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

Optimal. Leaf size=120 \[ -\frac {17 a^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 \sqrt [3]{b}}+\frac {17 a^2 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} \sqrt [3]{b}}-\frac {13}{18} a x \left (a+b x^3\right )^{2/3}-\frac {1}{6} x \left (a-b x^3\right ) \left (a+b x^3\right )^{2/3} \]

[Out]

-13/18*a*x*(b*x^3+a)^(2/3)-1/6*x*(-b*x^3+a)*(b*x^3+a)^(2/3)-17/18*a^2*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(1/3)+1
7/27*a^2*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 = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {416, 388, 239} \[ -\frac {17 a^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 \sqrt [3]{b}}+\frac {17 a^2 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} \sqrt [3]{b}}-\frac {13}{18} a x \left (a+b x^3\right )^{2/3}-\frac {1}{6} x \left (a-b x^3\right ) \left (a+b x^3\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13*a*x*(a + b*x^3)^(2/3))/18 - (x*(a - b*x^3)*(a + b*x^3)^(2/3))/6 + (17*a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a +
b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(1/3)) - (17*a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(18*b^(1/3))

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 \frac {\left (a-b x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx &=-\frac {1}{6} x \left (a-b x^3\right ) \left (a+b x^3\right )^{2/3}+\frac {\int \frac {7 a^2 b-13 a b^2 x^3}{\sqrt [3]{a+b x^3}} \, dx}{6 b}\\ &=-\frac {13}{18} a x \left (a+b x^3\right )^{2/3}-\frac {1}{6} x \left (a-b x^3\right ) \left (a+b x^3\right )^{2/3}+\frac {1}{9} \left (17 a^2\right ) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx\\ &=-\frac {13}{18} a x \left (a+b x^3\right )^{2/3}-\frac {1}{6} x \left (a-b x^3\right ) \left (a+b x^3\right )^{2/3}+\frac {17 a^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} \sqrt [3]{b}}-\frac {17 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 \sqrt [3]{b}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 141, normalized size = 1.18 \[ \frac {17 a^2 \left (\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )\right )}{54 \sqrt [3]{b}}+\left (a+b x^3\right )^{2/3} \left (\frac {b x^4}{6}-\frac {8 a x}{9}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x^3)^(2/3)*((-8*a*x)/9 + (b*x^4)/6) + (17*a^2*(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/S
qrt[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)]))/(54*b^(1/3))

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fricas [A]  time = 0.81, size = 399, normalized size = 3.32 \[ \left [\frac {51 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} b x^{3} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} x\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 2 \, a\right ) - 34 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 17 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (3 \, b^{2} x^{4} - 16 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b}, -\frac {102 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) + 34 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 17 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b^{2} x^{4} - 16 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(51*sqrt(1/3)*a^2*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) - 34*a
^2*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 17*a^2*(-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*(3*b^2*x^4 - 16*a*b*x)*(b*x^3 + a)^(2/3))/b, -1/54*(102*sqrt(
1/3)*a^2*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) +
34*a^2*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 17*a^2*(-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*(3*b^2*x^4 - 16*a*b*x)*(b*x^3 + a)^(2/3))/b]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 1.40, size = 436, normalized size = 3.63 \[ -\frac {1}{6} \, {\left (\frac {2 \, \sqrt {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 {1}{3}}} - \frac {\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 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}}\right )} a^{2} - \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 {4}{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 {4}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{2}}\right )} a b - \frac {1}{54} \, {\left (\frac {4 \, \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 {7}{3}}} - \frac {2 \, 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 {7}{3}}} + \frac {4 \, a^{2} \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 {7 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(1/3) - 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*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3))*a^2 - 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^(4/3) - a*log(b^(2/3) + (b*x^3
+ a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(4/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) - 6*(b
*x^3 + a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2))*a*b - 1/54*(4*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(
b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 2*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^(7/3) + 4*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) - 3*(7*(b*x^3 + a)^(2/3)*a^2*b/x^2 - 4*(b*x^3 +
a)^(5/3)*a^2/x^5)/(b^4 - 2*(b*x^3 + a)*b^3/x^3 + (b*x^3 + a)^2*b^2/x^6))*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 7.60, size = 121, normalized size = 1.01 \[ \frac {a^{\frac {5}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 a^{\frac {2}{3}} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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