Integrand size = 31, antiderivative size = 79 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {1}{2 x}+\frac {x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (547+347 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac {189 \arctan (x)}{8}-\frac {1119 \arctan \left (\frac {x}{\sqrt {2}}\right )}{32 \sqrt {2}} \]
-1/2/x+1/8*x*(11*x^2+9)/(x^4+3*x^2+2)^2-1/32*x*(347*x^2+547)/(x^4+3*x^2+2) +189/8*arctan(x)-1119/64*arctan(1/2*x*2^(1/2))*2^(1/2)
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {1}{64} \left (-\frac {2 \left (64+1250 x^2+2499 x^4+1684 x^6+363 x^8\right )}{x \left (2+3 x^2+x^4\right )^2}+1512 \arctan (x)-1119 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )\right ) \]
((-2*(64 + 1250*x^2 + 2499*x^4 + 1684*x^6 + 363*x^8))/(x*(2 + 3*x^2 + x^4) ^2) + 1512*ArcTan[x] - 1119*Sqrt[2]*ArcTan[x/Sqrt[2]])/64
Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2198, 25, 2198, 25, 2195, 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.
\(\displaystyle \int \frac {5 x^6+3 x^4+x^2+4}{x^2 \left (x^4+3 x^2+2\right )^3} \, dx\) |
\(\Big \downarrow \) 2198 |
\(\displaystyle \frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}-\frac {1}{8} \int -\frac {55 x^4-29 x^2+16}{x^2 \left (x^4+3 x^2+2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{8} \int \frac {55 x^4-29 x^2+16}{x^2 \left (x^4+3 x^2+2\right )^2}dx+\frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}\) |
\(\Big \downarrow \) 2198 |
\(\displaystyle \frac {1}{8} \left (-\frac {1}{4} \int -\frac {-347 x^4+441 x^2+32}{x^2 \left (x^4+3 x^2+2\right )}dx-\frac {x \left (347 x^2+547\right )}{4 \left (x^4+3 x^2+2\right )}\right )+\frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \int \frac {-347 x^4+441 x^2+32}{x^2 \left (x^4+3 x^2+2\right )}dx-\frac {x \left (347 x^2+547\right )}{4 \left (x^4+3 x^2+2\right )}\right )+\frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}\) |
\(\Big \downarrow \) 2195 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \int \left (-\frac {1119}{x^2+2}+\frac {16}{x^2}+\frac {756}{x^2+1}\right )dx-\frac {x \left (347 x^2+547\right )}{4 \left (x^4+3 x^2+2\right )}\right )+\frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (756 \arctan (x)-\frac {1119 \arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {16}{x}\right )-\frac {x \left (347 x^2+547\right )}{4 \left (x^4+3 x^2+2\right )}\right )+\frac {x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}\) |
(x*(9 + 11*x^2))/(8*(2 + 3*x^2 + x^4)^2) + (-1/4*(x*(547 + 347*x^2))/(2 + 3*x^2 + x^4) + (-16/x + 756*ArcTan[x] - (1119*ArcTan[x/Sqrt[2]])/Sqrt[2])/ 4)/8
3.1.97.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d*x)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[x^m*Pq, a + b*x^2 + c*x^4, x], d = Coeff[Pol ynomialRemainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Polynomial Remainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4) ^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[x^m*(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[(2*a*(p + 1)*(b^2 - 4*a*c)*Qx)/x^m + (b^2*d*(2* p + 3) - 2*a*c*d*(4*p + 5) - a*b*e)/x^m + c*(4*p + 7)*(b*d - 2*a*e)*x^(2 - m), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && GtQ[Expon[Pq, x ^2], 1] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && ILtQ[m/2, 0]
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {-\frac {363}{32} x^{8}-\frac {421}{8} x^{6}-\frac {2499}{32} x^{4}-\frac {625}{16} x^{2}-2}{x \left (x^{4}+3 x^{2}+2\right )^{2}}-\frac {1119 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{64}+\frac {189 \arctan \left (x \right )}{8}\) | \(56\) |
default | \(-\frac {\frac {207}{16} x^{3}+\frac {233}{8} x}{2 \left (x^{2}+2\right )^{2}}-\frac {1119 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{64}-\frac {1}{2 x}+\frac {-\frac {35}{8} x^{3}-\frac {37}{8} x}{\left (x^{2}+1\right )^{2}}+\frac {189 \arctan \left (x \right )}{8}\) | \(58\) |
(-363/32*x^8-421/8*x^6-2499/32*x^4-625/16*x^2-2)/x/(x^4+3*x^2+2)^2-1119/64 *arctan(1/2*x*2^(1/2))*2^(1/2)+189/8*arctan(x)
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.37 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {726 \, x^{8} + 3368 \, x^{6} + 4998 \, x^{4} + 1119 \, \sqrt {2} {\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 2500 \, x^{2} - 1512 \, {\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )} \arctan \left (x\right ) + 128}{64 \, {\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )}} \]
-1/64*(726*x^8 + 3368*x^6 + 4998*x^4 + 1119*sqrt(2)*(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*x)*arctan(1/2*sqrt(2)*x) + 2500*x^2 - 1512*(x^9 + 6*x^7 + 13* x^5 + 12*x^3 + 4*x)*arctan(x) + 128)/(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*x)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {- 363 x^{8} - 1684 x^{6} - 2499 x^{4} - 1250 x^{2} - 64}{32 x^{9} + 192 x^{7} + 416 x^{5} + 384 x^{3} + 128 x} + \frac {189 \operatorname {atan}{\left (x \right )}}{8} - \frac {1119 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{64} \]
(-363*x**8 - 1684*x**6 - 2499*x**4 - 1250*x**2 - 64)/(32*x**9 + 192*x**7 + 416*x**5 + 384*x**3 + 128*x) + 189*atan(x)/8 - 1119*sqrt(2)*atan(sqrt(2)* x/2)/64
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {1119}{64} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {363 \, x^{8} + 1684 \, x^{6} + 2499 \, x^{4} + 1250 \, x^{2} + 64}{32 \, {\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )}} + \frac {189}{8} \, \arctan \left (x\right ) \]
-1119/64*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/32*(363*x^8 + 1684*x^6 + 2499*x ^4 + 1250*x^2 + 64)/(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*x) + 189/8*arctan(x )
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {1119}{64} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {347 \, x^{7} + 1588 \, x^{5} + 2291 \, x^{3} + 1058 \, x}{32 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} - \frac {1}{2 \, x} + \frac {189}{8} \, \arctan \left (x\right ) \]
-1119/64*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/32*(347*x^7 + 1588*x^5 + 2291*x ^3 + 1058*x)/(x^4 + 3*x^2 + 2)^2 - 1/2/x + 189/8*arctan(x)
Time = 8.84 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {189\,\mathrm {atan}\left (x\right )}{8}-\frac {1119\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{64}-\frac {\frac {363\,x^8}{32}+\frac {421\,x^6}{8}+\frac {2499\,x^4}{32}+\frac {625\,x^2}{16}+2}{x^9+6\,x^7+13\,x^5+12\,x^3+4\,x} \]