\(\int \frac {x^6 (a+b x^3)^{4/3}}{c+d x^3} \, dx\) [724]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {a x^7 \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {7}{3},-\frac {4}{3},1,\frac {10}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

1/7*a*x^7*(b*x^3+a)^(1/3)*AppellF1(7/3,-4/3,1,10/3,-b*x^3/a,-d*x^3/c)/c/(1 
+b*x^3/a)^(1/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(65)=130\).

Time = 8.46 (sec) , antiderivative size = 343, normalized size of antiderivative = 5.28 \[ \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {x \left (2 \left (a+b x^3\right ) \left (2 a^2 d^2+3 a b d \left (-8 c+3 d x^3\right )+b^2 \left (20 c^2-8 c d x^3+5 d^2 x^6\right )\right )-\frac {\left (20 b^3 c^3-30 a b^2 c^2 d+8 a^2 b c d^2+a^3 d^3\right ) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c}+\frac {16 a^2 c^2 \left (10 b^2 c^2-12 a b c d+a^2 d^2\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{\left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{80 b d^3 \left (a+b x^3\right )^{2/3}} \] Input:

Integrate[(x^6*(a + b*x^3)^(4/3))/(c + d*x^3),x]
 

Output:

(x*(2*(a + b*x^3)*(2*a^2*d^2 + 3*a*b*d*(-8*c + 3*d*x^3) + b^2*(20*c^2 - 8* 
c*d*x^3 + 5*d^2*x^6)) - ((20*b^3*c^3 - 30*a*b^2*c^2*d + 8*a^2*b*c*d^2 + a^ 
3*d^3)*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 + (16*a^2*c^2*(10*b^2*c^2 - 12*a*b*c*d + a^2*d^2)*AppellF 
1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/((c + d*x^3)*(-4*a*c*Appe 
llF1[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)])))))/(80*b*d^3*(a + b*x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.35 (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.

\(\displaystyle \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {a \sqrt [3]{a+b x^3} \int \frac {x^6 \left (\frac {b x^3}{a}+1\right )^{4/3}}{d x^3+c}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {a x^7 \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {7}{3},-\frac {4}{3},1,\frac {10}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \sqrt [3]{\frac {b x^3}{a}+1}}\)

Input:

Int[(x^6*(a + b*x^3)^(4/3))/(c + d*x^3),x]
 

Output:

(a*x^7*(a + b*x^3)^(1/3)*AppellF1[7/3, -4/3, 1, 10/3, -((b*x^3)/a), -((d*x 
^3)/c)])/(7*c*(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]

\[\int \frac {x^{6} \left (b \,x^{3}+a \right )^{\frac {4}{3}}}{d \,x^{3}+c}d x\]

Input:

int(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\text {Timed out} \] Input:

integrate(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(x**6*(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(x**6*(a + b*x**3)**(4/3)/(c + d*x**3), x)
 

Maxima [F]

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

integrate(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)*x^6/(d*x^3 + c), x)
 

Giac [F]

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

integrate(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)*x^6/(d*x^3 + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\int \frac {x^6\,{\left (b\,x^3+a\right )}^{4/3}}{d\,x^3+c} \,d x \] Input:

int((x^6*(a + b*x^3)^(4/3))/(c + d*x^3),x)
 

Output:

int((x^6*(a + b*x^3)^(4/3))/(c + d*x^3), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d^{2} x -24 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c d x +9 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,d^{2} x^{4}+20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c^{2} x -8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c d \,x^{4}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d^{2} x^{7}-2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{3} c \,d^{2}+24 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b \,c^{2} d -20 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{2} c^{3}-2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{3} d^{3}-16 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b c \,d^{2}+60 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{2} c^{2} d -40 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{3} c^{3}}{40 b \,d^{3}} \] Input:

int(x^6*(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

(2*(a + b*x**3)**(1/3)*a**2*d**2*x - 24*(a + b*x**3)**(1/3)*a*b*c*d*x + 9* 
(a + b*x**3)**(1/3)*a*b*d**2*x**4 + 20*(a + b*x**3)**(1/3)*b**2*c**2*x - 8 
*(a + b*x**3)**(1/3)*b**2*c*d*x**4 + 5*(a + b*x**3)**(1/3)*b**2*d**2*x**7 
- 2*int((a + b*x**3)**(1/3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a**3 
*c*d**2 + 24*int((a + b*x**3)**(1/3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6 
),x)*a**2*b*c**2*d - 20*int((a + b*x**3)**(1/3)/(a*c + a*d*x**3 + b*c*x**3 
 + b*d*x**6),x)*a*b**2*c**3 - 2*int(((a + b*x**3)**(1/3)*x**3)/(a*c + a*d* 
x**3 + b*c*x**3 + b*d*x**6),x)*a**3*d**3 - 16*int(((a + b*x**3)**(1/3)*x** 
3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a**2*b*c*d**2 + 60*int(((a + 
b*x**3)**(1/3)*x**3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b**2*c**2 
*d - 40*int(((a + b*x**3)**(1/3)*x**3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x* 
*6),x)*b**3*c**3)/(40*b*d**3)