Integrand size = 13, antiderivative size = 96 \[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=-\frac {3 \sqrt [3]{x+x^3}}{2 x}-\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )-\frac {1}{2} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{4} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \] Output:
-3/2*(x^3+x)^(1/3)/x-1/2*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))*3^(1/2)-1/2 *ln(-x+(x^3+x)^(1/3))+1/4*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))
Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\frac {\sqrt [3]{x+x^3} \left (-6 \sqrt [3]{1+x^2}-2 \sqrt {3} x^{2/3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-2 x^{2/3} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )+x^{2/3} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{4 x \sqrt [3]{1+x^2}} \] Input:
Integrate[(x + x^3)^(1/3)/x^2,x]
Output:
((x + x^3)^(1/3)*(-6*(1 + x^2)^(1/3) - 2*Sqrt[3]*x^(2/3)*ArcTan[(Sqrt[3]*x ^(2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))] - 2*x^(2/3)*Log[-x^(2/3) + (1 + x^2 )^(1/3)] + x^(2/3)*Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3) ]))/(4*x*(1 + x^2)^(1/3))
Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1926, 1938, 266, 807, 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.
\(\displaystyle \int \frac {\sqrt [3]{x^3+x}}{x^2} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \int \frac {x}{\left (x^3+x\right )^{2/3}}dx-\frac {3 \sqrt [3]{x^3+x}}{2 x}\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {\sqrt [3]{x}}{\left (x^2+1\right )^{2/3}}dx}{\left (x^3+x\right )^{2/3}}-\frac {3 \sqrt [3]{x^3+x}}{2 x}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {3 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x}{\left (x^2+1\right )^{2/3}}d\sqrt [3]{x}}{\left (x^3+x\right )^{2/3}}-\frac {3 \sqrt [3]{x^3+x}}{2 x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {3 x^{2/3} \left (x^2+1\right )^{2/3} \int \frac {x^{2/3}}{(x+1)^{2/3}}dx^{2/3}}{2 \left (x^3+x\right )^{2/3}}-\frac {3 \sqrt [3]{x^3+x}}{2 x}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {3 x^{2/3} \left (x^2+1\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^{2/3}-\sqrt [3]{x+1}\right )\right )}{2 \left (x^3+x\right )^{2/3}}-\frac {3 \sqrt [3]{x^3+x}}{2 x}\) |
Input:
Int[(x + x^3)^(1/3)/x^2,x]
Output:
(-3*(x + x^3)^(1/3))/(2*x) + (3*x^(2/3)*(1 + x^2)^(2/3)*(-(ArcTan[(1 + (2* x^(2/3))/(1 + x)^(1/3))/Sqrt[3]]/Sqrt[3]) - Log[x^(2/3) - (1 + x)^(1/3)]/2 ))/(2*(x + x^3)^(2/3))
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p *((n - j)/(c^n*(m + j*p + 1))) Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Integer sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*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]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 2.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.18
method | result | size |
meijerg | \(-\frac {3 \operatorname {hypergeom}\left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], -x^{2}\right )}{2 x^{\frac {2}{3}}}\) | \(17\) |
pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right ) x +\ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}}{x^{2}}\right ) x -2 \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x}{x}\right ) x -6 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}}{4 x}\) | \(98\) |
trager | \(-\frac {3 \left (x^{3}+x \right )^{\frac {1}{3}}}{2 x}+\frac {\ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right )}{2}-\frac {3 \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x -87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+8\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )}{2}+\frac {3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+72 \left (x^{3}+x \right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x +57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}+48 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-3\right )}{2}\) | \(432\) |
risch | \(\text {Expression too large to display}\) | \(729\) |
Input:
int((x^3+x)^(1/3)/x^2,x,method=_RETURNVERBOSE)
Output:
-3/2/x^(2/3)*hypergeom([-1/3,-1/3],[2/3],-x^2)
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=-\frac {2 \, \sqrt {3} x \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + x \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) + 6 \, {\left (x^{3} + x\right )}^{\frac {1}{3}}}{4 \, x} \] Input:
integrate((x^3+x)^(1/3)/x^2,x, algorithm="fricas")
Output:
-1/4*(2*sqrt(3)*x*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^ 2 + 507) - 1274*sqrt(3)*(x^3 + x)^(2/3))/(2205*x^2 + 2197)) + x*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) + 6*(x^3 + x)^(1/3))/x
\[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\int \frac {\sqrt [3]{x \left (x^{2} + 1\right )}}{x^{2}}\, dx \] Input:
integrate((x**3+x)**(1/3)/x**2,x)
Output:
Integral((x*(x**2 + 1))**(1/3)/x**2, x)
\[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\int { \frac {{\left (x^{3} + x\right )}^{\frac {1}{3}}}{x^{2}} \,d x } \] Input:
integrate((x^3+x)^(1/3)/x^2,x, algorithm="maxima")
Output:
integrate((x^3 + x)^(1/3)/x^2, x)
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {3}{2} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + \frac {1}{4} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:
integrate((x^3+x)^(1/3)/x^2,x, algorithm="giac")
Output:
1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) - 3/2*(1/x^2 + 1 )^(1/3) + 1/4*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 1/2*log(abs ((1/x^2 + 1)^(1/3) - 1))
Timed out. \[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\int \frac {{\left (x^3+x\right )}^{1/3}}{x^2} \,d x \] Input:
int((x + x^3)^(1/3)/x^2,x)
Output:
int((x + x^3)^(1/3)/x^2, x)
\[ \int \frac {\sqrt [3]{x+x^3}}{x^2} \, dx=\int \frac {\left (x^{2}+1\right )^{\frac {1}{3}}}{x^{\frac {5}{3}}}d x \] Input:
int((x^3+x)^(1/3)/x^2,x)
Output:
int((x**(1/3)*(x**2 + 1)**(1/3))/x**2,x)