Integrand size = 27, antiderivative size = 45 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}+\frac {9 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}} \] Output:
(-7*x^2+5)/(8*x^4+16*x^2+24)+9/16*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5-7 x^2}{8 \left (3+2 x^2+x^4\right )}+\frac {9 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}} \] Input:
Integrate[(-x + 2*x^3 + 4*x^5)/(3 + 2*x^2 + x^4)^2,x]
Output:
(5 - 7*x^2)/(8*(3 + 2*x^2 + x^4)) + (9*ArcTan[(1 + x^2)/Sqrt[2]])/(8*Sqrt[ 2])
Time = 0.43 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2028, 2194, 25, 2191, 27, 1083, 217}
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.
\(\displaystyle \int \frac {4 x^5+2 x^3-x}{\left (x^4+2 x^2+3\right )^2} \, dx\) |
\(\Big \downarrow \) 2028 |
\(\displaystyle \int \frac {x \left (4 x^4+2 x^2-1\right )}{\left (x^4+2 x^2+3\right )^2}dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int -\frac {-4 x^4-2 x^2+1}{\left (x^4+2 x^2+3\right )^2}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {-4 x^4-2 x^2+1}{\left (x^4+2 x^2+3\right )^2}dx^2\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{2} \left (\frac {5-7 x^2}{4 \left (x^4+2 x^2+3\right )}-\frac {1}{8} \int -\frac {18}{x^4+2 x^2+3}dx^2\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{4} \int \frac {1}{x^4+2 x^2+3}dx^2+\frac {5-7 x^2}{4 \left (x^4+2 x^2+3\right )}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {5-7 x^2}{4 \left (x^4+2 x^2+3\right )}-\frac {9}{2} \int \frac {1}{-x^4-8}d\left (2 x^2+2\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {9 \arctan \left (\frac {2 x^2+2}{2 \sqrt {2}}\right )}{4 \sqrt {2}}+\frac {5-7 x^2}{4 \left (x^4+2 x^2+3\right )}\right )\) |
Input:
Int[(-x + 2*x^3 + 4*x^5)/(3 + 2*x^2 + x^4)^2,x]
Output:
((5 - 7*x^2)/(4*(3 + 2*x^2 + x^4)) + (9*ArcTan[(2 + 2*x^2)/(2*Sqrt[2])])/( 4*Sqrt[2]))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ {a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] && !(E qQ[p, 1] && EqQ[u, 1])
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 = 38, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {-\frac {7 x^{2}}{8}+\frac {5}{8}}{x^{4}+2 x^{2}+3}+\frac {9 \arctan \left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2}\right ) \sqrt {2}}{16}\) | \(38\) |
default | \(\frac {-\frac {7 x^{2}}{4}+\frac {5}{4}}{2 x^{4}+4 x^{2}+6}+\frac {9 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}\) | \(41\) |
Input:
int((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x,method=_RETURNVERBOSE)
Output:
(-7/8*x^2+5/8)/(x^4+2*x^2+3)+9/16*arctan(1/2*(x^2+1)*2^(1/2))*2^(1/2)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {9 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - 14 \, x^{2} + 10}{16 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \] Input:
integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="fricas")
Output:
1/16*(9*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 14*x^2 + 10)/(x^4 + 2*x^2 + 3)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5 - 7 x^{2}}{8 x^{4} + 16 x^{2} + 24} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \] Input:
integrate((4*x**5+2*x**3-x)/(x**4+2*x**2+3)**2,x)
Output:
(5 - 7*x**2)/(8*x**4 + 16*x**2 + 24) + 9*sqrt(2)*atan(sqrt(2)*x**2/2 + sqr t(2)/2)/16
\[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\int { \frac {4 \, x^{5} + 2 \, x^{3} - x}{{\left (x^{4} + 2 \, x^{2} + 3\right )}^{2}} \,d x } \] Input:
integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="maxima")
Output:
-1/8*(7*x^2 - 5)/(x^4 + 2*x^2 + 3) + 9/4*integrate(x/(x^4 + 2*x^2 + 3), x)
Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {7 \, x^{2} - 5}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \] Input:
integrate((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x, algorithm="giac")
Output:
9/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 1/8*(7*x^2 - 5)/(x^4 + 2*x^2 + 3)
Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {9\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}-\frac {\frac {7\,x^2}{8}-\frac {5}{8}}{x^4+2\,x^2+3} \] Input:
int((2*x^3 - x + 4*x^5)/(2*x^2 + x^4 + 3)^2,x)
Output:
(9*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/16 - ((7*x^2)/8 - 5/8)/(2*x^ 2 + x^4 + 3)
Time = 0.14 (sec) , antiderivative size = 256, normalized size of antiderivative = 5.69 \[ \int \frac {-x+2 x^3+4 x^5}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {-9 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}-2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right ) x^{4}-18 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}-2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right ) x^{2}-27 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}-2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right )-9 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}+2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right ) x^{4}-18 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}+2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right ) x^{2}-27 \sqrt {\sqrt {3}+1}\, \sqrt {\sqrt {3}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {3}-1}\, \sqrt {2}+2 x}{\sqrt {\sqrt {3}+1}\, \sqrt {2}}\right )+7 x^{4}+31}{16 x^{4}+32 x^{2}+48} \] Input:
int((4*x^5+2*x^3-x)/(x^4+2*x^2+3)^2,x)
Output:
( - 9*sqrt(sqrt(3) + 1)*sqrt(sqrt(3) - 1)*atan((sqrt(sqrt(3) - 1)*sqrt(2) - 2*x)/(sqrt(sqrt(3) + 1)*sqrt(2)))*x**4 - 18*sqrt(sqrt(3) + 1)*sqrt(sqrt( 3) - 1)*atan((sqrt(sqrt(3) - 1)*sqrt(2) - 2*x)/(sqrt(sqrt(3) + 1)*sqrt(2)) )*x**2 - 27*sqrt(sqrt(3) + 1)*sqrt(sqrt(3) - 1)*atan((sqrt(sqrt(3) - 1)*sq rt(2) - 2*x)/(sqrt(sqrt(3) + 1)*sqrt(2))) - 9*sqrt(sqrt(3) + 1)*sqrt(sqrt( 3) - 1)*atan((sqrt(sqrt(3) - 1)*sqrt(2) + 2*x)/(sqrt(sqrt(3) + 1)*sqrt(2)) )*x**4 - 18*sqrt(sqrt(3) + 1)*sqrt(sqrt(3) - 1)*atan((sqrt(sqrt(3) - 1)*sq rt(2) + 2*x)/(sqrt(sqrt(3) + 1)*sqrt(2)))*x**2 - 27*sqrt(sqrt(3) + 1)*sqrt (sqrt(3) - 1)*atan((sqrt(sqrt(3) - 1)*sqrt(2) + 2*x)/(sqrt(sqrt(3) + 1)*sq rt(2))) + 7*x**4 + 31)/(16*(x**4 + 2*x**2 + 3))