3.1.29 \(\int \frac {a-b x^3}{(a+b x^3)^{4/3}} \, dx\) [29]

Optimal. Leaf size=85 \[ \frac {2 x}{\sqrt [3]{a+b x^3}}-\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}} \]

[Out]

2*x/(b*x^3+a)^(1/3)+1/2*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(1/3)-1/3*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 = 0.01, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {393, 245} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {2 x}{\sqrt [3]{a+b x^3}}+\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)/(a + b*x^3)^(4/3),x]

[Out]

(2*x)/(a + b*x^3)^(1/3) - ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) + Log[-(b^(1
/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))

Rule 245

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 393

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

Rubi steps

\begin {align*} \int \frac {a-b x^3}{\left (a+b x^3\right )^{4/3}} \, dx &=\frac {2 x}{\sqrt [3]{a+b x^3}}-\int \frac {1}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac {2 x}{\sqrt [3]{a+b x^3}}-\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 142, normalized size = 1.67 \begin {gather*} \frac {2 x}{\sqrt [3]{a+b x^3}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 \sqrt [3]{b}}-\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/(a + b*x^3)^(1/3) - ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))]/(Sqrt[3]*b^(1/3)) + Lo
g[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(3*b^(1/3)) - Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^
(2/3)]/(6*b^(1/3))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {-b \,x^{3}+a}{\left (b \,x^{3}+a \right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 130, normalized size = 1.53 \begin {gather*} \frac {1}{6} \, b {\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 {4}{3}}} + \frac {6 \, x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} b} - \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 {4}{3}}} + \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}}\right )} + \frac {x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (66) = 132\).
time = 4.35, size = 372, normalized size = 4.38 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (b^{2} x^{3} + a b\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 12 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2 \, {\left (b x^{3} + a\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (b x^{3} + a\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{6 \, {\left (b^{2} x^{3} + a b\right )}}, \frac {12 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2 \, {\left (b x^{3} + a\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (b x^{3} + a\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {6 \, \sqrt {\frac {1}{3}} {\left (b^{2} x^{3} + a b\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}}}{6 \, {\left (b^{2} x^{3} + a b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(1/3)*(b^2*x^3 + a*b)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)
*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) + 12*(b*x^3 +
 a)^(2/3)*b*x + 2*(b*x^3 + a)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (b*x^3 + a)*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))/(b^2*x^3 + a*b), 1/6*(12*(b*x^3 + a)^(2/3)*b*x
 + 2*(b*x^3 + a)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (b*x^3 + a)*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) + 6*sqrt(1/3)*(b^2*x^3 + a*b)*arctan(sqrt(1/3)*(b^(1/3)*x +
2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3))/(b^2*x^3 + a*b)]

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Sympy [C] Result contains complex when optimal does not.
time = 3.39, size = 70, normalized size = 0.82 \begin {gather*} \frac {x \Gamma \left (\frac {1}{3}\right )}{3 \sqrt [3]{a} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {4}{3}\right )} - \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a-b\,x^3}{{\left (b\,x^3+a\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^3)/(a + b*x^3)^(4/3),x)

[Out]

int((a - b*x^3)/(a + b*x^3)^(4/3), x)

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