Integrand size = 20, antiderivative size = 59 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac {1+2 x^2}{2 \left (1+2 x^2+2 x^4\right )}+\arctan \left (1+2 x^2\right ) \] Output:
1/16*(4*x^2+3)/(2*x^4+2*x^2+1)^2+(2*x^2+1)/(4*x^4+4*x^2+2)+arctan(2*x^2+1)
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {11+36 x^2+48 x^4+32 x^6}{16 \left (1+2 x^2+2 x^4\right )^2}+\arctan \left (1+2 x^2\right ) \] Input:
Integrate[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]
Output:
(11 + 36*x^2 + 48*x^4 + 32*x^6)/(16*(1 + 2*x^2 + 2*x^4)^2) + ArcTan[1 + 2* x^2]
Time = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2027, 2194, 2191, 27, 1086, 1082, 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.
\(\displaystyle \int \frac {x^5+x}{\left (2 x^4+2 x^2+1\right )^3} \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {x \left (x^4+1\right )}{\left (2 x^4+2 x^2+1\right )^3}dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int \frac {x^4+1}{\left (2 x^4+2 x^2+1\right )^3}dx^2\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \int \frac {16}{\left (2 x^4+2 x^2+1\right )^2}dx^2+\frac {4 x^2+3}{8 \left (2 x^4+2 x^2+1\right )^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{\left (2 x^4+2 x^2+1\right )^2}dx^2+\frac {4 x^2+3}{8 \left (2 x^4+2 x^2+1\right )^2}\right )\) |
\(\Big \downarrow \) 1086 |
\(\displaystyle \frac {1}{2} \left (2 \left (\int \frac {1}{2 x^4+2 x^2+1}dx^2+\frac {2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}\right )+\frac {4 x^2+3}{8 \left (2 x^4+2 x^2+1\right )^2}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (2 \left (\frac {2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}-\int \frac {1}{-x^4-1}d\left (2 x^2+1\right )\right )+\frac {4 x^2+3}{8 \left (2 x^4+2 x^2+1\right )^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (2 \left (\arctan \left (2 x^2+1\right )+\frac {2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}\right )+\frac {4 x^2+3}{8 \left (2 x^4+2 x^2+1\right )^2}\right )\) |
Input:
Int[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]
Output:
((3 + 4*x^2)/(8*(1 + 2*x^2 + 2*x^4)^2) + 2*((1 + 2*x^2)/(2*(1 + 2*x^2 + 2* x^4)) + ArcTan[1 + 2*x^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] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && ILtQ[p, -1]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[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 = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {2 x^{6}+3 x^{4}+\frac {9}{4} x^{2}+\frac {11}{16}}{\left (2 x^{4}+2 x^{2}+1\right )^{2}}+\arctan \left (2 x^{2}+1\right )\) | \(41\) |
risch | \(\frac {2 x^{6}+3 x^{4}+\frac {9}{4} x^{2}+\frac {11}{16}}{\left (2 x^{4}+2 x^{2}+1\right )^{2}}+\arctan \left (2 x^{2}+1\right )\) | \(43\) |
parallelrisch | \(-\frac {-64 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{6}-8 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right )-5-32 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{2}+8 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right )+24 x^{8}+32 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{2}-64 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{4}+16 x^{6}+64 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{4}+32 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{8}+64 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{6}-32 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{8}-12 x^{2}}{16 \left (2 x^{4}+2 x^{2}+1\right )^{2}}\) | \(168\) |
Input:
int((x^5+x)/(2*x^4+2*x^2+1)^3,x,method=_RETURNVERBOSE)
Output:
2*(x^6+3/2*x^4+9/8*x^2+11/32)/(2*x^4+2*x^2+1)^2+arctan(2*x^2+1)
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 16 \, {\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (2 \, x^{2} + 1\right ) + 11}{16 \, {\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )}} \] Input:
integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="fricas")
Output:
1/16*(32*x^6 + 48*x^4 + 36*x^2 + 16*(4*x^8 + 8*x^6 + 8*x^4 + 4*x^2 + 1)*ar ctan(2*x^2 + 1) + 11)/(4*x^8 + 8*x^6 + 8*x^4 + 4*x^2 + 1)
