\(\int \frac {c+b x^2+c k^2 x^4}{\sqrt {(1-x^2) (1-k^2 x^2)} (-1+k^2 x^4)} \, dx\) [1575]

Optimal result

Integrand size = 49, antiderivative size = 108 \[ \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {(-b-2 c k) \arctan \left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 (-1+k) k}+\frac {(b-2 c k) \arctan \left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \]

[Out]

1/4*(-2*c*k-b)*arctan((-1+k)*x/(1+(-k^2-1)*x^2+k^2*x^4)^(1/2))/(-1+k)/k+1/2*(-2*c*k+b)*arctan((1+k)*x/(1+k*x^2
+(1+(-k^2-1)*x^2+k^2*x^4)^(1/2)))/k/(1+k)

Rubi [C] (verified)

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

Time = 0.71 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.68, number of steps used = 14, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {1976, 6857, 1117, 1224, 1712, 209} \[ \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=-\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right ) (b+2 c k)}{4 (1-k) k}+\frac {\arctan \left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right ) (b-2 c k)}{4 k (k+1)}+\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} (b-2 c k) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{3/2} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}-\frac {\left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} (b+2 c k) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{8 k^{3/2} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\frac {c \left (k x^2+1\right ) \sqrt {\frac {k^2 x^4-\left (k^2+1\right ) x^2+1}{\left (k x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(k+1)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}} \]

[In]

Int[(c + b*x^2 + c*k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

-1/4*((b + 2*c*k)*ArcTan[((1 - k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]])/((1 - k)*k) + ((b - 2*c*k)*ArcTan[((1
+ k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]])/(4*k*(1 + k)) + (c*(1 + k*x^2)*Sqrt[(1 - (1 + k^2)*x^2 + k^2*x^4)
/(1 + k*x^2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (1 + k)^2/(4*k)])/(2*Sqrt[k]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4])
+ ((b - 2*c*k)*(1 + k*x^2)*Sqrt[(1 - (1 + k^2)*x^2 + k^2*x^4)/(1 + k*x^2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (
1 + k)^2/(4*k)])/(8*k^(3/2)*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]) - ((b + 2*c*k)*(1 + k*x^2)*Sqrt[(1 - (1 + k^2)*
x^2 + k^2*x^4)/(1 + k*x^2)^2]*EllipticF[2*ArcTan[Sqrt[k]*x], (1 + k)^2/(4*k)])/(8*k^(3/2)*Sqrt[1 - (1 + k^2)*x
^2 + k^2*x^4])

Rule 209

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

Rule 1117

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1224

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[
a + b*x^2 + c*x^4], x], x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; Fr
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1712

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {c+b x^2+c k^2 x^4}{\left (-1+k^2 x^4\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx \\ & = \int \left (\frac {c}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}+\frac {2 c+b x^2}{\left (-1+k^2 x^4\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right ) \, dx \\ & = c \int \frac {1}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx+\int \frac {2 c+b x^2}{\left (-1+k^2 x^4\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx \\ & = \frac {c \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\int \left (-\frac {b+2 c k}{2 k \left (1-k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}+\frac {b-2 c k}{2 k \left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right ) \, dx \\ & = \frac {c \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(b-2 c k) \int \frac {1}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k}-\frac {(b+2 c k) \int \frac {1}{\left (1-k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k} \\ & = \frac {c \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(b-2 c k) \int \frac {1}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}+\frac {(b-2 c k) \int \frac {1-k x^2}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}-\frac {(b+2 c k) \int \frac {1}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}-\frac {(b+2 c k) \int \frac {1+k x^2}{\left (1-k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k} \\ & = \frac {c \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(b-2 c k) \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{8 k^{3/2} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}-\frac {(b+2 c k) \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{8 k^{3/2} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(b-2 c k) \text {Subst}\left (\int \frac {1}{1-\left (-1-2 k-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k}-\frac {(b+2 c k) \text {Subst}\left (\int \frac {1}{1-\left (-1+2 k-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k} \\ & = -\frac {(b+2 c k) \arctan \left (\frac {(1-k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 (1-k) k}+\frac {(b-2 c k) \arctan \left (\frac {(1+k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 k (1+k)}+\frac {c \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{2 \sqrt {k} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(b-2 c k) \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{8 k^{3/2} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}-\frac {(b+2 c k) \left (1+k x^2\right ) \sqrt {\frac {1-\left (1+k^2\right ) x^2+k^2 x^4}{\left (1+k x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {k} x\right ),\frac {(1+k)^2}{4 k}\right )}{8 k^{3/2} \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 11.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.86 \[ \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (2 c k \operatorname {EllipticF}\left (\arcsin (x),k^2\right )+(b-2 c k) \operatorname {EllipticPi}\left (-k,\arcsin (x),k^2\right )-(b+2 c k) \operatorname {EllipticPi}\left (k,\arcsin (x),k^2\right )\right )}{2 k \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]

[In]

Integrate[(c + b*x^2 + c*k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

(Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*(2*c*k*EllipticF[ArcSin[x], k^2] + (b - 2*c*k)*EllipticPi[-k, ArcSin[x], k^2]
- (b + 2*c*k)*EllipticPi[k, ArcSin[x], k^2]))/(2*k*Sqrt[(-1 + x^2)*(-1 + k^2*x^2)])

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99

method	result	size
elliptic	\(\frac {\left (\frac {\left (2 c k -b \right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{4 k \left (1+k \right )}+\frac {\left (2 c k +b \right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{4 k \left (-1+k \right )}\right ) \sqrt {2}}{2}\)	\(107\)
default	\(-\frac {-2 \left (-2 c k +b \right ) \left (\ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right )\right ) \sqrt {-\left (-1+k \right )^{2}}+\left (2 c k +b \right ) \left (\ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )+\ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )+2 \ln \left (2\right )\right ) \sqrt {-\left (1+k \right )^{2}}}{8 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}\, k}\)	\(253\)
pseudoelliptic	\(-\frac {-2 \left (-2 c k +b \right ) \left (\ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right )\right ) \sqrt {-\left (-1+k \right )^{2}}+\left (2 c k +b \right ) \left (\ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )+\ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )+2 \ln \left (2\right )\right ) \sqrt {-\left (1+k \right )^{2}}}{8 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}\, k}\)	\(253\)

[In]

int((c*k^2*x^4+b*x^2+c)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/4*(2*c*k-b)/k*2^(1/2)/(1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(1+k))+1/4*(2*c*k+b)/k*2^(1/2)/(-1+k
)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(-1+k)))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.61 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99 \[ \int \frac {c+b x^2+c k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {{\left (2 \, c k^{2} - {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) + {\left (2 \, c k^{2} + {\left (b + 2 \, c\right )} k + b\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{4 \, {\left (k^{3} - k\right )}} \]

[In]

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

[Out]

1/4*((2*c*k^2 - (b + 2*c)*k + b)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)/((k + 1)*x)) + (2*c*k^2 + (b + 2*c)*
k + b)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)/((k - 1)*x)))/(k^3 - k)

Sympy [F]

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

[In]

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

[Out]

Integral((b*x**2 + c*k**2*x**4 + c)/(sqrt((x - 1)*(x + 1)*(k*x - 1)*(k*x + 1))*(k*x**2 - 1)*(k*x**2 + 1)), x)

Maxima [F]

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

[In]

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

[Out]

integrate((c*k^2*x^4 + b*x^2 + c)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*k^2*x^4 + b*x^2 + c)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

Mupad [F(-1)]

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

[In]

int((c + b*x^2 + c*k^2*x^4)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

int((c + b*x^2 + c*k^2*x^4)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)), x)