Integrand size = 15, antiderivative size = 120 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}+\frac {a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}} \]
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*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/ 3))*3^(1/2))/b^(5/3)*3^(1/2)
Time = 0.36 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.40 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {x^2 \sqrt [3]{a+b x^3} \left (a+3 b x^3\right )}{18 b}+\frac {a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{27 b^{5/3}}-\frac {a^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{5/3}} \]
(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))
Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 843, 853}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^4 \sqrt [3]{a+b x^3} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{6} a \int \frac {x^4}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\left (b x^3+a\right )^{2/3}}dx}{3 b}\right )+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {1}{6} a \left (\frac {x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {2 a \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}\right )}{3 b}\right )+\frac {1}{6} x^5 \sqrt [3]{a+b x^3}\) |
(x^5*(a + b*x^3)^(1/3))/6 + (a*((x^2*(a + b*x^3)^(1/3))/(3*b) - (2*a*(-(Ar cTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(2/3))) - L og[b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*b^(2/3))))/(3*b)))/6
3.6.16.3.1 Defintions of rubi rules used
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] + Simp[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] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
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] - Simp[ 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]
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim p[-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]
Time = 4.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22
method | result | size |
pseudoelliptic | \(\frac {9 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {5}{3}}+3 a \,x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {2}{3}}-2 a^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+2 a^{2} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-a^{2} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{54 b^{\frac {5}{3}}}\) | \(146\) |
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)
Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=-\frac {2 \, \sqrt {3} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 2 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) - 3 \, {\left (3 \, b^{3} x^{5} + a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{3}} \]
-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)*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) - 3*(3*b^3*x^5 + a*b^2*x^2)*(b*x^3 + a)^(1/3))/b^3
Result contains complex when optimal does not.
Time = 1.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.32 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \]
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))
Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=-\frac {\sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {5}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{54 \, b^{\frac {5}{3}}} + \frac {a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {5}{3}}} + \frac {\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{x^{4}}}{18 \, {\left (b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}\right )}} \]
-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)
\[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4} \,d x } \]
Timed out. \[ \int x^4 \sqrt [3]{a+b x^3} \, dx=\int x^4\,{\left (b\,x^3+a\right )}^{1/3} \,d x \]