Integrand size = 17, antiderivative size = 54 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=\sqrt [4]{-x^3+x^4}+\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \]
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (2 \sqrt [4]{-1+x} x^{3/4}+\arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-\text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{2 \left ((-1+x) x^3\right )^{3/4}} \]
((-1 + x)^(3/4)*x^(9/4)*(2*(-1 + x)^(1/4)*x^(3/4) + ArcTan[((-1 + x)/x)^(- 1/4)] - ArcTanh[((-1 + x)/x)^(-1/4)]))/(2*((-1 + x)*x^3)^(3/4))
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1927, 1938, 73, 770, 756, 216, 219}
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.
\(\displaystyle \int \frac {\sqrt [4]{x^4-x^3}}{x} \, dx\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {1}{4} \int \frac {x^2}{\left (x^4-x^3\right )^{3/4}}dx\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \int \frac {1}{(x-1)^{3/4} \sqrt [4]{x}}dx}{4 \left (x^4-x^3\right )^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \int \frac {1}{\sqrt [4]{x}}d\sqrt [4]{x-1}}{\left (x^4-x^3\right )^{3/4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \int \frac {1}{2-x}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}}{\left (x^4-x^3\right )^{3/4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \int \frac {1}{\sqrt {x-1}+1}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\left (x^4-x^3\right )^{3/4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt [4]{x^4-x^3}-\frac {(x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\left (x^4-x^3\right )^{3/4}}\) |
(-x^3 + x^4)^(1/4) - ((-1 + x)^(3/4)*x^(9/4)*(ArcTan[(-1 + x)^(1/4)/x^(1/4 )]/2 + ArcTanh[(-1 + x)^(1/4)/x^(1/4)]/2))/(-x^3 + x^4)^(3/4)
3.7.93.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
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 + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + 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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.50
method | result | size |
meijerg | \(\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x\right )}{3 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}}}\) | \(27\) |
pseudoelliptic | \(\frac {\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}-x}{x}\right )}{4}-\frac {\arctan \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )}{2}-\frac {\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+x}{x}\right )}{4}+\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}\) | \(65\) |
trager | \(\left (x^{4}-x^{3}\right )^{\frac {1}{4}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{4}-\frac {\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )}{4}\) | \(158\) |
risch | \(\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+\frac {\left (\frac {\ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-2 x^{3}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}-4 x +1}{\left (x -1\right )^{2}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (x -1\right )^{2}}\right )}{4}\right ) \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} \left (x \left (x -1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x -1\right )}\) | \(390\) |
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx={\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]
(x^4 - x^3)^(1/4) - 1/2*arctan((x^4 - x^3)^(1/4)/x) - 1/4*log((x + (x^4 - x^3)^(1/4))/x) + 1/4*log(-(x - (x^4 - x^3)^(1/4))/x)
\[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x}\, dx \]
\[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=x {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - \frac {1}{2} \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
x*(-1/x + 1)^(1/4) - 1/2*arctan((-1/x + 1)^(1/4)) - 1/4*log((-1/x + 1)^(1/ 4) + 1) + 1/4*log(abs((-1/x + 1)^(1/4) - 1))
Timed out. \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x} \, dx=\int \frac {{\left (x^4-x^3\right )}^{1/4}}{x} \,d x \]