3.24.55 \(\int \frac {-i+\sqrt {k} x}{(i+\sqrt {k} x) \sqrt {(1-x^2) (1-k^2 x^2)}} \, dx\) [2355]

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

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 92, normalized size of antiderivative = 0.49, number of steps used = 8, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {1976, 1755, 12, 1261, 738, 212, 1712, 210} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{k+1}+\frac {i \tanh ^{-1}\left (\frac {-k (k+1) x^2+k+1}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{k+1} \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]

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

Rule 12

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

Rule 210

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

Rule 212

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

Rule 738

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

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

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 1755

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[Px, x,
0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2], D = Coeff[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^
2 - e^2*x^2)*Sqrt[a + b*x^2 + c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt[a + b*
x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px, x], 3] && NeQ[c*d^4 + b*d^2*e
^2 + a*e^4, 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]

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 ) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 3.00, size = 155, normalized size = 0.83 \begin {gather*} \frac {-2 i \sqrt {-1+x^2} \sqrt {-1+k^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-1+x^2}}\right )+(1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\text {ArcSin}(x)\left |k^2\right .\right )-2 (1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\text {ArcSin}(x)\left |k^2\right .\right )}{(1+k) \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\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[-1 + x^2]*Sqrt[-1 + k^2*x^2]*ArcTanh[Sqrt[-1 + k^2*x^2]/(Sqrt[k]*Sqrt[-1 + x^2])] + (1 + k)*Sqrt[
1 - x^2]*Sqrt[1 - k^2*x^2]*EllipticF[ArcSin[x], k^2] - 2*(1 + k)*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*EllipticPi[-k
, ArcSin[x], k^2])/((1 + k)*Sqrt[(-1 + x^2)*(-1 + k^2*x^2)])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.27, size = 295, normalized size = 1.58

method	result	size
elliptic	\(-\frac {\left (-\sqrt {k}\, x +i\right ) \left (k \,x^{2}+1\right ) \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, \left (\frac {i \ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{\sqrt {\left (1+k \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{1+k}\right )}{\left (i+\sqrt {k}\, x \right ) \left (\sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, k \,x^{2}-2 i x k \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-\sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\right )}\)	\(259\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {2 i \left (k \,x^{2}+1\right ) \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, \left (-\frac {i \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{2 \sqrt {\left (1+k \right )^{2}}}\right )}{\left (i+\sqrt {k}\, x \right ) \left (-k x \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+i \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\right )}\)	\(295\)

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]

(-x^2+1)^(1/2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticF(x,k)+2*I*(k*x^2+1)*((x^2-1)*(k^2*x^2
-1)*k)^(1/2)/(I+k^(1/2)*x)/(-k*x*((x^2-1)*(k^2*x^2-1))^(1/2)+I*((x^2-1)*(k^2*x^2-1)*k)^(1/2))*(-I*(-x^2+1)^(1/
2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticPi(x,-k,k)-1/2/((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*} \text {Failed to integrate} \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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Fricas [A]
time = 0.68, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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