\(\int \frac {1}{(a+b x^3)^{5/3} (c+d x^3)^3} \, dx\) [172]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F(-1)]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^3 \left (a+b x^3\right )^{2/3}} \] Output:

x*(1+b*x^3/a)^(2/3)*AppellF1(1/3,5/3,3,4/3,-b*x^3/a,-d*x^3/c)/a/c^3/(b*x^3 
+a)^(2/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(531\) vs. \(2(62)=124\).

Time = 11.00 (sec) , antiderivative size = 531, normalized size of antiderivative = 8.56 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\frac {\frac {b d \left (-9 b^2 c^2-16 a b c d+5 a^2 d^2\right ) x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{(-b c+a d)^3}-\frac {4 c \left (4 a c x \left (3 a^3 d^3 \left (6 c+5 d x^3\right )+a b^2 c d \left (54 c^2+35 c d x^3-16 d^2 x^6\right )-9 b^3 c^2 \left (2 c^2+3 c d x^3+d^2 x^6\right )+a^2 b d^2 \left (-54 c^2-43 c d x^3+5 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^4 \left (9 b^3 c^2 \left (c+d x^3\right )^2-a^3 d^3 \left (8 c+5 d x^3\right )+a b^2 c d^2 x^3 \left (19 c+16 d x^3\right )+a^2 b d^2 \left (19 c^2+8 c d x^3-5 d^2 x^6\right )\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{(b c-a d)^3 \left (c+d x^3\right )^2 \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}}{72 a c^3 \left (a+b x^3\right )^{2/3}} \] Input:

Integrate[1/((a + b*x^3)^(5/3)*(c + d*x^3)^3),x]
 

Output:

((b*d*(-9*b^2*c^2 - 16*a*b*c*d + 5*a^2*d^2)*x^4*(1 + (b*x^3)/a)^(2/3)*Appe 
llF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(-(b*c) + a*d)^3 - (4* 
c*(4*a*c*x*(3*a^3*d^3*(6*c + 5*d*x^3) + a*b^2*c*d*(54*c^2 + 35*c*d*x^3 - 1 
6*d^2*x^6) - 9*b^3*c^2*(2*c^2 + 3*c*d*x^3 + d^2*x^6) + a^2*b*d^2*(-54*c^2 
- 43*c*d*x^3 + 5*d^2*x^6))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x 
^3)/c)] + x^4*(9*b^3*c^2*(c + d*x^3)^2 - a^3*d^3*(8*c + 5*d*x^3) + a*b^2*c 
*d^2*x^3*(19*c + 16*d*x^3) + a^2*b*d^2*(19*c^2 + 8*c*d*x^3 - 5*d^2*x^6))*( 
3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*Appel 
lF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))/((b*c - a*d)^3*(c + d 
*x^3)^2*(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*Ap 
pellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))))/(72*a*c^3*(a + b* 
x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.30 (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.

\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{5/3} \left (d x^3+c\right )^3}dx}{a \left (a+b x^3\right )^{2/3}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^3 \left (a+b x^3\right )^{2/3}}\)

Input:

Int[1/((a + b*x^3)^(5/3)*(c + d*x^3)^3),x]
 

Output:

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

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]

\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (d \,x^{3}+c \right )^{3}}d x\]

Input:

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x)
 

Output:

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**3+a)**(5/3)/(d*x**3+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="maxima")
 

Output:

integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)^3), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="giac")
 

Output:

integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{5/3}\,{\left (d\,x^3+c\right )}^3} \,d x \] Input:

int(1/((a + b*x^3)^(5/3)*(c + d*x^3)^3),x)
 

Output:

int(1/((a + b*x^3)^(5/3)*(c + d*x^3)^3), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^3} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,c^{3}+3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,c^{2} d \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a c \,d^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,d^{3} x^{9}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,c^{3} x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,c^{2} d \,x^{6}+3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b c \,d^{2} x^{9}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,d^{3} x^{12}}d x \] Input:

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x)
 

Output:

int(1/((a + b*x**3)**(2/3)*a*c**3 + 3*(a + b*x**3)**(2/3)*a*c**2*d*x**3 + 
3*(a + b*x**3)**(2/3)*a*c*d**2*x**6 + (a + b*x**3)**(2/3)*a*d**3*x**9 + (a 
 + b*x**3)**(2/3)*b*c**3*x**3 + 3*(a + b*x**3)**(2/3)*b*c**2*d*x**6 + 3*(a 
 + b*x**3)**(2/3)*b*c*d**2*x**9 + (a + b*x**3)**(2/3)*b*d**3*x**12),x)