3.24.67 \(\int \frac {-x+x^2}{\sqrt {(1-x) x (1-k^2 x)} (1-2 x+k^2 x^2)} \, dx\) [2367]

3.24.67.1 Optimal result
3.24.67.2 Mathematica [A] (verified)
3.24.67.3 Rubi [C] (verified)
3.24.67.4 Maple [C] (verified)
3.24.67.5 Fricas [A] (verification not implemented)
3.24.67.6 Sympy [F(-1)]
3.24.67.7 Maxima [F(-2)]
3.24.67.8 Giac [F]
3.24.67.9 Mupad [F(-1)]

3.24.67.1 Optimal result

Integrand size = 41, antiderivative size = 189 \[ \int \frac {-x+x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1-2 x+k^2 x^2\right )} \, dx=-\frac {i \arctan \left (\frac {\sqrt {-2+k^2-2 i \sqrt {-1+k^2}} \sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}{-1+k^2 x}\right )}{2 \sqrt {-1+k^2} \sqrt {-2+k^2-2 i \sqrt {-1+k^2}}}+\frac {i \arctan \left (\frac {\sqrt {-2+k^2+2 i \sqrt {-1+k^2}} \sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}{-1+k^2 x}\right )}{2 \sqrt {-1+k^2} \sqrt {-2+k^2+2 i \sqrt {-1+k^2}}} \]

output
-1/2*I*arctan((-2+k^2-2*I*(k^2-1)^(1/2))^(1/2)*(x+(-k^2-1)*x^2+k^2*x^3)^(1 
/2)/(k^2*x-1))/(k^2-1)^(1/2)/(-2+k^2-2*I*(k^2-1)^(1/2))^(1/2)+1/2*I*arctan 
((-2+k^2+2*I*(k^2-1)^(1/2))^(1/2)*(x+(-k^2-1)*x^2+k^2*x^3)^(1/2)/(k^2*x-1) 
)/(k^2-1)^(1/2)/(-2+k^2+2*I*(k^2-1)^(1/2))^(1/2)
 
3.24.67.2 Mathematica [A] (verified)

Time = 5.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.80 \[ \int \frac {-x+x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1-2 x+k^2 x^2\right )} \, dx=-\frac {i \left (\frac {\arctan \left (\frac {\sqrt {-2+k^2-2 i \sqrt {-1+k^2}} (-1+x) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{\sqrt {-2+k^2-2 i \sqrt {-1+k^2}}}-\frac {\arctan \left (\frac {\sqrt {-2+k^2+2 i \sqrt {-1+k^2}} (-1+x) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{\sqrt {-2+k^2+2 i \sqrt {-1+k^2}}}\right )}{2 \sqrt {-1+k^2}} \]

input
Integrate[(-x + x^2)/(Sqrt[(1 - x)*x*(1 - k^2*x)]*(1 - 2*x + k^2*x^2)),x]
 
output
((-1/2*I)*(ArcTan[(Sqrt[-2 + k^2 - (2*I)*Sqrt[-1 + k^2]]*(-1 + x)*x)/Sqrt[ 
(-1 + x)*x*(-1 + k^2*x)]]/Sqrt[-2 + k^2 - (2*I)*Sqrt[-1 + k^2]] - ArcTan[( 
Sqrt[-2 + k^2 + (2*I)*Sqrt[-1 + k^2]]*(-1 + x)*x)/Sqrt[(-1 + x)*x*(-1 + k^ 
2*x)]]/Sqrt[-2 + k^2 + (2*I)*Sqrt[-1 + k^2]]))/Sqrt[-1 + k^2]
 
3.24.67.3 Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.66 (sec) , antiderivative size = 910, normalized size of antiderivative = 4.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2027, 2467, 25, 2035, 7279, 2009}

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.

\(\displaystyle \int \frac {x^2-x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (k^2 x^2-2 x+1\right )} \, dx\)

\(\Big \downarrow \) 2027

\(\displaystyle \int \frac {(x-1) x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (k^2 x^2-2 x+1\right )}dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int -\frac {(1-x) \sqrt {x}}{\left (k^2 x^2-2 x+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {(1-x) \sqrt {x}}{\left (k^2 x^2-2 x+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {(1-x) x}{\left (k^2 x^2-2 x+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 7279

