\(\int \sqrt {-x+x^4} \, dx\) [459]

Optimal result

Integrand size = 11, antiderivative size = 37 \[ \int \sqrt {-x+x^4} \, dx=\frac {1}{3} x \sqrt {-x+x^4}-\frac {1}{3} \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2029, 2054, 212} \[ \int \sqrt {-x+x^4} \, dx=\frac {1}{3} x \sqrt {x^4-x}-\frac {1}{3} \text {arctanh}\left (\frac {x^2}{\sqrt {x^4-x}}\right ) \]

[In]

Int[Sqrt[-x + x^4],x]

[Out]

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

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 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 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \sqrt {-x+x^4} \, dx=\frac {\sqrt {x \left (-1+x^3\right )} \left (x^{3/2}-\frac {\log \left (x^{3/2}+\sqrt {-1+x^3}\right )}{\sqrt {-1+x^3}}\right )}{3 \sqrt {x}} \]

[In]

Integrate[Sqrt[-x + x^4],x]

[Out]

(Sqrt[x*(-1 + x^3)]*(x^(3/2) - Log[x^(3/2) + Sqrt[-1 + x^3]]/Sqrt[-1 + x^3]))/(3*Sqrt[x])

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97

method	result	size
trager	\(\frac {x \sqrt {x^{4}-x}}{3}+\frac {\ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{6}\)	\(36\)
risch	\(\frac {x^{2} \left (x^{3}-1\right )}{3 \sqrt {x \left (x^{3}-1\right )}}+\frac {\ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{6}\)	\(43\)
meijerg	\(\frac {i \sqrt {\operatorname {signum}\left (x^{3}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{\frac {3}{2}} \sqrt {-x^{3}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{\frac {3}{2}}\right )\right )}{6 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{3}-1\right )}}\)	\(54\)
default	\(\frac {x \sqrt {x^{4}-x}}{3}-\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}+\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}\)	\(56\)
pseudoelliptic	\(\frac {x \sqrt {x^{4}-x}}{3}-\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}+\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}\)	\(56\)
elliptic	\(\frac {x \sqrt {x^{4}-x}}{3}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (x -1\right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (\operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )-\operatorname {EllipticPi}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \frac {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (x -1\right ) \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\)	\(301\)

[In]

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

[Out]

1/3*x*(x^4-x)^(1/2)+1/6*ln(2*x^3-2*x*(x^4-x)^(1/2)-1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \sqrt {-x+x^4} \, dx=\frac {1}{3} \, \sqrt {x^{4} - x} x + \frac {1}{6} \, \log \left (2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x - 1\right ) \]

[In]

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

[Out]

1/3*sqrt(x^4 - x)*x + 1/6*log(2*x^3 - 2*sqrt(x^4 - x)*x - 1)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(x**4 - x), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(x^4 - x), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \sqrt {-x+x^4} \, dx=\frac {1}{3} \, \sqrt {x^{4} - x} x - \frac {1}{6} \, \log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right ) + \frac {1}{6} \, \log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right ) \]

[In]

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

[Out]

1/3*sqrt(x^4 - x)*x - 1/6*log(sqrt(-1/x^3 + 1) + 1) + 1/6*log(abs(sqrt(-1/x^3 + 1) - 1))

Mupad [B] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \sqrt {-x+x^4} \, dx=\frac {2\,x\,\sqrt {x^4-x}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{2};\ \frac {3}{2};\ x^3\right )}{3\,\sqrt {1-x^3}} \]

[In]

int((x^4 - x)^(1/2),x)

[Out]

(2*x*(x^4 - x)^(1/2)*hypergeom([-1/2, 1/2], 3/2, x^3))/(3*(1 - x^3)^(1/2))