3.48 \(\int (a-b x^3)^2 (a+b x^3)^{4/3} \, dx\)

Optimal. Leaf size=94 \[ \frac{57 a^3 x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{44 \sqrt [3]{\frac{b x^3}{a}+1}}-\frac{9}{44} a x \left (a+b x^3\right )^{7/3}-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3} \]

[Out]

(-9*a*x*(a + b*x^3)^(7/3))/44 - (x*(a - b*x^3)*(a + b*x^3)^(7/3))/11 + (57*a^3*x*(a + b*x^3)^(1/3)*Hypergeomet
ric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(44*(1 + (b*x^3)/a)^(1/3))

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Rubi [A]  time = 0.033539, antiderivative size = 94, normalized size of antiderivative = 1., 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} \[ \frac{57 a^3 x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{44 \sqrt [3]{\frac{b x^3}{a}+1}}-\frac{9}{44} a x \left (a+b x^3\right )^{7/3}-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a*x*(a + b*x^3)^(7/3))/44 - (x*(a - b*x^3)*(a + b*x^3)^(7/3))/11 + (57*a^3*x*(a + b*x^3)^(1/3)*Hypergeomet
ric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(44*(1 + (b*x^3)/a)^(1/3))

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]

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 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 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])

Rubi steps

\begin{align*} \int \left (a-b x^3\right )^2 \left (a+b x^3\right )^{4/3} \, dx &=-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3}+\frac{\int \left (a+b x^3\right )^{4/3} \left (12 a^2 b-18 a b^2 x^3\right ) \, dx}{11 b}\\ &=-\frac{9}{44} a x \left (a+b x^3\right )^{7/3}-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3}+\frac{1}{44} \left (57 a^2\right ) \int \left (a+b x^3\right )^{4/3} \, dx\\ &=-\frac{9}{44} a x \left (a+b x^3\right )^{7/3}-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3}+\frac{\left (57 a^3 \sqrt [3]{a+b x^3}\right ) \int \left (1+\frac{b x^3}{a}\right )^{4/3} \, dx}{44 \sqrt [3]{1+\frac{b x^3}{a}}}\\ &=-\frac{9}{44} a x \left (a+b x^3\right )^{7/3}-\frac{1}{11} x \left (a-b x^3\right ) \left (a+b x^3\right )^{7/3}+\frac{57 a^3 x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac{4}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{44 \sqrt [3]{1+\frac{b x^3}{a}}}\\ \end{align*}

Mathematica [A]  time = 0.0481449, size = 97, normalized size = 1.03 \[ \frac{x \left (-78 a^2 b^2 x^6+114 a^4 \left (\frac{b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )+53 a^3 b x^3+106 a^4-5 a b^3 x^9+20 b^4 x^{12}\right )}{220 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(106*a^4 + 53*a^3*b*x^3 - 78*a^2*b^2*x^6 - 5*a*b^3*x^9 + 20*b^4*x^12 + 114*a^4*(1 + (b*x^3)/a)^(2/3)*Hyperg
eometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)]))/(220*(a + b*x^3)^(2/3))

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Maple [F]  time = 0.215, size = 0, normalized size = 0. \begin{align*} \int \left ( -b{x}^{3}+a \right ) ^{2} \left ( b{x}^{3}+a \right ) ^{{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^3+a)^2*(b*x^3+a)^(4/3),x)

[Out]

int((-b*x^3+a)^2*(b*x^3+a)^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{4}{3}}{\left (b x^{3} - a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2*(b*x^3+a)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)*(b*x^3 - a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} x^{9} - a b^{2} x^{6} - a^{2} b x^{3} + a^{3}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2*(b*x^3+a)^(4/3),x, algorithm="fricas")

[Out]

integral((b^3*x^9 - a*b^2*x^6 - a^2*b*x^3 + a^3)*(b*x^3 + a)^(1/3), x)

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Sympy [C]  time = 5.11501, size = 168, normalized size = 1.79 \begin{align*} \frac{a^{\frac{10}{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{a^{\frac{7}{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{a^{\frac{4}{3}} 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 )} + \frac{\sqrt [3]{a} b^{3} x^{10} \Gamma \left (\frac{10}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \frac{10}{3} \\ \frac{13}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{13}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**3+a)**2*(b*x**3+a)**(4/3),x)

[Out]

a**(10/3)*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) - a**(7/3)*b*x**4*g
amma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) - a**(4/3)*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)) + a**(1/3)*b**3*x**10*gamma(10/3)*hyper(
(-1/3, 10/3), (13/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(13/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{4}{3}}{\left (b x^{3} - a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2*(b*x^3+a)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)*(b*x^3 - a)^2, x)