\(\int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx\) [462]

   Optimal result
   Rubi [B] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=-2 \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \]

[Out]

-2*arctan(x/(x^4-x^3)^(1/4))+2*arctanh(x/(x^4-x^3)^(1/4))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(37)=74\).

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.24, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2081, 65, 246, 218, 212, 209} \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\frac {2 \sqrt [4]{x^4-x^3} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{x^4-x^3} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}} \]

[In]

Int[(-x^3 + x^4)^(1/4)/((-1 + x)*x),x]

[Out]

(2*(-x^3 + x^4)^(1/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/((-1 + x)^(1/4)*x^(3/4)) + (2*(-x^3 + x^4)^(1/4)*ArcTanh
[(-1 + x)^(1/4)/x^(1/4)])/((-1 + x)^(1/4)*x^(3/4))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[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]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-x^3+x^4} \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\left (2 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {\left (2 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {2 \sqrt [4]{-x^3+x^4} \arctan \left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}}+\frac {2 \sqrt [4]{-x^3+x^4} \text {arctanh}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\sqrt [4]{-1+x} x^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\frac {2 (-1+x)^{3/4} x^{9/4} \left (-\arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+\text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{\left ((-1+x) x^3\right )^{3/4}} \]

[In]

Integrate[(-x^3 + x^4)^(1/4)/((-1 + x)*x),x]

[Out]

(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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.83 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73

method result size
meijerg \(-\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x\right )}{3 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}}}\) \(27\)
pseudoelliptic \(2 \,\operatorname {arctanh}\left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )\) \(34\)
trager \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )\) \(144\)

[In]

int((x^4-x^3)^(1/4)/(x-1)/x,x,method=_RETURNVERBOSE)

[Out]

-4/3*signum(x-1)^(1/4)/(-signum(x-1))^(1/4)*x^(3/4)*hypergeom([3/4,3/4],[7/4],x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="fricas")

[Out]

2*arctan((x^4 - x^3)^(1/4)/x) + log((x + (x^4 - x^3)^(1/4))/x) - log(-(x - (x^4 - x^3)^(1/4))/x)

Sympy [F]

\[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x \left (x - 1\right )}\, dx \]

[In]

integrate((x**4-x**3)**(1/4)/(-1+x)/x,x)

[Out]

Integral((x**3*(x - 1))**(1/4)/(x*(x - 1)), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{{\left (x - 1\right )} x} \,d x } \]

[In]

integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)/((x - 1)*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

integrate((x^4-x^3)^(1/4)/(-1+x)/x,x, algorithm="giac")

[Out]

2*arctan((-1/x + 1)^(1/4)) + log((-1/x + 1)^(1/4) + 1) - log(abs((-1/x + 1)^(1/4) - 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-x^3+x^4}}{(-1+x) x} \, dx=\int \frac {{\left (x^4-x^3\right )}^{1/4}}{x\,\left (x-1\right )} \,d x \]

[In]

int((x^4 - x^3)^(1/4)/(x*(x - 1)),x)

[Out]

int((x^4 - x^3)^(1/4)/(x*(x - 1)), x)