\(\int \frac {(a+b x^3)^{4/3}}{x^3 (c+d x^3)} \, dx\) [727]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^3 \left (c+d x^3\right )} \, dx=-\frac {a \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},1,\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

-1/2*a*(b*x^3+a)^(1/3)*AppellF1(-2/3,-4/3,1,1/3,-b*x^3/a,-d*x^3/c)/c/x^2/( 
1+b*x^3/a)^(1/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(341\) vs. \(2(65)=130\).

Time = 10.29 (sec) , antiderivative size = 341, normalized size of antiderivative = 5.25 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^3 \left (c+d x^3\right )} \, dx=-\frac {b (-2 b c+a d) x^6 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {4 a c \left (-4 a c \left (a c-2 b c x^3+3 a d x^3+b d x^6\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}}{8 c^2 x^2 \left (a+b x^3\right )^{2/3}} \] Input:

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

Output:

-1/8*(b*(-2*b*c + a*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)] + (4*a*c*(-4*a*c*(a*c - 2*b*c*x^3 + 3*a*d*x^ 
3 + b*d*x^6)*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + 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*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^2*x^ 
2*(a + b*x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 65, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {a \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},1,\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac {b x^3}{a}+1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 1012

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])
 

rule 1013

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^3 \left (c+d x^3\right )} \, dx=\frac {-2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b +4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2 a b \,d^{2} x^{9}+b^{2} c d \,x^{9}+2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+a b \,c^{2} x^{3}}d x \right ) a^{4} d^{2} x^{2}-4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2 a b \,d^{2} x^{9}+b^{2} c d \,x^{9}+2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+a b \,c^{2} x^{3}}d x \right ) a^{3} b c d \,x^{2}-3 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2 a b \,d^{2} x^{9}+b^{2} c d \,x^{9}+2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+a b \,c^{2} x^{3}}d x \right ) a^{2} b^{2} c^{2} x^{2}+2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{2 a b \,d^{2} x^{6}+b^{2} c d \,x^{6}+2 a^{2} d^{2} x^{3}+3 a b c d \,x^{3}+b^{2} c^{2} x^{3}+2 a^{2} c d +a b \,c^{2}}d x \right ) a \,b^{3} c d \,x^{2}+\left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{2 a b \,d^{2} x^{6}+b^{2} c d \,x^{6}+2 a^{2} d^{2} x^{3}+3 a b c d \,x^{3}+b^{2} c^{2} x^{3}+2 a^{2} c d +a b \,c^{2}}d x \right ) b^{4} c^{2} x^{2}}{x^{2} \left (2 a d +b c \right )} \] Input:

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

Output:

( - 2*(a + b*x**3)**(1/3)*a*b + 4*int((a + b*x**3)**(1/3)/(2*a**2*c*d*x**3 
 + 2*a**2*d**2*x**6 + a*b*c**2*x**3 + 3*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b 
**2*c**2*x**6 + b**2*c*d*x**9),x)*a**4*d**2*x**2 - 4*int((a + b*x**3)**(1/ 
3)/(2*a**2*c*d*x**3 + 2*a**2*d**2*x**6 + a*b*c**2*x**3 + 3*a*b*c*d*x**6 + 
2*a*b*d**2*x**9 + b**2*c**2*x**6 + b**2*c*d*x**9),x)*a**3*b*c*d*x**2 - 3*i 
nt((a + b*x**3)**(1/3)/(2*a**2*c*d*x**3 + 2*a**2*d**2*x**6 + a*b*c**2*x**3 
 + 3*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + b**2*c*d*x**9),x)*a 
**2*b**2*c**2*x**2 + 2*int(((a + b*x**3)**(1/3)*x**3)/(2*a**2*c*d + 2*a**2 
*d**2*x**3 + a*b*c**2 + 3*a*b*c*d*x**3 + 2*a*b*d**2*x**6 + b**2*c**2*x**3 
+ b**2*c*d*x**6),x)*a*b**3*c*d*x**2 + int(((a + b*x**3)**(1/3)*x**3)/(2*a* 
*2*c*d + 2*a**2*d**2*x**3 + a*b*c**2 + 3*a*b*c*d*x**3 + 2*a*b*d**2*x**6 + 
b**2*c**2*x**3 + b**2*c*d*x**6),x)*b**4*c**2*x**2)/(x**2*(2*a*d + b*c))