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \left (\frac {1-\left (2-k^2\right ) x}{k^2 \left (k^2 x^2-2 x+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}-\frac {1}{k^2 \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (\frac {\left (-k^2+2 \sqrt {1-k^2}+2\right ) \text {arctanh}\left (\frac {\sqrt {1-k^2} \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 k^2 \left (-k^2+\sqrt {1-k^2}+1\right )}-\frac {\left (-k^2-2 \sqrt {1-k^2}+2\right ) \text {arctanh}\left (\frac {\sqrt {1-k^2} \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 k^2 \left (-k^2-\sqrt {1-k^2}+1\right )}+\frac {\left (-k^2+2 \sqrt {1-k^2}+2\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{4 k^{5/2} \left (k+\sqrt {1-k^2}+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}+\frac {\left (-k^2-2 \sqrt {1-k^2}+2\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{4 k^{5/2} \left (k-\sqrt {1-k^2}+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}-\frac {(k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{2 k^{5/2} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}+\frac {\left (-k+\sqrt {1-k^2}+1\right ) \left (k^2-2 \left (\sqrt {1-k^2}+1\right )\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticPi}\left (\frac {k+1}{2 k},2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{5/2} \left (\sqrt {1-k^2}+1\right ) \left (k+\sqrt {1-k^2}+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}-\frac {\left (-k^2-2 \sqrt {1-k^2}+2\right ) \left (-k-\sqrt {1-k^2}+1\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticPi}\left (\frac {k+1}{2 k},2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{5/2} \left (1-\sqrt {1-k^2}\right ) \left (k-\sqrt {1-k^2}+1\right ) \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

input
Int[(-x + x^2)/(Sqrt[(1 - x)*x*(1 - k^2*x)]*(1 - 2*x + k^2*x^2)),x]
 
output
(-2*Sqrt[x]*Sqrt[1 - (1 + k^2)*x + k^2*x^2]*(-1/4*((2 - k^2 - 2*Sqrt[1 - k 
^2])*ArcTanh[(Sqrt[1 - k^2]*Sqrt[x])/Sqrt[1 - (1 + k^2)*x + k^2*x^2]])/(k^ 
2*(1 - k^2 - Sqrt[1 - k^2])) + ((2 - k^2 + 2*Sqrt[1 - k^2])*ArcTanh[(Sqrt[ 
1 - k^2]*Sqrt[x])/Sqrt[1 - (1 + k^2)*x + k^2*x^2]])/(4*k^2*(1 - k^2 + Sqrt 
[1 - k^2])) - ((1 + k*x)*Sqrt[(1 - (1 + k^2)*x + k^2*x^2)/(1 + k*x)^2]*Ell 
ipticF[2*ArcTan[Sqrt[k]*Sqrt[x]], (1 + k)^2/(4*k)])/(2*k^(5/2)*Sqrt[1 - (1 
 + k^2)*x + k^2*x^2]) + ((2 - k^2 - 2*Sqrt[1 - k^2])*(1 + k*x)*Sqrt[(1 - ( 
1 + k^2)*x + k^2*x^2)/(1 + k*x)^2]*EllipticF[2*ArcTan[Sqrt[k]*Sqrt[x]], (1 
 + k)^2/(4*k)])/(4*k^(5/2)*(1 + k - Sqrt[1 - k^2])*Sqrt[1 - (1 + k^2)*x + 
k^2*x^2]) + ((2 - k^2 + 2*Sqrt[1 - k^2])*(1 + k*x)*Sqrt[(1 - (1 + k^2)*x + 
 k^2*x^2)/(1 + k*x)^2]*EllipticF[2*ArcTan[Sqrt[k]*Sqrt[x]], (1 + k)^2/(4*k 
)])/(4*k^(5/2)*(1 + k + Sqrt[1 - k^2])*Sqrt[1 - (1 + k^2)*x + k^2*x^2]) - 
((2 - k^2 - 2*Sqrt[1 - k^2])*(1 - k - Sqrt[1 - k^2])*(1 + k*x)*Sqrt[(1 - ( 
1 + k^2)*x + k^2*x^2)/(1 + k*x)^2]*EllipticPi[(1 + k)/(2*k), 2*ArcTan[Sqrt 
[k]*Sqrt[x]], (1 + k)^2/(4*k)])/(8*k^(5/2)*(1 - Sqrt[1 - k^2])*(1 + k - Sq 
rt[1 - k^2])*Sqrt[1 - (1 + k^2)*x + k^2*x^2]) + ((1 - k + Sqrt[1 - k^2])*( 
k^2 - 2*(1 + Sqrt[1 - k^2]))*(1 + k*x)*Sqrt[(1 - (1 + k^2)*x + k^2*x^2)/(1 
 + k*x)^2]*EllipticPi[(1 + k)/(2*k), 2*ArcTan[Sqrt[k]*Sqrt[x]], (1 + k)^2/ 
(4*k)])/(8*k^(5/2)*(1 + Sqrt[1 - k^2])*(1 + k + Sqrt[1 - k^2])*Sqrt[1 - (1 
 + k^2)*x + k^2*x^2])))/Sqrt[(1 - x)*x*(1 - k^2*x)]
 

