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

Optimal. Leaf size=60 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {k^2+1} \sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}{(x-1) \left (k^2 x-1\right )}\right )}{\sqrt {k^2+1}} \]

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Rubi [C]  time = 2.35, antiderivative size = 262, normalized size of antiderivative = 4.37, number of steps used = 16, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {6718, 6688, 6725, 714, 115, 934, 12, 168, 537} \begin {gather*} \frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x} \Pi \left (-\frac {1}{\sqrt {-k^2}};\sin ^{-1}\left (\sqrt {-k^2} \sqrt {-x}\right )|\frac {1}{k^2}\right )}{\sqrt {-k^2} \sqrt {x-x^2} \sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x} \Pi \left (\frac {1}{\sqrt {-k^2}};\sin ^{-1}\left (\sqrt {-k^2} \sqrt {-x}\right )|\frac {1}{k^2}\right )}{\sqrt {-k^2} \sqrt {x-x^2} \sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 115

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (GtQ[-(b/d), 0] || LtQ[-(b/f), 0])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 714

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Int[(d + e*x)^m/(Sqrt[b*x]*Sqrt[1
+ (c*x)/b]), x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4] && LtQ[
c, 0] && RationalQ[b]

Rule 934

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x])/Sqrt[a + b*x + c*x^2], Int[1/((d +
e*x)*Sqrt[f + g*x]*Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x]), x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6718

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n*z^q)^FracP
art[p])/(v^(m*FracPart[p])*w^(n*FracPart[p])*z^(q*FracPart[p])), Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; Free
Q[{a, m, n, p, q}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !FreeQ[w, x] &&  !FreeQ[z, x]

Rule 6725

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*} \int \frac {-1+k^2 x^2}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (1+k^2 x^2\right )} \, dx &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {-1+k^2 x^2}{\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \left (1+k^2 x^2\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {-1+k^2 x^2}{\sqrt {1-k^2 x} \sqrt {x-x^2} \left (1+k^2 x^2\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \left (\frac {1}{\sqrt {1-k^2 x} \sqrt {x-x^2}}-\frac {2}{\sqrt {1-k^2 x} \sqrt {x-x^2} \left (1+k^2 x^2\right )}\right ) \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-k^2 x} \sqrt {x-x^2}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-k^2 x} \sqrt {x-x^2} \left (1+k^2 x^2\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \left (\frac {1}{2 \sqrt {1-k^2 x} \left (1-\sqrt {-k^2} x\right ) \sqrt {x-x^2}}+\frac {1}{2 \sqrt {1-k^2 x} \left (1+\sqrt {-k^2} x\right ) \sqrt {x-x^2}}\right ) \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-k^2 x} \left (1-\sqrt {-k^2} x\right ) \sqrt {x-x^2}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-k^2 x} \left (1+\sqrt {-k^2} x\right ) \sqrt {x-x^2}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (\sqrt {2} \sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {2} \sqrt {2-2 x} \sqrt {-x} \sqrt {1-k^2 x} \left (1-\sqrt {-k^2} x\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}-\frac {\left (\sqrt {2} \sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {2} \sqrt {2-2 x} \sqrt {-x} \sqrt {1-k^2 x} \left (1+\sqrt {-k^2} x\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (\sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {2-2 x} \sqrt {-x} \sqrt {1-k^2 x} \left (1-\sqrt {-k^2} x\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}-\frac {\left (\sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {2-2 x} \sqrt {-x} \sqrt {1-k^2 x} \left (1+\sqrt {-k^2} x\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (2 \sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+2 x^2} \sqrt {1+k^2 x^2} \left (1-\sqrt {-k^2} x^2\right )} \, dx,x,\sqrt {-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}+\frac {\left (2 \sqrt {2-2 x} \sqrt {1-x} \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+2 x^2} \sqrt {1+k^2 x^2} \left (1+\sqrt {-k^2} x^2\right )} \, dx,x,\sqrt {-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x} \Pi \left (-\frac {1}{\sqrt {-k^2}};\sin ^{-1}\left (\sqrt {-k^2} \sqrt {-x}\right )|\frac {1}{k^2}\right )}{\sqrt {-k^2} \sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}+\frac {2 (1-x) \sqrt {-x} \sqrt {x} \sqrt {1-k^2 x} \Pi \left (\frac {1}{\sqrt {-k^2}};\sin ^{-1}\left (\sqrt {-k^2} \sqrt {-x}\right )|\frac {1}{k^2}\right )}{\sqrt {-k^2} \sqrt {(1-x) x \left (1-k^2 x\right )} \sqrt {x-x^2}}\\ \end {align*}

