3.20.66 \(\int \frac {1+k^4 x^4}{\sqrt {(1-x) x (1-k^2 x)} (-1+k^4 x^4)} \, dx\) [1966]

3.20.66.1 Optimal result
3.20.66.2 Mathematica [A] (verified)
3.20.66.3 Rubi [C] (verified)
3.20.66.4 Maple [A] (verified)
3.20.66.5 Fricas [B] (verification not implemented)
3.20.66.6 Sympy [F]
3.20.66.7 Maxima [F]
3.20.66.8 Giac [F]
3.20.66.9 Mupad [F(-1)]

3.20.66.1 Optimal result

Integrand size = 40, antiderivative size = 139 \[ \int \frac {1+k^4 x^4}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^4 x^4\right )} \, dx=-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{2 (-1+k)}-\frac {\arctan \left (\frac {(1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{2 (1+k)}-\frac {\arctan \left (\frac {\sqrt {1+k^2} \sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}{(-1+x) \left (-1+k^2 x\right )}\right )}{\sqrt {1+k^2}} \]

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

Time = 12.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \frac {1+k^4 x^4}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^4 x^4\right )} \, dx=-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{2 (-1+k)}-\frac {\arctan \left (\frac {(1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{2 (1+k)}-\frac {\arctan \left (\frac {\sqrt {1+k^2} x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{\sqrt {1+k^2}} \]

input
Integrate[(1 + k^4*x^4)/(Sqrt[(1 - x)*x*(1 - k^2*x)]*(-1 + k^4*x^4)),x]
 
output
-1/2*ArcTan[((-1 + k)*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]]/(-1 + k) - ArcTan[ 
((1 + k)*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]]/(2*(1 + k)) - ArcTan[(Sqrt[1 + 
k^2]*x)/Sqrt[(-1 + x)*x*(-1 + k^2*x)]]/Sqrt[1 + k^2]
 
3.20.66.3 Rubi [C] (verified)

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

Time = 2.07 (sec) , antiderivative size = 493, normalized size of antiderivative = 3.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2467, 25, 2035, 7276, 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.

\(\displaystyle \int \frac {k^4 x^4+1}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (k^4 x^4-1\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int -\frac {k^4 x^4+1}{\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (1-k^4 x^4\right )}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {k^4 x^4+1}{\sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (1-k^4 x^4\right )}dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \frac {k^4 x^4+1}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (1-k^4 x^4\right )}d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \int \left (\frac {2}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (1-k^4 x^4\right )}-\frac {1}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )d\sqrt {x}}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1} \left (\frac {\arctan \left (\frac {(1-k) \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 (1-k)}+\frac {\arctan \left (\frac {(k+1) \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{4 (k+1)}+\frac {\sqrt {-k^2} \arctan \left (\frac {\sqrt {k^2+1} \sqrt {x}}{\sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{2 \sqrt {-k^2 \left (k^2+1\right )}}-\frac {(k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{4 \sqrt {k} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}+\frac {\left (k-\sqrt {-k^2}\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{3/2} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}+\frac {\left (\sqrt {-k^2}+k\right ) (k x+1) \sqrt {\frac {k^2 x^2-\left (k^2+1\right ) x+1}{(k x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} \sqrt {x}\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{3/2} \sqrt {k^2 x^2-\left (k^2+1\right ) x+1}}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\)

input
Int[(1 + k^4*x^4)/(Sqrt[(1 - x)*x*(1 - k^2*x)]*(-1 + k^4*x^4)),x]
 
output
(-2*Sqrt[x]*Sqrt[1 - (1 + k^2)*x + k^2*x^2]*(ArcTan[((1 - k)*Sqrt[x])/Sqrt 
[1 - (1 + k^2)*x + k^2*x^2]]/(4*(1 - k)) + ArcTan[((1 + k)*Sqrt[x])/Sqrt[1 
 - (1 + k^2)*x + k^2*x^2]]/(4*(1 + k)) + (Sqrt[-k^2]*ArcTan[(Sqrt[1 + k^2] 
*Sqrt[x])/Sqrt[1 - (1 + k^2)*x + k^2*x^2]])/(2*Sqrt[-(k^2*(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*Sqrt[k]*Sqrt[1 - (1 + k^2)*x + k^2 
*x^2]) + ((k - Sqrt[-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)])/(8*k^(3/2) 
*Sqrt[1 - (1 + k^2)*x + k^2*x^2]) + ((k + Sqrt[-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)])/(8*k^(3/2)*Sqrt[1 - (1 + k^2)*x + k^2*x^2])))/Sqrt[(1 - x 
)*x*(1 - k^2*x)]
 

