Integrand size = 15, antiderivative size = 154 \[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {1}{3} x \sqrt [3]{a x^3+b x^6}-\frac {a x^2 \left (a+b x^3\right )^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3} \left (a x^3+b x^6\right )^{2/3}}-\frac {a x^2 \left (a+b x^3\right )^{2/3} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3} \left (a x^3+b x^6\right )^{2/3}} \] Output:
1/3*x*(b*x^6+a*x^3)^(1/3)-1/9*a*x^2*(b*x^3+a)^(2/3)*arctan(1/3*(1+2*b^(1/3 )*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(2/3)/(b*x^6+a*x^3)^(2/3)-1/6*a*x^ 2*(b*x^3+a)^(2/3)*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)/(b*x^6+a*x^3)^(2/3 )
Time = 0.44 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.10 \[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {x^2 \left (a+b x^3\right )^{2/3} \left (6 b^{2/3} x^2 \sqrt [3]{a+b x^3}-2 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )\right )}{18 b^{2/3} \left (x^3 \left (a+b x^3\right )\right )^{2/3}} \] Input:
Integrate[(a*x^3 + b*x^6)^(1/3),x]
Output:
(x^2*(a + b*x^3)^(2/3)*(6*b^(2/3)*x^2*(a + b*x^3)^(1/3) - 2*Sqrt[3]*a*ArcT an[(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)]))/(18*b^(2/3)*(x^3*(a + b*x^3))^(2/3))
Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1910, 1938, 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.
\(\displaystyle \int \sqrt [3]{a x^3+b x^6} \, dx\) |
\(\Big \downarrow \) 1910 |
\(\displaystyle \frac {1}{3} a \int \frac {x^3}{\left (b x^6+a x^3\right )^{2/3}}dx+\frac {1}{3} x \sqrt [3]{a x^3+b x^6}\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {a x^2 \left (a+b x^3\right )^{2/3} \int \frac {x}{\left (b x^3+a\right )^{2/3}}dx}{3 \left (a x^3+b x^6\right )^{2/3}}+\frac {1}{3} x \sqrt [3]{a x^3+b x^6}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {a x^2 \left (a+b x^3\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}\right )}{3 \left (a x^3+b x^6\right )^{2/3}}+\frac {1}{3} x \sqrt [3]{a x^3+b x^6}\) |
Input:
Int[(a*x^3 + b*x^6)^(1/3),x]
Output:
(x*(a*x^3 + b*x^6)^(1/3))/3 + (a*x^2*(a + b*x^3)^(2/3)*(-(ArcTan[(1 + (2*b ^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(2/3))) - Log[b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*b^(2/3))))/(3*(a*x^3 + b*x^6)^(2/3))
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]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1)) Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && !IntegerQ[p] && LtQ[0, j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])) Int[x^(m + j* p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !Inte gerQ[p] && NeQ[n, j] && PosQ[n - j]
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(\frac {6 \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}} x \,b^{\frac {2}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x^{2}+2 \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x^{2}}\right ) a -2 \ln \left (\frac {-b^{\frac {1}{3}} x^{2}+\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}}}{x^{2}}\right ) a +\ln \left (\frac {b^{\frac {2}{3}} x^{4}+b^{\frac {1}{3}} \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}} x^{2}+\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}}}{x^{4}}\right ) a}{18 b^{\frac {2}{3}}}\) | \(145\) |
Input:
int((b*x^6+a*x^3)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/18*(6*(x^3*(b*x^3+a))^(1/3)*x*b^(2/3)+2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1 /3)*x^2+2*(x^3*(b*x^3+a))^(1/3))/b^(1/3)/x^2)*a-2*ln((-b^(1/3)*x^2+(x^3*(b *x^3+a))^(1/3))/x^2)*a+ln((b^(2/3)*x^4+b^(1/3)*(x^3*(b*x^3+a))^(1/3)*x^2+( x^3*(b*x^3+a))^(2/3))/x^4)*a)/b^(2/3)
Time = 0.07 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.39 \[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {6 \, \sqrt {\frac {1}{3}} a b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{b^{2} x^{2}}\right ) + 6 \, {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} b^{2} x - 2 \, \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x^{2} - {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} b}{x^{2}}\right ) + \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{4} - {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x^{2} - {\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}} b}{x^{4}}\right )}{18 \, b^{2}} \] Input:
integrate((b*x^6+a*x^3)^(1/3),x, algorithm="fricas")
Output:
1/18*(6*sqrt(1/3)*a*b*sqrt(-(-b^2)^(1/3))*arctan(-sqrt(1/3)*((-b^2)^(1/3)* b*x^2 - 2*(b*x^6 + a*x^3)^(1/3)*(-b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x^2 )) + 6*(b*x^6 + a*x^3)^(1/3)*b^2*x - 2*(-b^2)^(2/3)*a*log(-((-b^2)^(2/3)*x ^2 - (b*x^6 + a*x^3)^(1/3)*b)/x^2) + (-b^2)^(2/3)*a*log(-((-b^2)^(1/3)*b*x ^4 - (b*x^6 + a*x^3)^(1/3)*(-b^2)^(2/3)*x^2 - (b*x^6 + a*x^3)^(2/3)*b)/x^4 ))/b^2
\[ \int \sqrt [3]{a x^3+b x^6} \, dx=\int \sqrt [3]{a x^{3} + b x^{6}}\, dx \] Input:
integrate((b*x**6+a*x**3)**(1/3),x)
Output:
Integral((a*x**3 + b*x**6)**(1/3), x)
\[ \int \sqrt [3]{a x^3+b x^6} \, dx=\int { {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((b*x^6+a*x^3)^(1/3),x, algorithm="maxima")
Output:
integrate((b*x^6 + a*x^3)^(1/3), x)
Time = 2.77 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.69 \[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {1}{18} \, {\left (\frac {6 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} x^{3}}{a} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}} + \frac {\log \left ({\left (b + \frac {a}{x^{3}}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {2}{3}}} - \frac {2 \, \log \left ({\left | {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {2}{3}}}\right )} a \] Input:
integrate((b*x^6+a*x^3)^(1/3),x, algorithm="giac")
Output:
1/18*(6*(b + a/x^3)^(1/3)*x^3/a + 2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b + a/x ^3)^(1/3) + b^(1/3))/b^(1/3))/b^(2/3) + log((b + a/x^3)^(2/3) + (b + a/x^3 )^(1/3)*b^(1/3) + b^(2/3))/b^(2/3) - 2*log(abs((b + a/x^3)^(1/3) - b^(1/3) ))/b^(2/3))*a
Time = 18.64 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.27 \[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {x\,{\left (b\,x^6+a\,x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -\frac {b\,x^3}{a}\right )}{2\,{\left (\frac {b\,x^3}{a}+1\right )}^{1/3}} \] Input:
int((a*x^3 + b*x^6)^(1/3),x)
Output:
(x*(a*x^3 + b*x^6)^(1/3)*hypergeom([-1/3, 2/3], 5/3, -(b*x^3)/a))/(2*((b*x ^3)/a + 1)^(1/3))
\[ \int \sqrt [3]{a x^3+b x^6} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2}}{3}+\frac {\left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a}{3} \] Input:
int((b*x^6+a*x^3)^(1/3),x)
Output:
((a + b*x**3)**(1/3)*x**2 + int(((a + b*x**3)**(1/3)*x)/(a + b*x**3),x)*a) /3