\(\int \frac {x^4}{(a+b x^3)^{2/3}} \, dx\) [568]

Optimal result

Integrand size = 15, antiderivative size = 97 \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b}+\frac {2 a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}} \]

[Out]

1/3*x^2*(b*x^3+a)^(1/3)/b+1/3*a*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)+2/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a
)^(1/3))*3^(1/2))/b^(5/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {327, 337} \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {2 a \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 b} \]

[In]

Int[x^4/(a + b*x^3)^(2/3),x]

[Out]

(x^2*(a + b*x^3)^(1/3))/(3*b) + (2*a*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*b^(5/3)
) + (a*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(3*b^(5/3))

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {(2 a) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx}{3 b} \\ & = \frac {x^2 \sqrt [3]{a+b x^3}}{3 b}+\frac {2 a \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46 \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {3 b^{2/3} x^2 \sqrt [3]{a+b x^3}+2 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+2 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{9 b^{5/3}} \]

[In]

Integrate[x^4/(a + b*x^3)^(2/3),x]

[Out]

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

Maple [A] (verified)

Time = 4.01 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26

method	result	size
pseudoelliptic	\(\frac {3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2} b^{\frac {2}{3}}-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a +2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a -\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a}{9 b^{\frac {5}{3}}}\)	\(122\)

[In]

int(x^4/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

1/9*(3*(b*x^3+a)^(1/3)*x^2*b^(2/3)-2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)*a+2*l
n((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a-ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)*a)/b^(5/3)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (74) = 148\).

Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.70 \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} x^{2} - 2 \, \sqrt {3} a {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) + 2 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{9 \, b^{3}} \]

[In]

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

[Out]

1/9*(3*(b*x^3 + a)^(1/3)*b^2*x^2 - 2*sqrt(3)*a*(b^2)^(1/6)*b*arctan(1/3*(sqrt(3)*(b^2)^(1/3)*b*x + 2*sqrt(3)*(
b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) + 2*a*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b
)/x) - a*(b^2)^(2/3)*log(((b^2)^(1/3)*b*x^2 + (b*x^3 + a)^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a)^(2/3)*b)/x^2))/b^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {8}{3}\right )} \]

[In]

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

[Out]

x**5*gamma(5/3)*hyper((2/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(8/3))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.41 \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=-\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{9 \, b^{\frac {5}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{9 \, b^{\frac {5}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{9 \, b^{\frac {5}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{3 \, {\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x} \]

[In]

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

[Out]

-2/9*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(5/3) - 1/9*a*log(b^(2/3) + (b*
x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(5/3) + 2/9*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(5/3)
- 1/3*(b*x^3 + a)^(1/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x)

Giac [F]

\[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {x^{4}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]

[In]

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

[Out]

integrate(x^4/(b*x^3 + a)^(2/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {x^4}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]

[In]

int(x^4/(a + b*x^3)^(2/3),x)

[Out]

int(x^4/(a + b*x^3)^(2/3), x)