Integrand size = 31, antiderivative size = 81 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {19 x^2}{2}-\frac {17 x^4}{4}+\frac {5 x^6}{6}+\frac {25 \left (3+5 x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {455 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {19}{2} \log \left (3+2 x^2+x^4\right ) \]
19/2*x^2-17/4*x^4+5/6*x^6+25/8*(5*x^2+3)/(x^4+2*x^2+3)+19/2*ln(x^4+2*x^2+3 )-455/16*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {1}{48} \left (456 x^2-204 x^4+40 x^6+\frac {150 \left (3+5 x^2\right )}{3+2 x^2+x^4}-1365 \sqrt {2} \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )+456 \log \left (3+2 x^2+x^4\right )\right ) \]
(456*x^2 - 204*x^4 + 40*x^6 + (150*(3 + 5*x^2))/(3 + 2*x^2 + x^4) - 1365*S qrt[2]*ArcTan[(1 + x^2)/Sqrt[2]] + 456*Log[3 + 2*x^2 + x^4])/48
Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2194, 2191, 27, 2188, 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.
\(\displaystyle \int \frac {x^7 \left (5 x^6+3 x^4+x^2+4\right )}{\left (x^4+2 x^2+3\right )^2} \, dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (5 x^6+3 x^4+x^2+4\right )}{\left (x^4+2 x^2+3\right )^2}dx^2\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \int -\frac {2 \left (-20 x^8+28 x^6-100 x^2+75\right )}{x^4+2 x^2+3}dx^2+\frac {25 \left (5 x^2+3\right )}{4 \left (x^4+2 x^2+3\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {25 \left (5 x^2+3\right )}{4 \left (x^4+2 x^2+3\right )}-\frac {1}{4} \int \frac {-20 x^8+28 x^6-100 x^2+75}{x^4+2 x^2+3}dx^2\right )\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle \frac {1}{2} \left (\frac {25 \left (5 x^2+3\right )}{4 \left (x^4+2 x^2+3\right )}-\frac {1}{4} \int \left (-20 x^4+68 x^2+\frac {303-152 x^2}{x^4+2 x^2+3}-76\right )dx^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {455 \arctan \left (\frac {x^2+1}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {20 x^6}{3}-34 x^4+76 x^2+76 \log \left (x^4+2 x^2+3\right )\right )+\frac {25 \left (5 x^2+3\right )}{4 \left (x^4+2 x^2+3\right )}\right )\) |
((25*(3 + 5*x^2))/(4*(3 + 2*x^2 + x^4)) + (76*x^2 - 34*x^4 + (20*x^6)/3 - (455*ArcTan[(1 + x^2)/Sqrt[2]])/Sqrt[2] + 76*Log[3 + 2*x^2 + x^4])/4)/2
3.2.1.3.1 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_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*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 [(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* (2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 2 - 4*a*c, 0] && LtQ[p, -1]
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 = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {5 x^{6}}{6}-\frac {17 x^{4}}{4}+\frac {19 x^{2}}{2}+\frac {\frac {125 x^{2}}{8}+\frac {75}{8}}{x^{4}+2 x^{2}+3}+\frac {19 \ln \left (x^{4}+2 x^{2}+3\right )}{2}-\frac {455 \arctan \left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2}\right ) \sqrt {2}}{16}\) | \(66\) |
default | \(\frac {5 x^{6}}{6}-\frac {17 x^{4}}{4}+\frac {19 x^{2}}{2}+\frac {\frac {125 x^{2}}{4}+\frac {75}{4}}{2 x^{4}+4 x^{2}+6}+\frac {19 \ln \left (x^{4}+2 x^{2}+3\right )}{2}-\frac {455 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}\) | \(69\) |
5/6*x^6-17/4*x^4+19/2*x^2+(125/8*x^2+75/8)/(x^4+2*x^2+3)+19/2*ln(x^4+2*x^2 +3)-455/16*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {40 \, x^{10} - 124 \, x^{8} + 168 \, x^{6} + 300 \, x^{4} - 1365 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 2118 \, x^{2} + 456 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 450}{48 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]
1/48*(40*x^10 - 124*x^8 + 168*x^6 + 300*x^4 - 1365*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 2118*x^2 + 456*(x^4 + 2*x^2 + 3)*log(x^ 4 + 2*x^2 + 3) + 450)/(x^4 + 2*x^2 + 3)
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5 x^{6}}{6} - \frac {17 x^{4}}{4} + \frac {19 x^{2}}{2} + \frac {125 x^{2} + 75}{8 x^{4} + 16 x^{2} + 24} + \frac {19 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{2} - \frac {455 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \]
5*x**6/6 - 17*x**4/4 + 19*x**2/2 + (125*x**2 + 75)/(8*x**4 + 16*x**2 + 24) + 19*log(x**4 + 2*x**2 + 3)/2 - 455*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2) /2)/16
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5}{6} \, x^{6} - \frac {17}{4} \, x^{4} + \frac {19}{2} \, x^{2} - \frac {455}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {25 \, {\left (5 \, x^{2} + 3\right )}}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {19}{2} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]
5/6*x^6 - 17/4*x^4 + 19/2*x^2 - 455/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1 )) + 25/8*(5*x^2 + 3)/(x^4 + 2*x^2 + 3) + 19/2*log(x^4 + 2*x^2 + 3)
Time = 0.43 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5}{6} \, x^{6} - \frac {17}{4} \, x^{4} + \frac {19}{2} \, x^{2} - \frac {455}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {76 \, x^{4} + 27 \, x^{2} + 153}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {19}{2} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]
5/6*x^6 - 17/4*x^4 + 19/2*x^2 - 455/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1 )) - 1/8*(76*x^4 + 27*x^2 + 153)/(x^4 + 2*x^2 + 3) + 19/2*log(x^4 + 2*x^2 + 3)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \frac {x^7 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {19\,\ln \left (x^4+2\,x^2+3\right )}{2}+\frac {\frac {125\,x^2}{8}+\frac {75}{8}}{x^4+2\,x^2+3}-\frac {455\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}+\frac {19\,x^2}{2}-\frac {17\,x^4}{4}+\frac {5\,x^6}{6} \]