Integrand size = 43, antiderivative size = 236 \[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\frac {\arctan \left (\frac {\left (1-2 i \sqrt {k}-k\right ) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-1+\sqrt {k}\right ) \left (1+\sqrt {k}\right )}+\frac {\arctan \left (\frac {\left (1+2 i \sqrt {k}-k\right ) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-1+\sqrt {k}\right ) \left (1+\sqrt {k}\right )}+\frac {a \sqrt {k} \arctan \left (\frac {\left (2 \sqrt {k}-2 k^{3/2}\right ) x^2}{1-2 k x^2+k^2 x^4+\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 \left (-1+\sqrt {k}\right ) \left (1+\sqrt {k}\right )} \]
arctan((1-2*I*k^(1/2)-k)*x/(1+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(1/2)))/(-1+k ^(1/2))/(1+k^(1/2))+arctan((1+2*I*k^(1/2)-k)*x/(1+k*x^2+(1+(-k^2-1)*x^2+k^ 2*x^4)^(1/2)))/(-1+k^(1/2))/(1+k^(1/2))+1/2*a*k^(1/2)*arctan((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^(1/2))/(1+k^(1/2))
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.98 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.67 \[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\frac {-a \sqrt {k} \sqrt {-1+x^2} \sqrt {-1+k^2 x^2} \arctan \left (\frac {\sqrt {-1+k^2 x^2}}{\sqrt {k} \sqrt {-1+x^2}}\right )+(-1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticF}\left (\arcsin (x),k^2\right )-2 (-1+k) \sqrt {1-x^2} \sqrt {1-k^2 x^2} \operatorname {EllipticPi}\left (k,\arcsin (x),k^2\right )}{(-1+k) \sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
(-(a*Sqrt[k]*Sqrt[-1 + x^2]*Sqrt[-1 + k^2*x^2]*ArcTan[Sqrt[-1 + k^2*x^2]/( Sqrt[k]*Sqrt[-1 + x^2])]) + (-1 + k)*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*Ellip ticF[ArcSin[x], k^2] - 2*(-1 + k)*Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*Elliptic Pi[k, ArcSin[x], k^2])/((-1 + k)*Sqrt[(-1 + x^2)*(-1 + k^2*x^2)])
Time = 0.58 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.44, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {2048, 2254, 27, 25, 1576, 1154, 217, 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.
| \(\displaystyle \int \frac {a k x+k x^2+1}{\left (k x^2-1\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx\) |
| \(\Big \downarrow \) 2048 |
| \(\displaystyle \int \frac {a k x+k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
| \(\Big \downarrow \) 2254 |
| \(\displaystyle \int \frac {a k x}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+\int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a k \int -\frac {x}{\left (1-k x^2\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx+\int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-a k \int \frac {x}{\left (1-k x^2\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-\frac {1}{2} a k \int \frac {1}{\left (1-k x^2\right ) \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}dx^2\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle a k \int \frac {1}{-x^4-4 (1-k)^2 k}d\frac {(1-k)^2 \left (k x^2+1\right )}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}+\int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \int \frac {k x^2+1}{\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-\frac {a \sqrt {k} \arctan \left (\frac {(1-k) \left (k x^2+1\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (1-k)}\) |
| \(\Big \downarrow \) 2212 |
| \(\displaystyle \int \frac {1}{-\frac {(1-k)^2 x^2}{k^2 x^4-\left (k^2+1\right ) x^2+1}-1}d\frac {x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}-\frac {a \sqrt {k} \arctan \left (\frac {(1-k) \left (k x^2+1\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (1-k)}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {a \sqrt {k} \arctan \left (\frac {(1-k) \left (k x^2+1\right )}{2 \sqrt {k} \sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (1-k)}-\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{1-k}\) |
-(ArcTan[((1 - k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(1 - k)) - (a*Sqrt [k]*ArcTan[((1 - k)*(1 + k*x^2))/(2*Sqrt[k]*Sqrt[1 - (1 + k^2)*x^2 + k^2*x ^4])])/(2*(1 - k))
