\(\int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx\) [708]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}-2 \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.84, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2045, 2057, 65, 246, 218, 212, 209} \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=\frac {2 (x-1)^{3/4} x^{9/4} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}}+\frac {2 (x-1)^{3/4} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}}-\frac {4 \sqrt [4]{x^4-x^3}}{x} \]

[In]

Int[(-x^3 + x^4)^(1/4)/x^2,x]

[Out]

(-4*(-x^3 + x^4)^(1/4))/x + (2*(-1 + x)^(3/4)*x^(9/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/(-x^3 + x^4)^(3/4) + (2*
(-1 + x)^(3/4)*x^(9/4)*ArcTanh[(-1 + x)^(1/4)/x^(1/4)])/(-x^3 + x^4)^(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 2045

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] - Dist[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] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[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] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\int \frac {x^2}{\left (-x^3+x^4\right )^{3/4}} \, dx \\ & = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left ((-1+x)^{3/4} x^{9/4}\right ) \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{\left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (4 (-1+x)^{3/4} x^{9/4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (4 (-1+x)^{3/4} x^{9/4}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (2 (-1+x)^{3/4} x^{9/4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}+\frac {\left (2 (-1+x)^{3/4} x^{9/4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {2 (-1+x)^{3/4} x^{9/4} \arctan \left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}+\frac {2 (-1+x)^{3/4} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=-\frac {2 (-1+x)^{3/4} x^2 \left (2 \sqrt [4]{-1+x}+\sqrt [4]{x} \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-\sqrt [4]{x} \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{\left ((-1+x) x^3\right )^{3/4}} \]

[In]

Integrate[(-x^3 + x^4)^(1/4)/x^2,x]

[Out]

(-2*(-1 + x)^(3/4)*x^2*(2*(-1 + x)^(1/4) + x^(1/4)*ArcTan[((-1 + x)/x)^(-1/4)] - 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 = 1.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.49

method	result	size
meijerg	\(-\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x\right )}{\left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {1}{4}}}\)	\(27\)
pseudoelliptic	\(\frac {\left (-\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}-x}{x}\right )+\ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )\right ) x -4 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\)	\(72\)
trager	\(-\frac {4 \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )\)	\(162\)
risch	\(-\frac {4 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}+\frac {\left (\ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+2 x^{3}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+4 x -1}{\left (x -1\right )^{2}}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}}{\left (x -1\right )^{2}}\right )\right ) \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} \left (x \left (x -1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x -1\right )}\)	\(393\)

[In]

int((x^4-x^3)^(1/4)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=\frac {2 \, x \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + x \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - x \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x} \]

[In]

integrate((x^4-x^3)^(1/4)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x^{2}}\, dx \]

[In]

integrate((x**4-x**3)**(1/4)/x**2,x)

[Out]

Integral((x**3*(x - 1))**(1/4)/x**2, x)

Maxima [F]

\[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x^{2}} \,d x } \]

[In]

integrate((x^4-x^3)^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=-4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

integrate((x^4-x^3)^(1/4)/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx=-\frac {4\,{\left (x^4-x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ x\right )}{x\,{\left (1-x\right )}^{1/4}} \]

[In]

int((x^4 - x^3)^(1/4)/x^2,x)

[Out]

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