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

Optimal. Leaf size=94 \[ \frac {12 a^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3} \]

[Out]

-6/5*a*x*(b*x^3+a)^(1/3)-1/5*x*(-b*x^3+a)*(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)

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Rubi [A]  time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {416, 388, 246, 245} \[ \frac {12 a^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*x*(a + b*x^3)^(1/3))/5 - (x*(a - b*x^3)*(a + b*x^3)^(1/3))/5 + (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))

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 388

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 416

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*} \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx &=-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {\int \frac {6 a^2 b-12 a b^2 x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{5 b}\\ &=-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {1}{5} \left (12 a^2\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {\left (12 a^2 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{5 \left (a+b x^3\right )^{2/3}}\\ &=-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\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}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.80 \[ \frac {12 a^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )-7 a^2 x-6 a b x^4+b^2 x^7}{5 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a^2*x - 6*a*b*x^4 + b^2*x^7 + 12*a^2*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)
])/(5*(a + b*x^3)^(2/3))

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{6} - 2 \, a b x^{3} + a^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a-b\,x^3\right )}^2}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.76, size = 121, normalized size = 1.29 \[ \frac {a^{\frac {4}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 \sqrt [3]{a} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*g
amma(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))

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