3.27.48.3.1 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[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[(Pr_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_), x_Symbol] :> Module[{r = Expon[Pr, x], k}, Int[Sum[Coeff[Pr, x, 2*k]* x^(2*k), {k, 0, r/2}]*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[C oeff[Pr, x, 2*k + 1]*x^(2*k), {k, 0, (r - 1)/2}]*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && PolyQ[Pr, x] && !Poly Q[Pr, x^2]
Time = 2.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.72
| method | result | size |
| elliptic | \(-\frac {a \ln \left (\frac {-\frac {2 \left (k^{2}-2 k +1\right )}{k}+\left (-k^{2}+2 k -1\right ) \left (x^{2}-\frac {1}{k}\right )+2 \sqrt {-\frac {k^{2}-2 k +1}{k}}\, \sqrt {k^{2} \left (x^{2}-\frac {1}{k}\right )^{2}+\left (-k^{2}+2 k -1\right ) \left (x^{2}-\frac {1}{k}\right )-\frac {k^{2}-2 k +1}{k}}}{x^{2}-\frac {1}{k}}\right )}{2 \sqrt {-\frac {k^{2}-2 k +1}{k}}}+\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{-1+k}\) | \(171\) |
| default | \(\frac {\left (a k -2 \sqrt {k}\right ) \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )+\left (-a k -2 \sqrt {k}\right ) \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )-4 \sqrt {k}\, \ln \left (2\right )}{4 \sqrt {-\left (-1+k \right )^{2}}\, \sqrt {k}}\) | \(184\) |
| pseudoelliptic | \(\frac {\left (a k -2 \sqrt {k}\right ) \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )+\left (-a k -2 \sqrt {k}\right ) \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )-4 \sqrt {k}\, \ln \left (2\right )}{4 \sqrt {-\left (-1+k \right )^{2}}\, \sqrt {k}}\) | \(184\) |
-1/2*a/(-(k^2-2*k+1)/k)^(1/2)*ln((-2*(k^2-2*k+1)/k+(-k^2+2*k-1)*(x^2-1/k)+ 2*(-(k^2-2*k+1)/k)^(1/2)*(k^2*(x^2-1/k)^2+(-k^2+2*k-1)*(x^2-1/k)-(k^2-2*k+ 1)/k)^(1/2))/(x^2-1/k))+1/(-1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(- 1+k))
Leaf count of result is larger than twice the leaf count of optimal. 1793 vs. \(2 (196) = 392\).
Time = 1.13 (sec) , antiderivative size = 1793, normalized size of antiderivative = 7.60 \[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\text {Too large to display} \]
-1/8*sqrt(-(a^2*k + 4*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1))*(k^2 - 2 *k + 1) + 4)/(k^2 - 2*k + 1))*log(-2*(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)*(a ^3*k + (a^3*k^2 - 4*a*k)*x^2 + 2*(a^2*k^3 - 2*(a^2 + 2)*k^2 + (a^2 + 8)*k - 4)*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1))*x - 4*a) + (2*a^2*k^3*x^4 + 2*(a*k^3 - 2*a*k^2 + a*k)*x^3 + 2*a^2*k - 2*(a^2*k^3 + a^2*k)*x^2 + 2*( a*k^2 - 2*a*k + a)*x - (4*(k^4 - 2*k^3 + k^2)*x^4 + (a*k^5 - 4*a*k^4 + 6*a *k^3 - 4*a*k^2 + a*k)*x^3 - 4*(k^4 - 2*k^3 + 2*k^2 - 2*k + 1)*x^2 + 4*k^2 + (a*k^4 - 4*a*k^3 + 6*a*k^2 - 4*a*k + a)*x - 8*k + 4)*sqrt(a^2*k/(k^4 - 4 *k^3 + 6*k^2 - 4*k + 1)))*sqrt(-(a^2*k + 4*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1))*(k^2 - 2*k + 1) + 4)/(k^2 - 2*k + 1)))/(k^2*x^4 - 2*k*x^2 + 1 )) + 1/8*sqrt(-(a^2*k + 4*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1))*(k^2 - 2*k + 1) + 4)/(k^2 - 2*k + 1))*log(-2*(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1 )*(a^3*k + (a^3*k^2 - 4*a*k)*x^2 + 2*(a^2*k^3 - 2*(a^2 + 2)*k^2 + (a^2 + 8 )*k - 4)*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1))*x - 4*a) - (2*a^2*k^3 *x^4 + 2*(a*k^3 - 2*a*k^2 + a*k)*x^3 + 2*a^2*k - 2*(a^2*k^3 + a^2*k)*x^2 + 2*(a*k^2 - 2*a*k + a)*x - (4*(k^4 - 2*k^3 + k^2)*x^4 + (a*k^5 - 4*a*k^4 + 6*a*k^3 - 4*a*k^2 + a*k)*x^3 - 4*(k^4 - 2*k^3 + 2*k^2 - 2*k + 1)*x^2 + 4* k^2 + (a*k^4 - 4*a*k^3 + 6*a*k^2 - 4*a*k + a)*x - 8*k + 4)*sqrt(a^2*k/(k^4 - 4*k^3 + 6*k^2 - 4*k + 1)))*sqrt(-(a^2*k + 4*sqrt(a^2*k/(k^4 - 4*k^3 + 6 *k^2 - 4*k + 1))*(k^2 - 2*k + 1) + 4)/(k^2 - 2*k + 1)))/(k^2*x^4 - 2*k*...
\[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int \frac {a k x + k x^{2} + 1}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right )}\, dx \]
\[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int { \frac {a k x + k x^{2} + 1}{{\left (k x^{2} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
\[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int { \frac {a k x + k x^{2} + 1}{{\left (k x^{2} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
Timed out. \[ \int \frac {1+a k x+k x^2}{\left (-1+k x^2\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx=\int \frac {k\,x^2+a\,k\,x+1}{\left (k\,x^2-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]