Integrand size = 31, antiderivative size = 71 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=-\frac {2}{9 x^2}-\frac {25 \left (5+x^2\right )}{72 \left (3+2 x^2+x^4\right )}-\frac {71 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{216 \sqrt {2}}-\frac {13 \log (x)}{27}+\frac {13}{108} \log \left (3+2 x^2+x^4\right ) \]
-2/9/x^2-25/72*(x^2+5)/(x^4+2*x^2+3)-13/27*ln(x)+13/108*ln(x^4+2*x^2+3)-71 /432*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)
Result contains complex when optimal does not.
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.42 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=\frac {1}{864} \left (-\frac {192}{x^2}-\frac {300 \left (5+x^2\right )}{3+2 x^2+x^4}-416 \log (x)+\sqrt {2} \left (-71 i+52 \sqrt {2}\right ) \log \left (-i+\sqrt {2}-i x^2\right )+\sqrt {2} \left (71 i+52 \sqrt {2}\right ) \log \left (i+\sqrt {2}+i x^2\right )\right ) \]
(-192/x^2 - (300*(5 + x^2))/(3 + 2*x^2 + x^4) - 416*Log[x] + Sqrt[2]*(-71* I + 52*Sqrt[2])*Log[-I + Sqrt[2] - I*x^2] + Sqrt[2]*(71*I + 52*Sqrt[2])*Lo g[I + Sqrt[2] + I*x^2])/864
Time = 0.35 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2194, 2177, 27, 2159, 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.
\(\displaystyle \int \frac {5 x^6+3 x^4+x^2+4}{x^3 \left (x^4+2 x^2+3\right )^2} \, dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int \frac {5 x^6+3 x^4+x^2+4}{x^4 \left (x^4+2 x^2+3\right )^2}dx^2\) |
\(\Big \downarrow \) 2177 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \int \frac {2 \left (-25 x^4-20 x^2+48\right )}{9 x^4 \left (x^4+2 x^2+3\right )}dx^2-\frac {25 \left (x^2+5\right )}{36 \left (x^4+2 x^2+3\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{36} \int \frac {-25 x^4-20 x^2+48}{x^4 \left (x^4+2 x^2+3\right )}dx^2-\frac {25 \left (x^2+5\right )}{36 \left (x^4+2 x^2+3\right )}\right )\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{36} \int \left (\frac {52 x^2-19}{3 \left (x^4+2 x^2+3\right )}-\frac {52}{3 x^2}+\frac {16}{x^4}\right )dx^2-\frac {25 \left (x^2+5\right )}{36 \left (x^4+2 x^2+3\right )}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{36} \left (-\frac {71 \arctan \left (\frac {x^2+1}{\sqrt {2}}\right )}{3 \sqrt {2}}-\frac {16}{x^2}-\frac {52 \log \left (x^2\right )}{3}+\frac {26}{3} \log \left (x^4+2 x^2+3\right )\right )-\frac {25 \left (x^2+5\right )}{36 \left (x^4+2 x^2+3\right )}\right )\) |
((-25*(5 + x^2))/(36*(3 + 2*x^2 + x^4)) + (-16/x^2 - (71*ArcTan[(1 + x^2)/ Sqrt[2]])/(3*Sqrt[2]) - (52*Log[x^2])/3 + (26*Log[3 + 2*x^2 + x^4])/3)/36) /2
3.2.6.3.1 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[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x + c* x^2, x], R = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + (2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^ m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Qx)/(d + e*x )^m - ((2*p + 3)*(2*c*R - b*S))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a* e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] : > Simp[1/2 Subst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2) ^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && IntegerQ [(m - 1)/2]
Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {2}{9 x^{2}}-\frac {13 \ln \left (x \right )}{27}+\frac {-\frac {75 x^{2}}{4}-\frac {375}{4}}{54 x^{4}+108 x^{2}+162}+\frac {13 \ln \left (x^{4}+2 x^{2}+3\right )}{108}-\frac {71 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{432}\) | \(63\) |
risch | \(\frac {-\frac {41}{72} x^{4}-\frac {157}{72} x^{2}-\frac {2}{3}}{x^{2} \left (x^{4}+2 x^{2}+3\right )}-\frac {13 \ln \left (x \right )}{27}+\frac {13 \ln \left (5041 x^{4}+10082 x^{2}+15123\right )}{108}-\frac {71 \sqrt {2}\, \arctan \left (\frac {\left (71 x^{2}+71\right ) \sqrt {2}}{142}\right )}{432}\) | \(67\) |
-2/9/x^2-13/27*ln(x)+1/54*(-75/4*x^2-375/4)/(x^4+2*x^2+3)+13/108*ln(x^4+2* x^2+3)-71/432*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))
Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=-\frac {246 \, x^{4} + 71 \, \sqrt {2} {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 942 \, x^{2} - 52 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 208 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )} \log \left (x\right ) + 288}{432 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} \]
-1/432*(246*x^4 + 71*sqrt(2)*(x^6 + 2*x^4 + 3*x^2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 942*x^2 - 52*(x^6 + 2*x^4 + 3*x^2)*log(x^4 + 2*x^2 + 3) + 208*(x^ 6 + 2*x^4 + 3*x^2)*log(x) + 288)/(x^6 + 2*x^4 + 3*x^2)
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=\frac {- 41 x^{4} - 157 x^{2} - 48}{72 x^{6} + 144 x^{4} + 216 x^{2}} - \frac {13 \log {\left (x \right )}}{27} + \frac {13 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{108} - \frac {71 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{432} \]
(-41*x**4 - 157*x**2 - 48)/(72*x**6 + 144*x**4 + 216*x**2) - 13*log(x)/27 + 13*log(x**4 + 2*x**2 + 3)/108 - 71*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2) /2)/432
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=-\frac {71}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {41 \, x^{4} + 157 \, x^{2} + 48}{72 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac {13}{54} \, \log \left (x^{2}\right ) \]
-71/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/72*(41*x^4 + 157*x^2 + 4 8)/(x^6 + 2*x^4 + 3*x^2) + 13/108*log(x^4 + 2*x^2 + 3) - 13/54*log(x^2)
Time = 0.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=-\frac {71}{432} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {41 \, x^{4} + 157 \, x^{2} + 48}{72 \, {\left (x^{6} + 2 \, x^{4} + 3 \, x^{2}\right )}} + \frac {13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) - \frac {13}{54} \, \log \left (x^{2}\right ) \]
-71/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/72*(41*x^4 + 157*x^2 + 4 8)/(x^6 + 2*x^4 + 3*x^2) + 13/108*log(x^4 + 2*x^2 + 3) - 13/54*log(x^2)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (3+2 x^2+x^4\right )^2} \, dx=\frac {13\,\ln \left (x^4+2\,x^2+3\right )}{108}-\frac {13\,\ln \left (x\right )}{27}-\frac {\frac {41\,x^4}{72}+\frac {157\,x^2}{72}+\frac {2}{3}}{x^6+2\,x^4+3\,x^2}-\frac {71\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{432} \]