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

Optimal result

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

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))
 

Mathematica [A] (warning: unable to verify)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (2 \sqrt [4]{-1+x} x^{3/4} (-1+4 x)+3 \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-3 \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{16 \left ((-1+x) x^3\right )^{3/4}} \] Input:

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

Output:

((-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))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.63, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {1910, 1930, 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.

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

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{2} x \sqrt [4]{x^4-x^3}-\frac {1}{8} \int \frac {x^3}{\left (x^4-x^3\right )^{3/4}}dx\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \int \frac {x^2}{\left (x^4-x^3\right )^{3/4}}dx-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \int \frac {1}{(x-1)^{3/4} \sqrt [4]{x}}dx}{4 \left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \int \frac {1}{\sqrt [4]{x}}d\sqrt [4]{x-1}}{\left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \int \frac {1}{2-x}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}}{\left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \int \frac {1}{\sqrt {x-1}+1}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{8} \left (-\frac {3 (x-1)^{3/4} x^{9/4} \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\left (x^4-x^3\right )^{3/4}}-\sqrt [4]{x^4-x^3}\right )+\frac {1}{2} \sqrt [4]{x^4-x^3} x\)

Input:

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

Output:

(x*(-x^3 + x^4)^(1/4))/2 + (-(-x^3 + x^4)^(1/4) - (3*(-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))/8
 

Defintions of rubi rules used

rule 73

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]
 

rule 216

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])
 

rule 219

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])
 

rule 756

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]
 

rule 770

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]
 

rule 1910

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[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 1930

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] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(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]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1938

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]
 

Maple [C] (warning: unable to verify)

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

Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.43

method	result	size
meijerg	\(\frac {4 \operatorname {signum}\left (-1+x \right )^{\frac {1}{4}} x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], x\right )}{7 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{4}}}\)	\(27\)
pseudoelliptic	\(\frac {x^{6} \left (16 x \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-4 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-3 \ln \left (\frac {x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )-6 \arctan \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )+3 \ln \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )\right )}{32 {\left (x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}\right )}^{2} \left (x^{2}+\sqrt {x^{3} \left (-1+x \right )}\right )^{2} {\left (\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x \right )}^{2}}\)	\(127\)
trager	\(\left (-\frac {1}{8}+\frac {x}{2}\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\frac {3 \ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-2 x^{3}+x^{2}}{x^{2}}\right )}{32}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{32}\)	\(162\)
risch	\(\frac {\left (-1+4 x \right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{8}+\frac {\left (\frac {3 \ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-2 x^{3}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}-4 x +1}{\left (-1+x \right )^{2}}\right )}{32}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (-1+x \right )^{2}}\right )}{32}\right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} \left (x \left (-1+x \right )^{3}\right )^{\frac {1}{4}}}{x \left (-1+x \right )}\)	\(397\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.40 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {1}{8} \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x - 1\right )} + \frac {3}{16} \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {3}{4}}}{x^{3} - x^{2}}\right ) - \frac {3}{32} \, \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{32} \, \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {1}{8} \, {\left ({\left (-\frac {1}{x} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{2} - \frac {3}{16} \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{32} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{32} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \] Input:

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

Output:

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 = 7.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.43 \[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {4\,x\,{\left (x^4-x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ x\right )}{7\,{\left (1-x\right )}^{1/4}} \] Input:

int((x^4 - x^3)^(1/4),x)
 

Output:

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

Reduce [F]

\[ \int \sqrt [4]{-x^3+x^4} \, dx=\frac {x^{\frac {7}{4}} \left (x -1\right )^{\frac {1}{4}}}{2}-\frac {x^{\frac {3}{4}} \left (x -1\right )^{\frac {1}{4}}}{8}-\frac {3 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}}}{x^{\frac {5}{4}}-x^{\frac {1}{4}}}d x \right )}{32} \] Input:

int((x^4-x^3)^(1/4),x)
 

Output:

(16*x**(3/4)*(x - 1)**(1/4)*x - 4*x**(3/4)*(x - 1)**(1/4) - 3*int((x - 1)* 
*(1/4)/(x**(1/4)*x - x**(1/4)),x))/32