\(\int (a-b x^3)^2 \sqrt [3]{a+b x^3} \, dx\) [49]

Optimal result

Integrand size = 22, antiderivative size = 94 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=-\frac {3}{8} a x \left (a+b x^3\right )^{4/3}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}+\frac {3 a^2 x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{1+\frac {b x^3}{a}}} \]

[Out]

-3/8*a*x*(b*x^3+a)^(4/3)-1/8*x*(-b*x^3+a)*(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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {427, 396, 252, 251} \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {3 a^2 x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {3}{8} a x \left (a+b x^3\right )^{4/3}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3} \]

[In]

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

[Out]

(-3*a*x*(a + b*x^3)^(4/3))/8 - (x*(a - b*x^3)*(a + b*x^3)^(4/3))/8 + (3*a^2*x*(a + b*x^3)^(1/3)*Hypergeometric
2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(2*(1 + (b*x^3)/a)^(1/3))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

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

Rule 427

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] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[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] /; F
reeQ[{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] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}+\frac {\int \sqrt [3]{a+b x^3} \left (9 a^2 b-15 a b^2 x^3\right ) \, dx}{8 b} \\ & = -\frac {3}{8} a x \left (a+b x^3\right )^{4/3}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}+\frac {1}{2} \left (3 a^2\right ) \int \sqrt [3]{a+b x^3} \, dx \\ & = -\frac {3}{8} a x \left (a+b x^3\right )^{4/3}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}+\frac {\left (3 a^2 \sqrt [3]{a+b x^3}\right ) \int \sqrt [3]{1+\frac {b x^3}{a}} \, dx}{2 \sqrt [3]{1+\frac {b x^3}{a}}} \\ & = -\frac {3}{8} a x \left (a+b x^3\right )^{4/3}-\frac {1}{8} x \left (a-b x^3\right ) \left (a+b x^3\right )^{4/3}+\frac {3 a^2 x \sqrt [3]{a+b x^3} \, _2F_1\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}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.71 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {x \left (2 a^3-a^2 b x^3-2 a b^2 x^6+b^3 x^9+6 a^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{8 \left (a+b x^3\right )^{2/3}} \]

[In]

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

[Out]

(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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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 = 1.46 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34 \[ \int \left (a-b x^3\right )^2 \sqrt [3]{a+b x^3} \, dx=\frac {a^{\frac {7}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 a^{\frac {4}{3}} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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