Integrand size = 44, antiderivative size = 74 \[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=\frac {\arctan \left (\frac {a+2 \left (1+a^2-b\right ) x+a x^2}{\sqrt {2} \sqrt {1-b} \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}}\right )}{\sqrt {2} \sqrt {1-b}} \]
1/2*arctan(1/2*(a+2*(a^2-b+1)*x+a*x^2)*2^(1/2)/(1-b)^(1/2)/(2*a*x^3+x^4+2* b*x^2+2*a*x+1)^(1/2))*2^(1/2)/(1-b)^(1/2)
Time = 0.77 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-1+b} x}{1+2 a x+x^2-\sqrt {1+2 b x^2+x^4+2 a \left (x+x^3\right )}}\right )}{\sqrt {-1+b}} \]
-((Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[-1 + b]*x)/(1 + 2*a*x + x^2 - Sqrt[1 + 2* b*x^2 + x^4 + 2*a*(x + x^3)])])/Sqrt[-1 + b])
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2507}
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.
\(\displaystyle \int \frac {1-x^2}{\left (2 a x+x^2+1\right ) \sqrt {2 a x^3+2 a x+2 b x^2+x^4+1}} \, dx\) |
\(\Big \downarrow \) 2507 |
\(\displaystyle \frac {\arctan \left (\frac {2 x \left (a^2-b+1\right )+a x^2+a}{\sqrt {2} \sqrt {1-b} \sqrt {2 a x^3+2 a x+2 b x^2+x^4+1}}\right )}{\sqrt {2} \sqrt {1-b}}\) |
ArcTan[(a + 2*(1 + a^2 - b)*x + a*x^2)/(Sqrt[2]*Sqrt[1 - b]*Sqrt[1 + 2*a*x + 2*b*x^2 + 2*a*x^3 + x^4])]/(Sqrt[2]*Sqrt[1 - b])
3.4.27.3.1 Defintions of rubi rules used
Int[((f_) + (g_.)*(x_)^2)/(((d_) + (e_.)*(x_) + (d_.)*(x_)^2)*Sqrt[(a_) + ( b_.)*(x_) + (c_.)*(x_)^2 + (b_.)*(x_)^3 + (a_.)*(x_)^4]), x_Symbol] :> Simp [a*(f/(d*Rt[a^2*(2*a - c), 2]))*ArcTan[(a*b + (4*a^2 + b^2 - 2*a*c)*x + a*b *x^2)/(2*Rt[a^2*(2*a - c), 2]*Sqrt[a + b*x + c*x^2 + b*x^3 + a*x^4])], x] / ; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[b*d - a*e, 0] && EqQ[f + g, 0] && PosQ[a^2*(2*a - c)]
Time = 2.51 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {\ln \left (2\right )+\ln \left (\frac {\sqrt {2 b -2}\, \sqrt {2 a \,x^{3}+x^{4}+2 x^{2} b +2 a x +1}-a \,x^{2}+\left (-2 a^{2}+2 b -2\right ) x -a}{2 a x +x^{2}+1}\right )}{\sqrt {2 b -2}}\) | \(78\) |
pseudoelliptic | \(\frac {\ln \left (2\right )+\ln \left (\frac {\sqrt {2 b -2}\, \sqrt {2 a \,x^{3}+x^{4}+2 x^{2} b +2 a x +1}-a \,x^{2}+\left (-2 a^{2}+2 b -2\right ) x -a}{2 a x +x^{2}+1}\right )}{\sqrt {2 b -2}}\) | \(78\) |
elliptic | \(\text {Expression too large to display}\) | \(258804\) |
1/(2*b-2)^(1/2)*(ln(2)+ln(((2*b-2)^(1/2)*(2*a*x^3+x^4+2*b*x^2+2*a*x+1)^(1/ 2)-a*x^2+(-2*a^2+2*b-2)*x-a)/(2*a*x+x^2+1)))
