Integrand size = 22, antiderivative size = 37 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=-2 \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \] Output:
-2*arctan(x/(x^4-x^3)^(1/4))+2*arctanh(x/(x^4-x^3)^(1/4))
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\frac {2 (-1+x)^{3/4} x^{9/4} \left (-\arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+\text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{\left ((-1+x) x^3\right )^{3/4}} \] Input:
Integrate[(-x^3 + x^4)^(1/4)/((-1 + x)*x),x]
Output:
(2*(-1 + x)^(3/4)*x^(9/4)*(-ArcTan[((-1 + x)/x)^(-1/4)] + ArcTanh[((-1 + x )/x)^(-1/4)]))/((-1 + x)*x^3)^(3/4)
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2467, 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.
\(\displaystyle \int \frac {\sqrt [4]{x^4-x^3}}{(x-1) x} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^3} \int \frac {1}{(x-1)^{3/4} \sqrt [4]{x}}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \frac {1}{\sqrt [4]{x}}d\sqrt [4]{x-1}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \frac {1}{2-x}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \int \frac {1}{\sqrt {x-1}+1}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
Input:
Int[(-x^3 + x^4)^(1/4)/((-1 + x)*x),x]
Output:
(4*(-x^3 + x^4)^(1/4)*(ArcTan[(-1 + x)^(1/4)/x^(1/4)]/2 + ArcTanh[(-1 + x) ^(1/4)/x^(1/4)]/2))/((-1 + x)^(1/4)*x^(3/4))
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[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73
method | result | size |
meijerg | \(-\frac {4 \operatorname {signum}\left (-1+x \right )^{\frac {1}{4}} x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x\right )}{3 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{4}}}\) | \(27\) |
pseudoelliptic | \(2 \,\operatorname {arctanh}\left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )\) | \(34\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )-\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-2 x^{3}+x^{2}}{x^{2}}\right )\) | \(144\) |
Input:
int((x^4-x^3)^(1/4)/(-1+x)/x,x,method=_RETURNVERBOSE)
Output:
-4/3*signum(-1+x)^(1/4)/(-signum(-1+x))^(1/4)*x^(3/4)*hypergeom([3/4,3/4], [7/4],x)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (33) = 66\).
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=-2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {3}{4}}}{x^{3} - x^{2}}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="fricas")
Output:
-2*arctan((x^4 - x^3)^(3/4)/(x^3 - x^2)) + log((x + (x^4 - x^3)^(1/4))/x) - log(-(x - (x^4 - x^3)^(1/4))/x)
\[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x \left (x - 1\right )}\, dx \] Input:
integrate((x**4-x**3)**(1/4)/(-1+x)/x,x)
Output:
Integral((x**3*(x - 1))**(1/4)/(x*(x - 1)), x)
\[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{{\left (x - 1\right )} x} \,d x } \] Input:
integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="maxima")
Output:
integrate((x^4 - x^3)^(1/4)/((x - 1)*x), x)
Time = 0.40 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \] Input:
integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="giac")
Output:
2*arctan((-1/x + 1)^(1/4)) + log((-1/x + 1)^(1/4) + 1) - log(abs((-1/x + 1 )^(1/4) - 1))
Timed out. \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int \frac {{\left (x^4-x^3\right )}^{1/4}}{x\,\left (x-1\right )} \,d x \] Input:
int((x^4 - x^3)^(1/4)/(x*(x - 1)),x)
Output:
int((x^4 - x^3)^(1/4)/(x*(x - 1)), x)
\[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int \frac {x^{\frac {3}{4}} \left (x -1\right )^{\frac {1}{4}}}{x^{2}-x}d x \] Input:
int((x^4-x^3)^(1/4)/(-1+x)/x,x)
Output:
int((x**(3/4)*(x - 1)**(1/4))/(x**2 - x),x)