\(\int (a-b x^3)^2 \sqrt [3]{a+b x^3} \, dx\) [89]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=-\frac {1}{2} a x \left (a+b x^3\right )^{4/3}+\frac {1}{8} b x^4 \left (a+b x^3\right )^{4/3}+\frac {3 a^2 x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

-1/2*a*x*(b*x^3+a)^(4/3)+1/8*b*x^4*(b*x^3+a)^(4/3)+3/2*a^2*x*(b*x^3+a)^(1/ 
3)*hypergeom([-1/3, 1/3],[4/3],-b*x^3/a)/(1+b*x^3/a)^(1/3)
 

Mathematica [A] (verified)

Time = 5.75 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {x \left (2 a^3-a^2 b x^3-2 a b^2 x^6+b^3 x^9+6 a^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{8 \left (a+b x^3\right )^{2/3}} \] Input:

Integrate[(a - b*x^3)^2*(a + b*x^3)^(1/3),x]
 

Output:

(x*(2*a^3 - a^2*b*x^3 - 2*a*b^2*x^6 + b^3*x^9 + 6*a^3*(1 + (b*x^3)/a)^(2/3 
)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)]))/(8*(a + b*x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {933, 27, 913, 779, 778}

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.

\(\displaystyle \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx\)

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int 3 a b \left (3 a-5 b x^3\right ) \sqrt [3]{b x^3+a}dx}{8 b}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{8} a \int \left (3 a-5 b x^3\right ) \sqrt [3]{b x^3+a}dx-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {3}{8} a \left (4 a \int \sqrt [3]{b x^3+a}dx-x \left (a+b x^3\right )^{4/3}\right )-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {3}{8} a \left (\frac {4 a \sqrt [3]{a+b x^3} \int \sqrt [3]{\frac {b x^3}{a}+1}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}-x \left (a+b x^3\right )^{4/3}\right )-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {3}{8} a \left (\frac {4 a x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{\frac {b x^3}{a}+1}}-x \left (a+b x^3\right )^{4/3}\right )-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}\)

Input:

Int[(a - b*x^3)^2*(a + b*x^3)^(1/3),x]
 

Output:

-1/8*(x*(a - b*x^3)*(a + b*x^3)^(4/3)) + (3*a*(-(x*(a + b*x^3)^(4/3)) + (4 
*a*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(1 
 + (b*x^3)/a)^(1/3)))/8
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), 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, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

Maple [F]

\[\int \left (-b \,x^{3}+a \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}}d x\]

Input:

int((-b*x^3+a)^2*(b*x^3+a)^(1/3),x)
 

Output:

int((-b*x^3+a)^2*(b*x^3+a)^(1/3),x)
 

Fricas [F]

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

integrate((-b*x^3+a)^2*(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

integral((b^2*x^6 - 2*a*b*x^3 + a^2)*(b*x^3 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.42 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {a^{\frac {7}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 a^{\frac {4}{3}} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate((-b*x**3+a)**2*(b*x**3+a)**(1/3),x)
 

Output:

a**(7/3)*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a) 
/(3*gamma(4/3)) - 2*a**(4/3)*b*x**4*gamma(4/3)*hyper((-1/3, 4/3), (7/3,), 
b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**(1/3)*b**2*x**7*gamma(7/3)*h 
yper((-1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))
 

Maxima [F]

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

integrate((-b*x^3+a)^2*(b*x^3+a)^(1/3),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(1/3)*(b*x^3 - a)^2, x)
 

Giac [F]

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

integrate((-b*x^3+a)^2*(b*x^3+a)^(1/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^3 + a)^(1/3)*(b*x^3 - a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\int {\left (b\,x^3+a\right )}^{1/3}\,{\left (a-b\,x^3\right )}^2 \,d x \] Input:

int((a + b*x^3)^(1/3)*(a - b*x^3)^2,x)
 

Output:

int((a + b*x^3)^(1/3)*(a - b*x^3)^2, x)
 

Reduce [F]

\[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} x}{4}-\frac {3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{4}}{8}+\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{7}}{8}+\frac {3 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{3}}{4} \] Input:

int((-b*x^3+a)^2*(b*x^3+a)^(1/3),x)
 

Output:

(2*(a + b*x**3)**(1/3)*a**2*x - 3*(a + b*x**3)**(1/3)*a*b*x**4 + (a + b*x* 
*3)**(1/3)*b**2*x**7 + 6*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**3)/8