Time = 0.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.41 \[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=\left [\frac {\sqrt {2} \log \left (\frac {4 \, a^{3} x^{3} + {\left (a^{2} + 2 \, b - 2\right )} x^{4} + 4 \, a^{3} x + 2 \, {\left (2 \, a^{4} + 5 \, a^{2} - 2 \, {\left (2 \, a^{2} + 3\right )} b + 4 \, b^{2} + 2\right )} x^{2} + a^{2} - \frac {2 \, \sqrt {2} \sqrt {2 \, a x^{3} + x^{4} + 2 \, b x^{2} + 2 \, a x + 1} {\left ({\left (a b - a\right )} x^{2} + a b - 2 \, {\left (a^{2} - {\left (a^{2} + 2\right )} b + b^{2} + 1\right )} x - a\right )}}{\sqrt {b - 1}} + 2 \, b - 2}{4 \, a x^{3} + x^{4} + 2 \, {\left (2 \, a^{2} + 1\right )} x^{2} + 4 \, a x + 1}\right )}{4 \, \sqrt {b - 1}}, \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {1}{b - 1}} \arctan \left (\frac {\sqrt {2} \sqrt {2 \, a x^{3} + x^{4} + 2 \, b x^{2} + 2 \, a x + 1} {\left (b - 1\right )} \sqrt {-\frac {1}{b - 1}}}{a x^{2} + 2 \, {\left (a^{2} - b + 1\right )} x + a}\right )\right ] \]
[1/4*sqrt(2)*log((4*a^3*x^3 + (a^2 + 2*b - 2)*x^4 + 4*a^3*x + 2*(2*a^4 + 5 *a^2 - 2*(2*a^2 + 3)*b + 4*b^2 + 2)*x^2 + a^2 - 2*sqrt(2)*sqrt(2*a*x^3 + x ^4 + 2*b*x^2 + 2*a*x + 1)*((a*b - a)*x^2 + a*b - 2*(a^2 - (a^2 + 2)*b + b^ 2 + 1)*x - a)/sqrt(b - 1) + 2*b - 2)/(4*a*x^3 + x^4 + 2*(2*a^2 + 1)*x^2 + 4*a*x + 1))/sqrt(b - 1), 1/2*sqrt(2)*sqrt(-1/(b - 1))*arctan(sqrt(2)*sqrt( 2*a*x^3 + x^4 + 2*b*x^2 + 2*a*x + 1)*(b - 1)*sqrt(-1/(b - 1))/(a*x^2 + 2*( a^2 - b + 1)*x + a))]
\[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=- \int \frac {x^{2}}{2 a x \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1} + x^{2} \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1} + \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1}}\, dx - \int \left (- \frac {1}{2 a x \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1} + x^{2} \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1} + \sqrt {2 a x^{3} + 2 a x + 2 b x^{2} + x^{4} + 1}}\right )\, dx \]
-Integral(x**2/(2*a*x*sqrt(2*a*x**3 + 2*a*x + 2*b*x**2 + x**4 + 1) + x**2* sqrt(2*a*x**3 + 2*a*x + 2*b*x**2 + x**4 + 1) + sqrt(2*a*x**3 + 2*a*x + 2*b *x**2 + x**4 + 1)), x) - Integral(-1/(2*a*x*sqrt(2*a*x**3 + 2*a*x + 2*b*x* *2 + x**4 + 1) + x**2*sqrt(2*a*x**3 + 2*a*x + 2*b*x**2 + x**4 + 1) + sqrt( 2*a*x**3 + 2*a*x + 2*b*x**2 + x**4 + 1)), x)
\[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=\int { -\frac {x^{2} - 1}{\sqrt {2 \, a x^{3} + x^{4} + 2 \, b x^{2} + 2 \, a x + 1} {\left (2 \, a x + x^{2} + 1\right )}} \,d x } \]
\[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=\int { -\frac {x^{2} - 1}{\sqrt {2 \, a x^{3} + x^{4} + 2 \, b x^{2} + 2 \, a x + 1} {\left (2 \, a x + x^{2} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1-x^2}{\left (1+2 a x+x^2\right ) \sqrt {1+2 a x+2 b x^2+2 a x^3+x^4}} \, dx=-\int \frac {x^2-1}{\left (x^2+2\,a\,x+1\right )\,\sqrt {x^4+2\,a\,x^3+2\,b\,x^2+2\,a\,x+1}} \,d x \]