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 x^{6} + 48 x^{4} + 36 x^{2} + 11}{64 x^{8} + 128 x^{6} + 128 x^{4} + 64 x^{2} + 16} + \operatorname {atan}{\left (2 x^{2} + 1 \right )} \] Input:
integrate((x**5+x)/(2*x**4+2*x**2+1)**3,x)
Output:
(32*x**6 + 48*x**4 + 36*x**2 + 11)/(64*x**8 + 128*x**6 + 128*x**4 + 64*x** 2 + 16) + atan(2*x**2 + 1)
\[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\int { \frac {x^{5} + x}{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{3}} \,d x } \] Input:
integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="maxima")
Output:
1/16*(32*x^6 + 48*x^4 + 36*x^2 + 11)/(4*x^8 + 8*x^6 + 8*x^4 + 4*x^2 + 1) + 2*integrate(x/(2*x^4 + 2*x^2 + 1), x)
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 11}{16 \, {\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{2}} + \arctan \left (2 \, x^{2} + 1\right ) \] Input:
integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="giac")
Output:
1/16*(32*x^6 + 48*x^4 + 36*x^2 + 11)/(2*x^4 + 2*x^2 + 1)^2 + arctan(2*x^2 + 1)
Time = 22.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\mathrm {atan}\left (2\,x^2+1\right )+\frac {\frac {x^6}{2}+\frac {3\,x^4}{4}+\frac {9\,x^2}{16}+\frac {11}{64}}{x^8+2\,x^6+2\,x^4+x^2+\frac {1}{4}} \] Input:
int((x + x^5)/(2*x^2 + 2*x^4 + 1)^3,x)
Output:
atan(2*x^2 + 1) + ((9*x^2)/16 + (3*x^4)/4 + x^6/2 + 11/64)/(x^2 + 2*x^4 + 2*x^6 + x^8 + 1/4)
Time = 0.16 (sec) , antiderivative size = 366, normalized size of antiderivative = 6.20 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {-64 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}-2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{8}-128 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}-2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{6}-128 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}-2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{4}-64 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}-2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{2}-16 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}-2 x}{\sqrt {\sqrt {2}+1}}\right )-64 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}+2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{8}-128 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}+2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{6}-128 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}+2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{4}-64 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}+2 x}{\sqrt {\sqrt {2}+1}}\right ) x^{2}-16 \sqrt {\sqrt {2}+1}\, \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}-1}+2 x}{\sqrt {\sqrt {2}+1}}\right )-16 x^{8}+16 x^{4}+20 x^{2}+7}{64 x^{8}+128 x^{6}+128 x^{4}+64 x^{2}+16} \] Input:
int((x^5+x)/(2*x^4+2*x^2+1)^3,x)
Output:
( - 64*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) - 2*x)/ sqrt(sqrt(2) + 1))*x**8 - 128*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan((sq rt(sqrt(2) - 1) - 2*x)/sqrt(sqrt(2) + 1))*x**6 - 128*sqrt(sqrt(2) + 1)*sqr t(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) - 2*x)/sqrt(sqrt(2) + 1))*x**4 - 64 *sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) - 2*x)/sqrt(s qrt(2) + 1))*x**2 - 16*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan((sqrt(sqrt (2) - 1) - 2*x)/sqrt(sqrt(2) + 1)) - 64*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1 )*atan((sqrt(sqrt(2) - 1) + 2*x)/sqrt(sqrt(2) + 1))*x**8 - 128*sqrt(sqrt(2 ) + 1)*sqrt(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) + 2*x)/sqrt(sqrt(2) + 1)) *x**6 - 128*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) + 2*x)/sqrt(sqrt(2) + 1))*x**4 - 64*sqrt(sqrt(2) + 1)*sqrt(sqrt(2) - 1)*atan ((sqrt(sqrt(2) - 1) + 2*x)/sqrt(sqrt(2) + 1))*x**2 - 16*sqrt(sqrt(2) + 1)* sqrt(sqrt(2) - 1)*atan((sqrt(sqrt(2) - 1) + 2*x)/sqrt(sqrt(2) + 1)) - 16*x **8 + 16*x**4 + 20*x**2 + 7)/(16*(4*x**8 + 8*x**6 + 8*x**4 + 4*x**2 + 1))