4 - a*b*c**3*d*x 
**3 + 9*a*b*c**2*d**2*x**6 + 13*a*b*c*d**3*x**9 + 5*a*b*d**4*x**12 - 2*...