\(\int \sqrt [4]{-x^3+x^4} \, dx\) [833]

Optimal result

Integrand size = 13, antiderivative size = 63 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {1}{8} (-1+4 x) \sqrt [4]{-x^3+x^4}+\frac {3}{16} \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-\frac {3}{16} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \]

[Out]

1/8*(-1+4*x)*(x^4-x^3)^(1/4)+3/16*arctan(x/(x^4-x^3)^(1/4))-3/16*arctanh(x/(x^4-x^3)^(1/4))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {2029, 2049, 2057, 65, 246, 218, 212, 209} \[ \int \sqrt [4]{-x^3+x^4} \, dx=-\frac {3 (x-1)^{3/4} x^{9/4} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{16 \left (x^4-x^3\right )^{3/4}}-\frac {3 (x-1)^{3/4} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{16 \left (x^4-x^3\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^3} x-\frac {1}{8} \sqrt [4]{x^4-x^3} \]

[In]

Int[(-x^3 + x^4)^(1/4),x]

[Out]

-1/8*(-x^3 + x^4)^(1/4) + (x*(-x^3 + x^4)^(1/4))/2 - (3*(-1 + x)^(3/4)*x^(9/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])
/(16*(-x^3 + x^4)^(3/4)) - (3*(-1 + x)^(3/4)*x^(9/4)*ArcTanh[(-1 + x)^(1/4)/x^(1/4)])/(16*(-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 2029

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Dist[a
*(n - j)*(p/(n*p + 1)), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*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 {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {1}{8} \int \frac {x^3}{\left (-x^3+x^4\right )^{3/4}} \, dx \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {3}{32} \int \frac {x^2}{\left (-x^3+x^4\right )^{3/4}} \, dx \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {\left (3 (-1+x)^{3/4} x^{9/4}\right ) \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{32 \left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {\left (3 (-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 )}{8 \left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {\left (3 (-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 )}{8 \left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {\left (3 (-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 )}{16 \left (-x^3+x^4\right )^{3/4}}-\frac {\left (3 (-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 )}{16 \left (-x^3+x^4\right )^{3/4}} \\ & = -\frac {1}{8} \sqrt [4]{-x^3+x^4}+\frac {1}{2} x \sqrt [4]{-x^3+x^4}-\frac {3 (-1+x)^{3/4} x^{9/4} \arctan \left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{16 \left (-x^3+x^4\right )^{3/4}}-\frac {3 (-1+x)^{3/4} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{16 \left (-x^3+x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (2 \sqrt [4]{-1+x} x^{3/4} (-1+4 x)+3 \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-3 \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{16 \left ((-1+x) x^3\right )^{3/4}} \]

[In]

Integrate[(-x^3 + x^4)^(1/4),x]

[Out]

((-1 + x)^(3/4)*x^(9/4)*(2*(-1 + x)^(1/4)*x^(3/4)*(-1 + 4*x) + 3*ArcTan[((-1 + x)/x)^(-1/4)] - 3*ArcTanh[((-1
+ x)/x)^(-1/4)]))/(16*((-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 = 1.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.43

method	result	size
meijerg	\(\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], x\right )}{7 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}}}\)	\(27\)
pseudoelliptic	\(\frac {x^{6} \left (16 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} x +3 \ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}-x}{x}\right )-6 \arctan \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{x}\right )-3 \ln \left (\frac {\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+x}{x}\right )-4 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}\right )}{32 {\left (\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}-x \right )}^{2} \left (x^{2}+\sqrt {x^{3} \left (x -1\right )}\right )^{2} {\left (\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}+x \right )}^{2}}\)	\(127\)
trager	\(\left (-\frac {1}{8}+\frac {x}{2}\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{32}-\frac {3 \ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )}{32}\)	\(164\)
risch	\(\frac {\left (-1+4 x \right ) \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{8}+\frac {\left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (x -1\right )^{2}}\right )}{32}+\frac {3 \ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-2 x^{3}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}-4 x +1}{\left (x -1\right )^{2}}\right )}{32}\right ) \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} \left (x \left (x -1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x -1\right )}\)	\(397\)

[In]

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

[Out]

4/7*signum(x-1)^(1/4)/(-signum(x-1))^(1/4)*x^(7/4)*hypergeom([-1/4,7/4],[11/4],x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.27 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {1}{8} \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x - 1\right )} - \frac {3}{16} \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{32} \, \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{32} \, \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

1/8*(x^4 - x^3)^(1/4)*(4*x - 1) - 3/16*arctan((x^4 - x^3)^(1/4)/x) - 3/32*log((x + (x^4 - x^3)^(1/4))/x) + 3/3
2*log(-(x - (x^4 - x^3)^(1/4))/x)

Sympy [F]

\[ \int \sqrt [4]{-x^3+x^4} \, dx=\int \sqrt [4]{x^{4} - x^{3}}\, dx \]

[In]

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

[Out]

Integral((x**4 - x**3)**(1/4), x)

Maxima [F]

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

[In]

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

[Out]

integrate((x^4 - x^3)^(1/4), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {1}{8} \, {\left ({\left (-\frac {1}{x} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{2} - \frac {3}{16} \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{32} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{32} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

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

[Out]

1/8*((-1/x + 1)^(5/4) + 3*(-1/x + 1)^(1/4))*x^2 - 3/16*arctan((-1/x + 1)^(1/4)) - 3/32*log((-1/x + 1)^(1/4) +
1) + 3/32*log(abs((-1/x + 1)^(1/4) - 1))

Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.43 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {4\,x\,{\left (x^4-x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ x\right )}{7\,{\left (1-x\right )}^{1/4}} \]

[In]

int((x^4 - x^3)^(1/4),x)

[Out]

(4*x*(x^4 - x^3)^(1/4)*hypergeom([-1/4, 7/4], 11/4, x))/(7*(1 - x)^(1/4))