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Mathematica [C]  time = 2.49, size = 165, normalized size = 2.75 \begin {gather*} \frac {2 i \sqrt {\frac {1}{x-1}+1} (x-1)^{3/2} \sqrt {\frac {1-\frac {1}{k^2}}{x-1}+1} \left (\left (k^2-1\right ) F\left (i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )+(1-i k) \Pi \left (\frac {k-i}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )+(1+i k) \Pi \left (\frac {k+i}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )\right )}{\left (k^2+1\right ) \sqrt {(x-1) x \left (k^2 x-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*Sqrt[1 + (-1 + x)^(-1)]*Sqrt[1 + (1 - k^(-2))/(-1 + x)]*(-1 + x)^(3/2)*((-1 + k^2)*EllipticF[I*ArcSinh[
1/Sqrt[-1 + x]], 1 - k^(-2)] + (1 - I*k)*EllipticPi[(-I + k)/k, I*ArcSinh[1/Sqrt[-1 + x]], 1 - k^(-2)] + (1 +
I*k)*EllipticPi[(I + k)/k, I*ArcSinh[1/Sqrt[-1 + x]], 1 - k^(-2)]))/((1 + k^2)*Sqrt[(-1 + x)*x*(-1 + k^2*x)])

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IntegrateAlgebraic [A]  time = 0.13, size = 60, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[1 + k^2]*Sqrt[x + (-1 - k^2)*x^2 + k^2*x^3])/((-1 + x)*(-1 + k^2*x))])/Sqrt[1 + k^2]

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fricas [A]  time = 0.50, size = 92, normalized size = 1.53 \begin {gather*} \frac {\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 )}{\sqrt {k^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))/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^{2} - 1}{{\left (k^{2} x^{2} + 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.11, size = 341, normalized size = 5.68

method	result	size
default	\(-\frac {2 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {-1+x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticF \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}-\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}+\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}\)	\(341\)
elliptic	\(-\frac {2 \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticF \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}-\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {i}{k}\right )}+\frac {2 i \sqrt {-k^{2} x +1}\, \sqrt {-\frac {1}{\frac {1}{k^{2}}-1}+\frac {x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {i}{k}\right )}\)	\(345\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/k^2*(-(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))-2*I/k^3*(-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)/(1/k^2-I/k)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2-I/k),(1/k^
2/(1/k^2-1))^(1/2))+2*I/k^3*(-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)/(1/k^2+I/k)*EllipticPi((-(x-1/k^2)*k^2)^(1/2),1/k^2/(1/k^2+I/k),(1/k^2/(1/k^2-1))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.76, size = 64, normalized size = 1.07 \begin {gather*} \frac {\ln \left (\frac {k^2\,x^2-2\,x\,\left (k^2+1\right )+1+\sqrt {k^2+1}\,\sqrt {x\,\left (k^2\,x-1\right )\,\left (x-1\right )}\,2{}\mathrm {i}}{k^2\,x^2+1}\right )\,1{}\mathrm {i}}{\sqrt {k^2+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((k^2*x^2 - 2*x*(k^2 + 1) + (k^2 + 1)^(1/2)*(x*(k^2*x - 1)*(x - 1))^(1/2)*2i + 1)/(k^2*x^2 + 1))*1i)/(k^2
+ 1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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