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

Optimal. Leaf size=139 \[ -\frac {38 a^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{81 \sqrt [3]{b}}+\frac {76 a^3 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{81 \sqrt {3} \sqrt [3]{b}}+\frac {38}{81} a^2 x \left (a+b x^3\right )^{2/3}-\frac {8}{27} a x \left (a+b x^3\right )^{5/3}-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3} \]

[Out]

38/81*a^2*x*(b*x^3+a)^(2/3)-8/27*a*x*(b*x^3+a)^(5/3)-1/9*x*(-b*x^3+a)*(b*x^3+a)^(5/3)-38/81*a^3*ln(-b^(1/3)*x+
(b*x^3+a)^(1/3))/b^(1/3)+76/243*a^3*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(1/3)*3^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 139, 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, 195, 239} \[ \frac {38}{81} a^2 x \left (a+b x^3\right )^{2/3}-\frac {38 a^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{81 \sqrt [3]{b}}+\frac {76 a^3 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{81 \sqrt {3} \sqrt [3]{b}}-\frac {8}{27} a x \left (a+b x^3\right )^{5/3}-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(38*a^2*x*(a + b*x^3)^(2/3))/81 - (8*a*x*(a + b*x^3)^(5/3))/27 - (x*(a - b*x^3)*(a + b*x^3)^(5/3))/9 + (76*a^3
*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(81*Sqrt[3]*b^(1/3)) - (38*a^3*Log[-(b^(1/3)*x) + (a +
b*x^3)^(1/3)])/(81*b^(1/3))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, 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 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 \left (a-b x^3\right )^2 \left (a+b x^3\right )^{2/3} \, dx &=-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3}+\frac {\int \left (a+b x^3\right )^{2/3} \left (10 a^2 b-16 a b^2 x^3\right ) \, dx}{9 b}\\ &=-\frac {8}{27} a x \left (a+b x^3\right )^{5/3}-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3}+\frac {1}{27} \left (38 a^2\right ) \int \left (a+b x^3\right )^{2/3} \, dx\\ &=\frac {38}{81} a^2 x \left (a+b x^3\right )^{2/3}-\frac {8}{27} a x \left (a+b x^3\right )^{5/3}-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3}+\frac {1}{81} \left (76 a^3\right ) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac {38}{81} a^2 x \left (a+b x^3\right )^{2/3}-\frac {8}{27} a x \left (a+b x^3\right )^{5/3}-\frac {1}{9} x \left (a-b x^3\right ) \left (a+b x^3\right )^{5/3}+\frac {76 a^3 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3} \sqrt [3]{b}}-\frac {38 a^3 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{81 \sqrt [3]{b}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 151, normalized size = 1.09 \[ \frac {1}{243} \left (\frac {38 a^3 \left (\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )\right )}{\sqrt [3]{b}}+3 \left (a+b x^3\right )^{2/3} \left (5 a^2 x-24 a b x^4+9 b^2 x^7\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x^3)^(2/3)*(5*a^2*x - 24*a*b*x^4 + 9*b^2*x^7) + (38*a^3*(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*
x^3)^(1/3))/Sqrt[3]] - 2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b
^(1/3)*x)/(a + b*x^3)^(1/3)]))/b^(1/3))/243

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fricas [A]  time = 0.88, size = 421, normalized size = 3.03 \[ \left [\frac {114 \, \sqrt {\frac {1}{3}} a^{3} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} b x^{3} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} x\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 2 \, a\right ) - 76 \, a^{3} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 38 \, a^{3} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (9 \, b^{3} x^{7} - 24 \, a b^{2} x^{4} + 5 \, a^{2} b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{243 \, b}, -\frac {228 \, \sqrt {\frac {1}{3}} a^{3} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) + 76 \, a^{3} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 38 \, a^{3} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (9 \, b^{3} x^{7} - 24 \, a b^{2} x^{4} + 5 \, a^{2} b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{243 \, b}\right ] \]

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]

[1/243*(114*sqrt(1/3)*a^3*b*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*
((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 76
*a^3*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 38*a^3*(-b)^(2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a
)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 3*(9*b^3*x^7 - 24*a*b^2*x^4 + 5*a^2*b*x)*(b*x^3 + a)^(2/3))/b
, -1/243*(228*sqrt(1/3)*a^3*b*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1/3))*sqrt(
-(-b)^(1/3)/b)/x) + 76*a^3*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 38*a^3*(-b)^(2/3)*log(((-b)^
(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 3*(9*b^3*x^7 - 24*a*b^2*x^4 + 5*a^2*b*x
)*(b*x^3 + a)^(2/3))/b]

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

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)^(2/3)*(b*x^3 - a)^2, x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \left (-b \,x^{3}+a \right )^{2} \left (b \,x^{3}+a \right )^{\frac {2}{3}}\, dx \]

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 [B]  time = 1.53, size = 552, normalized size = 3.97 \[ -\frac {1}{9} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x^{2}}\right )} a^{2} - \frac {1}{27} \, {\left (\frac {2 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} + \frac {3 \, {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}}\right )} a b - \frac {1}{243} \, {\left (\frac {4 \, \sqrt {3} a^{3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} - \frac {2 \, a^{3} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, a^{3} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{2}}{x^{2}} + \frac {11 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{3} b}{x^{5}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{3}}{x^{8}}\right )}}{b^{5} - \frac {3 \, {\left (b x^{3} + a\right )} b^{4}}{x^{3}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{3}}{x^{6}} - \frac {{\left (b x^{3} + a\right )}^{3} b^{2}}{x^{9}}}\right )} b^{2} \]

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]

-1/9*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(1/3) - a*log(b^(2/3) + (b*x
^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(1/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3) + 3
*(b*x^3 + a)^(2/3)*a/((b - (b*x^3 + a)/x^3)*x^2))*a^2 - 1/27*(2*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b
*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b
^(4/3) + 2*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) + 3*((b*x^3 + a)^(2/3)*a^2*b/x^2 + 2*(b*x^3 + a)^(5
/3)*a^2/x^5)/(b^3 - 2*(b*x^3 + a)*b^2/x^3 + (b*x^3 + a)^2*b/x^6))*a*b - 1/243*(4*sqrt(3)*a^3*arctan(1/3*sqrt(3
)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 2*a^3*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^
3 + a)^(2/3)/x^2)/b^(7/3) + 4*a^3*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) + 3*(2*(b*x^3 + a)^(2/3)*a^3*b^2
/x^2 + 11*(b*x^3 + a)^(5/3)*a^3*b/x^5 - 4*(b*x^3 + a)^(8/3)*a^3/x^8)/(b^5 - 3*(b*x^3 + a)*b^4/x^3 + 3*(b*x^3 +
a)^2*b^3/x^6 - (b*x^3 + a)^3*b^2/x^9))*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.17, size = 126, normalized size = 0.91 \[ \frac {a^{\frac {8}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 a^{\frac {5}{3}} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {2}{3}} b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

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**(8/3)*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) - 2*a**(5/3)*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)) + a**(2/3)*b**2*x**7*gamma(7/3)
*hyper((-2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))

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