3.20.66.3.1 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 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 7276
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 
3.20.66.4 Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.72

method result size
default \(\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right )}{2+2 k}+\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )}{-2+2 k}+\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{x \sqrt {k^{2}+1}}\right )}{\sqrt {k^{2}+1}}\) \(100\)
pseudoelliptic \(\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{\left (1+k \right ) x}\right )}{2+2 k}+\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )}{-2+2 k}+\frac {\arctan \left (\frac {\sqrt {\left (-1+x \right ) x \left (k^{2} x -1\right )}}{x \sqrt {k^{2}+1}}\right )}{\sqrt {k^{2}+1}}\) \(100\)
elliptic \(-\frac {2 \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticF}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}-\frac {\sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-1\right )}-\frac {\sqrt {-k^{2} x +1}\, \sqrt {-\frac {\left (-1+x \right ) k^{2}}{k^{2}-1}}\, \sqrt {k^{2} x}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (k^{2} \textit {\_Z}^{2}+1\right )}{\sum }\operatorname {EllipticPi}\left (\sqrt {-k^{2} x +1}, \frac {k^{2} \underline {\hspace {1.25 ex}}\alpha +1}{k^{2}+1}, \sqrt {-\frac {1}{k^{2}-1}}\right )\right )}{\left (k^{2}+1\right ) \sqrt {x \left (k^{2} x^{2}-k^{2} x -x +1\right )}}-\frac {\sqrt {-k^{2} x +1}\, \sqrt {-\frac {\left (-1+x \right ) k^{2}}{k^{2}-1}}\, \sqrt {k^{2} x}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (k^{2} \textit {\_Z}^{2}+1\right )}{\sum }\frac {\operatorname {EllipticPi}\left (\sqrt {-k^{2} x +1}, \frac {k^{2} \underline {\hspace {1.25 ex}}\alpha +1}{k^{2}+1}, \sqrt {-\frac {1}{k^{2}-1}}\right )}{\underline {\hspace {1.25 ex}}\alpha }\right )}{k^{2} \left (k^{2}+1\right ) \sqrt {x \left (k^{2} x^{2}-k^{2} x -x +1\right )}}+\frac {\sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+1\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+1\right )}\) \(559\)

input
int((k^4*x^4+1)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^4*x^4-1),x,method=_RETURNVER 
BOSE)
 
output
arctan(((-1+x)*x*(k^2*x-1))^(1/2)/(1+k)/x)/(2+2*k)+arctan(((-1+x)*x*(k^2*x 
-1))^(1/2)/(-1+k)/x)/(-2+2*k)+1/(k^2+1)^(1/2)*arctan(((-1+x)*x*(k^2*x-1))^ 
(1/2)/x/(k^2+1)^(1/2))
 
3.20.66.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (122) = 244\).

Time = 0.30 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.06 \[ \int \frac {1+k^4 x^4}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^4 x^4\right )} \, dx=\frac {2 \, \sqrt {k^{2} + 1} {\left (k^{2} - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + 1\right )} x + 1\right )} \sqrt {k^{2} + 1}}{2 \, {\left ({\left (k^{4} + k^{2}\right )} x^{3} - {\left (k^{4} + 2 \, k^{2} + 1\right )} x^{2} + {\left (k^{2} + 1\right )} x\right )}}\right ) + {\left (k^{3} - k^{2} + k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} + k^{2}\right )} x^{3} - {\left (k^{3} + k^{2} + k + 1\right )} x^{2} + {\left (k + 1\right )} x\right )}}\right ) + {\left (k^{3} + k^{2} + k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} - k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} - k^{2}\right )} x^{3} - {\left (k^{3} - k^{2} + k - 1\right )} x^{2} + {\left (k - 1\right )} x\right )}}\right )}{4 \, {\left (k^{4} - 1\right )}} \]

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

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

input
integrate((k**4*x**4+1)/((1-x)*x*(-k**2*x+1))**(1/2)/(k**4*x**4-1),x)
 
output
Integral((k**4*x**4 + 1)/(sqrt(x*(x - 1)*(k**2*x - 1))*(k*x - 1)*(k*x + 1) 
*(k**2*x**2 + 1)), x)
 
3.20.66.7 Maxima [F]

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

input
integrate((k^4*x^4+1)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^4*x^4-1),x, algorithm= 
"maxima")
 
output
integrate((k^4*x^4 + 1)/((k^4*x^4 - 1)*sqrt((k^2*x - 1)*(x - 1)*x)), x)
 
3.20.66.8 Giac [F]

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

input
integrate((k^4*x^4+1)/((1-x)*x*(-k^2*x+1))^(1/2)/(k^4*x^4-1),x, algorithm= 
"giac")
 
output
integrate((k^4*x^4 + 1)/((k^4*x^4 - 1)*sqrt((k^2*x - 1)*(x - 1)*x)), x)
 
3.20.66.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1+k^4 x^4}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^4 x^4\right )} \, dx=\text {Hanged} \]

input
int((k^4*x^4 + 1)/((k^4*x^4 - 1)*(x*(k^2*x - 1)*(x - 1))^(1/2)),x)
 
output
\text{Hanged}