3.24.67.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2027
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2035
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7279
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 
3.24.67.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.59 (sec) , antiderivative size = 1121, normalized size of antiderivative = 5.93

method result size
default \(\text {Expression too large to display}\) \(1121\)
elliptic \(\text {Expression too large to display}\) \(1132\)

input
int((x^2-x)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^2*x^2-2*x+1),x,method=_RETURNVER 
BOSE)
 
output
-2/k^4*(-(x-1/k^2)*k^2)^(1/2)*((-1+x)/(1/k^2-1))^(1/2)*(k^2*x)^(1/2)/(k^2* 
x^3-k^2*x^2-x^2+x)^(1/2)*EllipticF((-(x-1/k^2)*k^2)^(1/2),(1/k^2/(1/k^2-1) 
)^(1/2))+1/k^2*(-2/(-k^2+1)*(-k^2*x+1)^(1/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^ 
(1/2)*(k^2*x)^(1/2)/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)*EllipticPi((-(x-1/k^2)*k 
^2)^(1/2),1/k^2/(1/k^2-(1+(-k^2+1)^(1/2))/k^2),(1/k^2/(1/k^2-1))^(1/2))+2/ 
(-k^2+1)*(-k^2*x+1)^(1/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^(1/2)*(k^2*x)^(1/2) 
/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/ 
k^2-(1+(-k^2+1)^(1/2))/k^2),(1/k^2/(1/k^2-1))^(1/2))/k^2-1/(-k^2+1)^(1/2)* 
(-k^2*x+1)^(1/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^(1/2)*(k^2*x)^(1/2)/(k^2*x^3 
-k^2*x^2-x^2+x)^(1/2)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2-(1+(- 
k^2+1)^(1/2))/k^2),(1/k^2/(1/k^2-1))^(1/2))+2/(-k^2+1)^(1/2)*(-k^2*x+1)^(1 
/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^(1/2)*(k^2*x)^(1/2)/(k^2*x^3-k^2*x^2-x^2+ 
x)^(1/2)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2-(1+(-k^2+1)^(1/2)) 
/k^2),(1/k^2/(1/k^2-1))^(1/2))/k^2-2/(-k^2+1)*(-k^2*x+1)^(1/2)*(-1/(1/k^2- 
1)+1/(1/k^2-1)*x)^(1/2)*(k^2*x)^(1/2)/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)*Ellipt 
icPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2+(-1+(-k^2+1)^(1/2))/k^2),(1/k^2/( 
1/k^2-1))^(1/2))+2/(-k^2+1)*(-k^2*x+1)^(1/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^ 
(1/2)*(k^2*x)^(1/2)/(k^2*x^3-k^2*x^2-x^2+x)^(1/2)*EllipticPi((-(x-1/k^2)*k 
^2)^(1/2),1/k^2/(1/k^2+(-1+(-k^2+1)^(1/2))/k^2),(1/k^2/(1/k^2-1))^(1/2))/k 
^2+1/(-k^2+1)^(1/2)*(-k^2*x+1)^(1/2)*(-1/(1/k^2-1)+1/(1/k^2-1)*x)^(1/2)...
 
