\(\int \frac {x^3 (a+b x^3)^{4/3}}{c+d x^3} \, dx\) [725]

Optimal result

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

1/4*a*x^4*(b*x^3+a)^(1/3)*AppellF1(4/3,-4/3,1,7/3,-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. \(280\) vs. \(2(65)=130\).

Time = 8.14 (sec) , antiderivative size = 280, normalized size of antiderivative = 4.31 \[ \int \frac {x^3 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {x \left (4 \left (a+b x^3\right ) \left (-5 b c+6 a d+2 b d x^3\right )+\frac {\left (10 b^2 c^2-15 a b c d+4 a^2 d^2\right ) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c}+\frac {16 a^2 c^2 (-5 b c+6 a d) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{\left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{40 d^2 \left (a+b x^3\right )^{2/3}} \] Input:

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

Output:

(x*(4*(a + b*x^3)*(-5*b*c + 6*a*d + 2*b*d*x^3) + ((10*b^2*c^2 - 15*a*b*c*d 
 + 4*a^2*d^2)*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*(-5*b*c + 6*a*d)*AppellF1[1/3, 2/3, 
1, 4/3, -((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)])))))/(40*d^2*(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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a d x -5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{4}-6 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} c d +5 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b \,c^{2}+4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} d^{2}-15 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b c d +10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c^{2}}{10 d^{2}} \] Input:

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

Output:

(6*(a + b*x**3)**(1/3)*a*d*x - 5*(a + b*x**3)**(1/3)*b*c*x + 2*(a + b*x**3 
)**(1/3)*b*d*x**4 - 6*int((a + b*x**3)**(1/3)/(a*c + a*d*x**3 + b*c*x**3 + 
 b*d*x**6),x)*a**2*c*d + 5*int((a + b*x**3)**(1/3)/(a*c + a*d*x**3 + b*c*x 
**3 + b*d*x**6),x)*a*b*c**2 + 4*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*d**2 - 15*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*c*d + 10*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**2*c**2)/(10*d* 
*2)