\(\int \frac {1}{x \sqrt [4]{x-x^2}} \, dx\) [770]

Optimal result

Integrand size = 15, antiderivative size = 36 \[ \int \frac {1}{x \sqrt [4]{x-x^2}} \, dx=-\frac {4 (1-x)^{3/4}}{\sqrt [4]{x}}+2 \sqrt {2} E\left (\left .\frac {1}{2} \arcsin (1-2 x)\right |2\right ) \] Output:

-4*(1-x)^(3/4)/x^(1/4)-2*2^(1/2)*EllipticE(sin(1/2*arcsin(-1+2*x)),2^(1/2) 
)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \sqrt [4]{x-x^2}} \, dx=-\frac {4 (-((-1+x) x))^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},1-x\right )}{3 x^{3/4}} \] Input:

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

Output:

(-4*(-((-1 + x)*x))^(3/4)*Hypergeometric2F1[3/4, 5/4, 7/4, 1 - x])/(3*x^(3 
/4))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(113\) vs. \(2(36)=72\).

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 3.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1137, 61, 73, 840, 842, 858, 807, 226}

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.

\(\displaystyle \int \frac {1}{x \sqrt [4]{x-x^2}} \, dx\)

\(\Big \downarrow \) 1137

\(\displaystyle \frac {\sqrt [4]{1-x} \sqrt [4]{x} \int \frac {1}{\sqrt [4]{1-x} x^{5/4}}dx}{\sqrt [4]{x-x^2}}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 840

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

\(\Big \downarrow \) 842

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 226

\(\displaystyle \frac {\sqrt [4]{1-x} \sqrt [4]{x} \left (8 \left (\frac {\sqrt [4]{1-\frac {1}{1-x}} \sqrt [4]{1-x} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {1-x}\right )\right |2\right )}{2 \sqrt [4]{x}}-\frac {x^{3/4}}{2 \sqrt [4]{1-x}}\right )-\frac {4 (1-x)^{3/4}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-x^2}}\)

Input:

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

Output:

((1 - x)^(1/4)*x^(1/4)*((-4*(1 - x)^(3/4))/x^(1/4) + 8*(-1/2*x^(3/4)/(1 - 
x)^(1/4) + ((1 - (1 - x)^(-1))^(1/4)*(1 - x)^(1/4)*EllipticE[ArcSin[Sqrt[1 
 - x]]/2, 2])/(2*x^(1/4)))))/(x - x^2)^(1/4)
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 226

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] 
))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 840

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[(a + b*x^4)^(3/4) 
/(2*b*x), x] + Simp[a/(2*b)   Int[1/(x^2*(a + b*x^4)^(1/4)), x], x] /; Free 
Q[{a, b}, x] && NegQ[b/a]
 

rule 842

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Simp[x*((1 + a/(b* 
x^4))^(1/4)/(a + b*x^4)^(1/4))   Int[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] 
/; FreeQ[{a, b}, x] && NegQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.36

method	result	size
meijerg	\(-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {3}{4}\right ], x\right )}{x^{\frac {1}{4}}}\)	\(13\)
risch	\(\frac {-4+4 x}{\left (-x \left (-1+x \right )\right )^{\frac {1}{4}}}-\frac {8 x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x\right )}{3}\)	\(27\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(-x^2 + x)^(3/4)/(x^3 - x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt [4]{x-x^2}} \, dx=\int \frac {1}{x\,{\left (x-x^2\right )}^{1/4}} \,d x \] Input:

int(1/(x*(x - x^2)^(1/4)),x)
 

Output:

int(1/(x*(x - x^2)^(1/4)), x)
 

Reduce [F]

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

int(1/x/(-x^2+x)^(1/4),x)
 

Output:

int(1/(x**(1/4)*( - x + 1)**(1/4)*x),x)