Integrand size = 38, antiderivative size = 59 \[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=-\frac {x \sqrt {1+3 x^2+2 x^4}}{2 \left (1+2 x^2+2 x^4\right )}-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \]
Time = 0.59 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=-\frac {x \sqrt {1+3 x^2+2 x^4}}{2 \left (1+2 x^2+2 x^4\right )}-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.13 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {\left (2 x^4-1\right ) \sqrt {2 x^4+3 x^2+1}}{\left (2 x^4+2 x^2+1\right )^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\sqrt {2 x^4+3 x^2+1}}{2 x^4+2 x^2+1}-\frac {2 \left (x^2+1\right ) \sqrt {2 x^4+3 x^2+1}}{\left (2 x^4+2 x^2+1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (x^2+1\right ) \sqrt {\frac {2 x^2+1}{x^2+1}} \operatorname {EllipticF}(\arctan (x),-1)}{2 \sqrt {2 x^4+3 x^2+1}}+\frac {i \left (x^2+1\right ) \operatorname {EllipticPi}\left (\frac {1}{2}-\frac {i}{2},\arctan \left (\sqrt {2} x\right ),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {x^2+1}{2 x^2+1}} \sqrt {2 x^4+3 x^2+1}}-\frac {i \left (x^2+1\right ) \operatorname {EllipticPi}\left (\frac {1}{2}+\frac {i}{2},\arctan \left (\sqrt {2} x\right ),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {x^2+1}{2 x^2+1}} \sqrt {2 x^4+3 x^2+1}}-\frac {i \sqrt {2 x^4+3 x^2+1} x}{-4 x^2-(2-2 i)}-\frac {i \sqrt {2 x^4+3 x^2+1} x}{4 x^2+(2+2 i)}\) |
((-I)*x*Sqrt[1 + 3*x^2 + 2*x^4])/((-2 + 2*I) - 4*x^2) - (I*x*Sqrt[1 + 3*x^ 2 + 2*x^4])/((2 + 2*I) + 4*x^2) - ((1 + x^2)*Sqrt[(1 + 2*x^2)/(1 + x^2)]*E llipticF[ArcTan[x], -1])/(2*Sqrt[1 + 3*x^2 + 2*x^4]) + ((I/2)*(1 + x^2)*El lipticPi[1/2 - I/2, ArcTan[Sqrt[2]*x], 1/2])/(Sqrt[2]*Sqrt[(1 + x^2)/(1 + 2*x^2)]*Sqrt[1 + 3*x^2 + 2*x^4]) - ((I/2)*(1 + x^2)*EllipticPi[1/2 + I/2, ArcTan[Sqrt[2]*x], 1/2])/(Sqrt[2]*Sqrt[(1 + x^2)/(1 + 2*x^2)]*Sqrt[1 + 3*x ^2 + 2*x^4])
3.8.60.3.1 Defintions of rubi rules used
Time = 5.48 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.25
method | result | size |
elliptic | \(\frac {\left (-\frac {\sqrt {2 x^{4}+3 x^{2}+1}\, \sqrt {2}}{4 x \left (\frac {2 x^{4}+3 x^{2}+1}{2 x^{2}}-\frac {1}{2}\right )}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2 x^{4}+3 x^{2}+1}}{x}\right )}{2}\right ) \sqrt {2}}{2}\) | \(74\) |
trager | \(-\frac {x \sqrt {2 x^{4}+3 x^{2}+1}}{2 \left (2 x^{4}+2 x^{2}+1\right )}+\frac {\ln \left (-\frac {-2 x^{4}+2 x \sqrt {2 x^{4}+3 x^{2}+1}-4 x^{2}-1}{2 x^{4}+2 x^{2}+1}\right )}{4}\) | \(81\) |
risch | \(-\frac {x \sqrt {2 x^{4}+3 x^{2}+1}}{2 \left (2 x^{4}+2 x^{2}+1\right )}-\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}-2 x \sqrt {2}+3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{4}+\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}+2 x \sqrt {2}-3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{4}\) | \(123\) |
default | \(\frac {\left (-2 x^{4}-2 x^{2}-1\right ) \operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}-2 x \sqrt {2}+3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )+\left (2 x^{4}+2 x^{2}+1\right ) \operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}+2 x \sqrt {2}-3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )-2 x \sqrt {2 x^{4}+3 x^{2}+1}}{-8 \left (\sqrt {2}-1\right ) x^{2}+2 \left (2 x^{2}+\sqrt {2}\right )^{2}}\) | \(158\) |
pseudoelliptic | \(\frac {\left (-2 x^{4}-2 x^{2}-1\right ) \operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}-2 x \sqrt {2}+3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )+\left (2 x^{4}+2 x^{2}+1\right ) \operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}+2 x \sqrt {2}-3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )-2 x \sqrt {2 x^{4}+3 x^{2}+1}}{-8 \left (\sqrt {2}-1\right ) x^{2}+2 \left (2 x^{2}+\sqrt {2}\right )^{2}}\) | \(158\) |
1/2*(-1/4*(2*x^4+3*x^2+1)^(1/2)*2^(1/2)/x/(1/2*(2*x^4+3*x^2+1)/x^2-1/2)-1/ 2*2^(1/2)*arctanh((2*x^4+3*x^2+1)^(1/2)/x))*2^(1/2)
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=\frac {{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )} \log \left (\frac {2 \, x^{4} + 4 \, x^{2} - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 1} x + 1}{2 \, x^{4} + 2 \, x^{2} + 1}\right ) - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 1} x}{4 \, {\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}} \]
1/4*((2*x^4 + 2*x^2 + 1)*log((2*x^4 + 4*x^2 - 2*sqrt(2*x^4 + 3*x^2 + 1)*x + 1)/(2*x^4 + 2*x^2 + 1)) - 2*sqrt(2*x^4 + 3*x^2 + 1)*x)/(2*x^4 + 2*x^2 + 1)
\[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=\int \frac {\sqrt {\left (x^{2} + 1\right ) \left (2 x^{2} + 1\right )} \left (2 x^{4} - 1\right )}{\left (2 x^{4} + 2 x^{2} + 1\right )^{2}}\, dx \]
\[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=\int { \frac {\sqrt {2 \, x^{4} + 3 \, x^{2} + 1} {\left (2 \, x^{4} - 1\right )}}{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{2}} \,d x } \]
\[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=\int { \frac {\sqrt {2 \, x^{4} + 3 \, x^{2} + 1} {\left (2 \, x^{4} - 1\right )}}{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx=\int \frac {\left (2\,x^4-1\right )\,\sqrt {2\,x^4+3\,x^2+1}}{{\left (2\,x^4+2\,x^2+1\right )}^2} \,d x \]