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

Optimal. Leaf size=187 \[ \frac {\tan ^{-1}\left (\frac {\left (-k-2 \sqrt {k}-1\right ) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2-1}\right )}{k+1}+\frac {\tan ^{-1}\left (\frac {\left (-k+2 \sqrt {k}-1\right ) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2-1}\right )}{k+1}+\frac {i \tanh ^{-1}\left (\frac {\left (2 k^{3/2}+2 \sqrt {k}\right ) x^2}{k^2 x^4+\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+2 k x^2+1}\right )}{k+1} \]

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Rubi [C]  time = 1.45, antiderivative size = 201, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6719, 6742, 419, 2113, 537, 571, 93, 208} \begin {gather*} \frac {2 i \sqrt {1-x^2} \sqrt {1-k^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {k} \sqrt {1-x^2}}{\sqrt {1-k^2 x^2}}\right )}{(k+1) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rule 419

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

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 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 2113

Int[1/(((a_) + (b_.)*(x_))*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[a, Int[1/((
a^2 - b^2*x^2)*Sqrt[c + d*x^2]*Sqrt[e + f*x^2]), x], x] - Dist[b, Int[x/((a^2 - b^2*x^2)*Sqrt[c + d*x^2]*Sqrt[
e + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 6742

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

Rubi steps

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

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Mathematica [C]  time = 0.36, size = 208, normalized size = 1.11 \begin {gather*} \frac {-2 i \sqrt {k} \sqrt {x^2-1} \sqrt {k^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {k (k+1)} \sqrt {x^2-1}}{\sqrt {k+1} \sqrt {k^2 x^2-1}}\right )+\sqrt {k+1} \sqrt {k (k+1)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )-2 \sqrt {k+1} \sqrt {k (k+1)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {k+1} \sqrt {k (k+1)} \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.63, size = 56, normalized size = 0.30 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {(1+k) x}{-1+2 i \sqrt {k} x+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{1+k} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.20, size = 250, normalized size = 1.34 \begin {gather*} \frac {i \, \log \left (\frac {{\left (-i \, k^{6} - 5 i \, k^{5} - 10 i \, k^{4} - 10 i \, k^{3} - 5 i \, k^{2} - i \, k\right )} x^{3} + {\left (i \, k^{5} + 5 i \, k^{4} + 10 i \, k^{3} + 10 i \, k^{2} + 5 i \, k + i\right )} x + \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left (k^{4} + 4 \, k^{3} - {\left (k^{5} + 4 \, k^{4} + 6 \, k^{3} + 4 \, k^{2} + k\right )} x^{2} - 2 \, {\left (-i \, k^{4} - 4 i \, k^{3} - 6 i \, k^{2} - 4 i \, k - i\right )} \sqrt {k} x + 6 \, k^{2} + 4 \, k + 1\right )} + 2 \, {\left ({\left (k^{5} + 3 \, k^{4} + 3 \, k^{3} + k^{2}\right )} x^{4} + k^{3} - {\left (k^{5} + 3 \, k^{4} + 4 \, k^{3} + 4 \, k^{2} + 3 \, k + 1\right )} x^{2} + 3 \, k^{2} + 3 \, k + 1\right )} \sqrt {k}}{4 \, {\left (k^{5} x^{4} + 2 \, k^{4} x^{2} + k^{3}\right )}}\right )}{k + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*log(1/4*((-I*k^6 - 5*I*k^5 - 10*I*k^4 - 10*I*k^3 - 5*I*k^2 - I*k)*x^3 + (I*k^5 + 5*I*k^4 + 10*I*k^3 + 10*I*k
^2 + 5*I*k + I)*x + sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)*(k^4 + 4*k^3 - (k^5 + 4*k^4 + 6*k^3 + 4*k^2 + k)*x^2 - 2
*(-I*k^4 - 4*I*k^3 - 6*I*k^2 - 4*I*k - I)*sqrt(k)*x + 6*k^2 + 4*k + 1) + 2*((k^5 + 3*k^4 + 3*k^3 + k^2)*x^4 +
k^3 - (k^5 + 3*k^4 + 4*k^3 + 4*k^2 + 3*k + 1)*x^2 + 3*k^2 + 3*k + 1)*sqrt(k))/(k^5*x^4 + 2*k^4*x^2 + k^3))/(k
+ 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 145, normalized size = 0.78

method	result	size
elliptic	\(\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{1+k}+\frac {i \ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{\sqrt {\left (1+k \right )^{2}}}\)	\(145\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-2 i \left (-\frac {i \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{2 \sqrt {\left (1+k \right )^{2}}}\right )\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(1+k))+I/((1+k)^2)^(1/2)*ln((2*k^2+4*k+2+(-k^3-2*k^2-k)*(x^2+1/
k)+2*((1+k)^2)^(1/2)*(k^3*(x^2+1/k)^2+(-k^3-2*k^2-k)*(x^2+1/k)+k^2+2*k+1)^(1/2))/(x^2+1/k))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {k}\,x-\mathrm {i}}{\left (\sqrt {k}\,x+1{}\mathrm {i}\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^(1/2)*x - 1i)/((k^(1/2)*x + 1i)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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