\(\int \frac {(a-b x^3)^2}{(a+b x^3)^{2/3}} \, dx\) [90]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=-\frac {7}{5} a x \sqrt [3]{a+b x^3}+\frac {1}{5} b x^4 \sqrt [3]{a+b x^3}+\frac {12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \] Output:

-7/5*a*x*(b*x^3+a)^(1/3)+1/5*b*x^4*(b*x^3+a)^(1/3)+12/5*a^2*x*(1+b*x^3/a)^ 
(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)^(2/3)
 

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {-7 a^2 x-6 a b x^4+b^2 x^7+12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \] Input:

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

Output:

(-7*a^2*x - 6*a*b*x^4 + b^2*x^7 + 12*a^2*x*(1 + (b*x^3)/a)^(2/3)*Hypergeom 
etric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*(a + b*x^3)^(2/3))
 

Rubi [A] (verified)

Time = 0.38 (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.

\(\displaystyle \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \frac {6 a b \left (a-2 b x^3\right )}{\left (b x^3+a\right )^{2/3}}dx}{5 b}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{5} a \int \frac {a-2 b x^3}{\left (b x^3+a\right )^{2/3}}dx-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {6}{5} a \left (2 a \int \frac {1}{\left (b x^3+a\right )^{2/3}}dx-x \sqrt [3]{a+b x^3}\right )-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {6}{5} a \left (\frac {2 a \left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{2/3}}dx}{\left (a+b x^3\right )^{2/3}}-x \sqrt [3]{a+b x^3}\right )-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {6}{5} a \left (\frac {2 a x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-x \sqrt [3]{a+b x^3}\right )-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}\)

Input:

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

Output:

-1/5*(x*(a - b*x^3)*(a + b*x^3)^(1/3)) + (6*a*(-(x*(a + b*x^3)^(1/3)) + (2 
*a*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)]) 
/(a + b*x^3)^(2/3)))/5
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.51 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {a^{\frac {4}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 \sqrt [3]{a} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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