Integrand size = 47, antiderivative size = 187 \[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\frac {\arctan \left (\frac {\left (-1-2 \sqrt {k}-k\right ) x}{-1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{1+k}+\frac {\arctan \left (\frac {\left (-1+2 \sqrt {k}-k\right ) x}{-1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{1+k}+\frac {i \text {arctanh}\left (\frac {\left (2 \sqrt {k}+2 k^{3/2}\right ) x^2}{1+2 k x^2+k^2 x^4+\left (-1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{1+k} \]
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)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 4.05 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.83 \[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\frac {-2 i \sqrt {-1+x^2} \sqrt {-1+k^2 x^2} \text {arctanh}\left (\frac {\sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-1+x^2}}\right )+(1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticF}\left (\arcsin (x),k^2\right )-2 (1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (-k,\arcsin (x),k^2\right )}{(1+k) \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
((-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[A rcSin[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)])
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.48, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.191, Rules used = {2048, 2279, 27, 25, 1576, 1154, 219, 2212, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {k} x-i}{\left (\sqrt {k} x+i\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle \int \frac {\sqrt {k} x-i}{\left (\sqrt {k} x+i\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 2279 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+\int \frac {2 i \sqrt {k} x}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+2 i \sqrt {k} \int -\frac {x}{\left (k x^2+1\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-2 i \sqrt {k} \int \frac {x}{\left (k x^2+1\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-i \sqrt {k} \int \frac {1}{\left (k x^2+1\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx^2\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+2 i \sqrt {k} \int \frac {1}{4 k (k+1)^2-x^4}d\frac {(k+1)^2 \left (1-k x^2\right )}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \int \frac {1-k x^2}{\left (-k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+\frac {i \text {arctanh}\left (\frac {(k+1) \left (1-k x^2\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{k+1}\) |
\(\Big \downarrow \) 2212 |
\(\displaystyle \int \frac {1}{-\frac {(k+1)^2 x^2}{k^2 x^4-\left (k^2+1\right ) x^2+1}-1}d\frac {x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\frac {i \text {arctanh}\left (\frac {(k+1) \left (1-k x^2\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{k+1}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\arctan \left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{k+1}+\frac {i \text {arctanh}\left (\frac {(k+1) \left (1-k x^2\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{k+1}\) |
-(ArcTan[((1 + k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(1 + k)) + (I*ArcT anh[((1 + k)*(1 - k*x^2))/(2*Sqrt[k]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4])])/ (1 + k)
3.24.55.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[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]
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] /; F reeQ[{a, b, c, d, e, n, p}, x]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Simp[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] & & EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
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]
Time = 1.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.48
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (2\right )+\ln \left (\frac {-i \sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+i \left (k^{2}-2 k +1\right ) x}{i k \,x^{2}-2 \sqrt {k}\, x -i}\right )}{\sqrt {-\left (1+k \right )^{2}}}\) | \(90\) |
elliptic | \(\frac {\left (-\sqrt {k}\, x +i\right ) \left (k \,x^{2}+1\right ) \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, \left (\frac {i \ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{\sqrt {\left (1+k \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{1+k}\right )}{\left (i+\sqrt {k}\, x \right ) \left (2 i k x \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-\sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, k \,x^{2}+\sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\right )}\) | \(257\) |
default | \(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {2 i \left (k \,x^{2}+1\right ) \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\, \left (-\frac {i \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \operatorname {EllipticPi}\left (x , -k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{2 \sqrt {\left (1+k \right )^{2}}}\right )}{\left (i+\sqrt {k}\, x \right ) \left (-k x \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+i \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right ) k}\right )}\) | \(295\) |
-(ln(2)+ln((-I*(-(1+k)^2)^(1/2)*((x^2-1)*(k^2*x^2-1))^(1/2)-2*k^(3/2)*x^2+ 2*k^(1/2)+I*(k^2-2*k+1)*x)/(I*k*x^2-2*k^(1/2)*x-I)))/(-(1+k)^2)^(1/2)
Time = 0.63 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.34 \[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\frac {i \, \log \left (\frac {{\left (-i \, k^{6} - 5 i \, k^{5} - 10 i \, k^{4} - 10 i \, k^{3} - 5 i \, k^{2} - i \, k\right )} x^{3} + {\left (i \, k^{5} + 5 i \, k^{4} + 10 i \, k^{3} + 10 i \, k^{2} + 5 i \, k + i\right )} x + \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left (k^{4} + 4 \, k^{3} - {\left (k^{5} + 4 \, k^{4} + 6 \, k^{3} + 4 \, k^{2} + k\right )} x^{2} - 2 \, {\left (-i \, k^{4} - 4 i \, k^{3} - 6 i \, k^{2} - 4 i \, k - i\right )} \sqrt {k} x + 6 \, k^{2} + 4 \, k + 1\right )} + 2 \, {\left ({\left (k^{5} + 3 \, k^{4} + 3 \, k^{3} + k^{2}\right )} x^{4} + k^{3} - {\left (k^{5} + 3 \, k^{4} + 4 \, k^{3} + 4 \, k^{2} + 3 \, k + 1\right )} x^{2} + 3 \, k^{2} + 3 \, k + 1\right )} \sqrt {k}}{4 \, {\left (k^{5} x^{4} + 2 \, k^{4} x^{2} + k^{3}\right )}}\right )}{k + 1} \]
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)
\[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int \frac {\sqrt {k} x - i}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (\sqrt {k} x + i\right )}\, dx \]
\[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int { \frac {\sqrt {k} x - i}{\sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}} {\left (\sqrt {k} x + i\right )}} \,d x } \]
\[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int { \frac {\sqrt {k} x - i}{\sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}} {\left (\sqrt {k} x + i\right )}} \,d x } \]
Timed out. \[ \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int \frac {\sqrt {k}\,x-\mathrm {i}}{\left (\sqrt {k}\,x+1{}\mathrm {i}\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]