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

Optimal. Leaf size=497 \[ \frac {i \left (i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}+i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)}-\frac {i \left (-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)} \]

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Rubi [A]  time = 0.16, antiderivative size = 45, normalized size of antiderivative = 0.09, number of steps used = 2, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2112, 203} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {k^2+1} x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )}{\sqrt {k^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[1 + k^2]*x)/Sqrt[(1 - x^2)*(1 - k^2*x^2)]]/Sqrt[1 + k^2])

Rule 203

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

Rule 2112

Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coef
f[v, x, 4], d = Coeff[1/u, x, 0], e = Coeff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Dist[A, Subst[Int[1/(d - (b*d -
a*e)*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x
^2, 2] && PolyQ[1/u, x^2, 2]

Rubi steps

\begin {align*} \int \frac {-1+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 &=-\operatorname {Subst}\left (\int \frac {1}{1-\left (-1-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {1+k^2} x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )}{\sqrt {1+k^2}}\\ \end {align*}

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Mathematica [C]  time = 0.47, size = 78, normalized size = 0.16 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (F\left (\sin ^{-1}(x)|k^2\right )-\Pi \left (-i k;\sin ^{-1}(x)|k^2\right )-\Pi \left (i k;\sin ^{-1}(x)|k^2\right )\right )}{\sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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IntegrateAlgebraic [A]  time = 2.84, size = 497, normalized size = 1.00 \begin {gather*} -\frac {i \left (-i \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}-i k^2 \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+\sqrt {2} \sqrt {k} \sqrt {1+k^2} \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}\right ) \tan ^{-1}\left (\frac {\sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}} x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-i+\sqrt {k}\right )^2 \left (i+\sqrt {k}\right )^2 (-i+k) (i+k)}+\frac {i \left (i \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+i k^2 \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+\sqrt {2} \sqrt {k} \sqrt {1+k^2} \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}\right ) \tan ^{-1}\left (\frac {\sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}} x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-i+\sqrt {k}\right )^2 \left (i+\sqrt {k}\right )^2 (-i+k) (i+k)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*((-I)*Sqrt[1 - 2*k + k^2 - (2*I)*Sqrt[2]*Sqrt[k]*Sqrt[1 + k^2]] - I*k^2*Sqrt[1 - 2*k + k^2 - (2*I)*Sqrt[
2]*Sqrt[k]*Sqrt[1 + k^2]] + Sqrt[2]*Sqrt[k]*Sqrt[1 + k^2]*Sqrt[1 - 2*k + k^2 - (2*I)*Sqrt[2]*Sqrt[k]*Sqrt[1 +
k^2]])*ArcTan[(Sqrt[1 - 2*k + k^2 - (2*I)*Sqrt[2]*Sqrt[k]*Sqrt[1 + k^2]]*x)/(1 + k*x^2 + Sqrt[1 + (-1 - k^2)*x
^2 + k^2*x^4])])/((-I + Sqrt[k])^2*(I + Sqrt[k])^2*(-I + k)*(I + k)) + (I*(I*Sqrt[1 - 2*k + k^2 + (2*I)*Sqrt[2
]*Sqrt[k]*Sqrt[1 + k^2]] + I*k^2*Sqrt[1 - 2*k + k^2 + (2*I)*Sqrt[2]*Sqrt[k]*Sqrt[1 + k^2]] + Sqrt[2]*Sqrt[k]*S
qrt[1 + k^2]*Sqrt[1 - 2*k + k^2 + (2*I)*Sqrt[2]*Sqrt[k]*Sqrt[1 + k^2]])*ArcTan[(Sqrt[1 - 2*k + k^2 + (2*I)*Sqr
t[2]*Sqrt[k]*Sqrt[1 + k^2]]*x)/(1 + k*x^2 + Sqrt[1 + (-1 - k^2)*x^2 + k^2*x^4])])/((-I + Sqrt[k])^2*(I + Sqrt[
k])^2*(-I + k)*(I + k))

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fricas [A]  time = 0.55, size = 62, normalized size = 0.12 \begin {gather*} -\frac {\arctan \left (\frac {2 \, \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} \sqrt {k^{2} + 1} x}{k^{2} x^{4} - 2 \, {\left (k^{2} + 1\right )} x^{2} + 1}\right )}{2 \, \sqrt {k^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 51, normalized size = 0.10

method	result	size
elliptic	\(\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}\, \sqrt {2}}{x \sqrt {2 k^{2}+2}}\right ) \sqrt {2}}{\sqrt {2 k^{2}+2}}\)	\(51\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (k^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {-\frac {\arctanh \left (\frac {\left (2 k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-k^{2}-1\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2} k^{4}+k^{4} x^{2}-2 k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+6 k^{2} x^{2}+\underline {\hspace {1.25 ex}}\alpha ^{2}-4 k^{2}+x^{2}-4\right )}{2 \left (k^{4}+6 k^{2}+1\right ) \sqrt {-k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\right )}{\sqrt {-k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {2 \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \underline {\hspace {1.25 ex}}\alpha ^{3} k^{2} \EllipticPi \left (x , -k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}, k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}}{\underline {\hspace {1.25 ex}}\alpha ^{3}}}{4 k^{2}}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {k^2\,x^4-1}{\left (k^2\,x^4+1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k^{2} x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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