3.24.67.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.47 \[ \int \frac {-x+x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1-2 x+k^2 x^2\right )} \, dx=\left [\frac {{\left (k^{2} - 1\right )} \log \left (\frac {k^{4} x^{4} + 4 \, k^{2} x^{3} - 2 \, {\left (3 \, k^{2} + 2\right )} x^{2} - 4 \, \sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 1\right )} + 4 \, x + 1}{k^{4} x^{4} - 4 \, k^{2} x^{3} + 2 \, {\left (k^{2} + 2\right )} x^{2} - 4 \, x + 1}\right ) - \sqrt {-k^{2} + 1} \log \left (\frac {k^{4} x^{4} - 4 \, {\left (2 \, k^{4} - k^{2}\right )} x^{3} + 2 \, {\left (4 \, k^{4} + k^{2} - 2\right )} x^{2} - 4 \, \sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, k^{2} x + 1\right )} \sqrt {-k^{2} + 1} - 4 \, {\left (2 \, k^{2} - 1\right )} x + 1}{k^{4} x^{4} - 4 \, k^{2} x^{3} + 2 \, {\left (k^{2} + 2\right )} x^{2} - 4 \, x + 1}\right )}{4 \, {\left (k^{4} - k^{2}\right )}}, \frac {{\left (k^{2} - 1\right )} \log \left (\frac {k^{4} x^{4} + 4 \, k^{2} x^{3} - 2 \, {\left (3 \, k^{2} + 2\right )} x^{2} - 4 \, \sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 1\right )} + 4 \, x + 1}{k^{4} x^{4} - 4 \, k^{2} x^{3} + 2 \, {\left (k^{2} + 2\right )} x^{2} - 4 \, x + 1}\right ) + 2 \, \sqrt {k^{2} - 1} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, k^{2} x + 1\right )} \sqrt {k^{2} - 1}}{2 \, {\left ({\left (k^{4} - k^{2}\right )} x^{3} - {\left (k^{4} - 1\right )} x^{2} + {\left (k^{2} - 1\right )} x\right )}}\right )}{4 \, {\left (k^{4} - k^{2}\right )}}\right ] \]

input
integrate((x^2-x)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^2*x^2-2*x+1),x, algorithm= 
"fricas")
 
output
[1/4*((k^2 - 1)*log((k^4*x^4 + 4*k^2*x^3 - 2*(3*k^2 + 2)*x^2 - 4*sqrt(k^2* 
x^3 - (k^2 + 1)*x^2 + x)*(k^2*x^2 - 1) + 4*x + 1)/(k^4*x^4 - 4*k^2*x^3 + 2 
*(k^2 + 2)*x^2 - 4*x + 1)) - sqrt(-k^2 + 1)*log((k^4*x^4 - 4*(2*k^4 - k^2) 
*x^3 + 2*(4*k^4 + k^2 - 2)*x^2 - 4*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k^2* 
x^2 - 2*k^2*x + 1)*sqrt(-k^2 + 1) - 4*(2*k^2 - 1)*x + 1)/(k^4*x^4 - 4*k^2* 
x^3 + 2*(k^2 + 2)*x^2 - 4*x + 1)))/(k^4 - k^2), 1/4*((k^2 - 1)*log((k^4*x^ 
4 + 4*k^2*x^3 - 2*(3*k^2 + 2)*x^2 - 4*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k 
^2*x^2 - 1) + 4*x + 1)/(k^4*x^4 - 4*k^2*x^3 + 2*(k^2 + 2)*x^2 - 4*x + 1)) 
+ 2*sqrt(k^2 - 1)*arctan(1/2*sqrt(k^2*x^3 - (k^2 + 1)*x^2 + x)*(k^2*x^2 - 
2*k^2*x + 1)*sqrt(k^2 - 1)/((k^4 - k^2)*x^3 - (k^4 - 1)*x^2 + (k^2 - 1)*x) 
))/(k^4 - k^2)]
 
3.24.67.6 Sympy [F(-1)]

Timed out. \[ \int \frac {-x+x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1-2 x+k^2 x^2\right )} \, dx=\text {Timed out} \]

input
integrate((x**2-x)/((1-x)*x*(-k**2*x+1))**(1/2)/(k**2*x**2-2*x+1),x)
 
output
Timed out
 
3.24.67.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {-x+x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1-2 x+k^2 x^2\right )} \, dx=\text {Exception raised: ValueError} \]

input
integrate((x^2-x)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^2*x^2-2*x+1),x, algorithm= 
"maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(k-1>0)', see `assume?` for more 
details)Is
 
3.24.67.8 Giac [F]

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

input
integrate((x^2-x)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^2*x^2-2*x+1),x, algorithm= 
"giac")
 
output
integrate((x^2 - x)/((k^2*x^2 - 2*x + 1)*sqrt((k^2*x - 1)*(x - 1)*x)), x)
 
3.24.67.9 Mupad [F(-1)]

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

input
int(-(x - x^2)/((k^2*x^2 - 2*x + 1)*(x*(k^2*x - 1)*(x - 1))^(1/2)),x)
 
output
int(-(x - x^2)/((k^2*x^2 - 2*x + 1)*(x*(k^2*x - 1)*(x - 1))^(1/2)), x)