\(\int x^4 \sqrt [3]{a+b x^3} \, dx\) [516]

Optimal result

Integrand size = 15, antiderivative size = 120 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}+\frac {a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}} \]

[Out]

1/18*a*x^2*(b*x^3+a)^(1/3)/b+1/6*x^5*(b*x^3+a)^(1/3)+1/18*a^2*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)+1/27*a^2*a
rctan(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 = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {285, 327, 337} \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {a^2 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}}+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}+\frac {a x^2 \sqrt [3]{a+b x^3}}{18 b} \]

[In]

Int[x^4*(a + b*x^3)^(1/3),x]

[Out]

(a*x^2*(a + b*x^3)^(1/3))/(18*b) + (x^5*(a + b*x^3)^(1/3))/6 + (a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3
))/Sqrt[3]])/(9*Sqrt[3]*b^(5/3)) + (a^2*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(18*b^(5/3))

Rule 285

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

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 {1}{6} x^5 \sqrt [3]{a+b x^3}+\frac {1}{6} a \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx \\ & = \frac {a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}-\frac {a^2 \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx}{9 b} \\ & = \frac {a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}+\frac {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} b^{5/3}}+\frac {a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.40 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {x^2 \sqrt [3]{a+b x^3} \left (a+3 b x^3\right )}{18 b}+\frac {a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{27 b^{5/3}}-\frac {a^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{5/3}} \]

[In]

Integrate[x^4*(a + b*x^3)^(1/3),x]

[Out]

(x^2*(a + b*x^3)^(1/3)*(a + 3*b*x^3))/(18*b) + (a^2*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3
))])/(9*Sqrt[3]*b^(5/3)) + (a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(27*b^(5/3)) - (a^2*Log[b^(2/3)*x^2 + b
^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*b^(5/3))

Maple [A] (verified)

Time = 4.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(\frac {9 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {5}{3}}+3 a \,x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {2}{3}}-2 a^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+2 a^{2} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-a^{2} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{54 b^{\frac {5}{3}}}\)	\(146\)

[In]

int(x^4*(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/54*(9*x^5*(b*x^3+a)^(1/3)*b^(5/3)+3*a*x^2*(b*x^3+a)^(1/3)*b^(2/3)-2*a^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)+2*a^2*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-a^2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)
^(1/3)*x+(b*x^3+a)^(2/3))/x^2))/b^(5/3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=-\frac {2 \, \sqrt {3} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 2 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) - 3 \, {\left (3 \, b^{3} x^{5} + a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{3}} \]

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

-1/54*(2*sqrt(3)*a^2*(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^2*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + a^2*(b^2)^(2/3)*l
og(((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) - 3*(3*b^3*x^5 + a*b^2*x^2
)*(b*x^3 + a)^(1/3))/b^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.32 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \]

[In]

integrate(x**4*(b*x**3+a)**(1/3),x)

[Out]

a**(1/3)*x**5*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(8/3))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=-\frac {\sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {5}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{54 \, b^{\frac {5}{3}}} + \frac {a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {5}{3}}} + \frac {\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{x^{4}}}{18 \, {\left (b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}\right )}} \]

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

-1/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(5/3) - 1/54*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^(5/3) + 1/27*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)
/b^(5/3) + 1/18*(2*(b*x^3 + a)^(1/3)*a^2*b/x + (b*x^3 + a)^(4/3)*a^2/x^4)/(b^3 - 2*(b*x^3 + a)*b^2/x^3 + (b*x^
3 + a)^2*b/x^6)

Giac [F]

\[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4} \,d x } \]

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*x^4, x)

Mupad [F(-1)]

Timed out. \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\int x^4\,{\left (b\,x^3+a\right )}^{1/3} \,d x \]

[In]

int(x^4*(a + b*x^3)^(1/3),x)

[Out]

int(x^4*(a + b*x^3)^(1/3), x)