\(\int \frac {x+x^5}{(1+2 x^2+2 x^4)^3} \, dx\) [336]

Optimal result

Integrand size = 20, antiderivative size = 59 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac {1+2 x^2}{2 \left (1+2 x^2+2 x^4\right )}+\arctan \left (1+2 x^2\right ) \]

[Out]

1/16*(4*x^2+3)/(2*x^4+2*x^2+1)^2+1/2*(2*x^2+1)/(2*x^4+2*x^2+1)+arctan(2*x^2+1)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1607, 1677, 1674, 12, 628, 631, 210} \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\arctan \left (2 x^2+1\right )+\frac {2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}+\frac {4 x^2+3}{16 \left (2 x^4+2 x^2+1\right )^2} \]

[In]

Int[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]

[Out]

(3 + 4*x^2)/(16*(1 + 2*x^2 + 2*x^4)^2) + (1 + 2*x^2)/(2*(1 + 2*x^2 + 2*x^4)) + ArcTan[1 + 2*x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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])

Rule 628

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] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1674

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[Pq, 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] + Dist[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
]

Rule 1677

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[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] && Inte
gerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (1+x^4\right )}{\left (1+2 x^2+2 x^4\right )^3} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{\left (1+2 x+2 x^2\right )^3} \, dx,x,x^2\right ) \\ & = \frac {3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac {1}{16} \text {Subst}\left (\int \frac {16}{\left (1+2 x+2 x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\text {Subst}\left (\int \frac {1}{\left (1+2 x+2 x^2\right )^2} \, dx,x,x^2\right ) \\ & = \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 )}+\text {Subst}\left (\int \frac {1}{1+2 x+2 x^2} \, dx,x,x^2\right ) \\ & = \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 )}-\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+2 x^2\right ) \\ & = \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 )}+\tan ^{-1}\left (1+2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {11+36 x^2+48 x^4+32 x^6}{16 \left (1+2 x^2+2 x^4\right )^2}+\arctan \left (1+2 x^2\right ) \]

[In]

Integrate[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]

[Out]

(11 + 36*x^2 + 48*x^4 + 32*x^6)/(16*(1 + 2*x^2 + 2*x^4)^2) + ArcTan[1 + 2*x^2]

Maple [A] (verified)

Time = 0.12 (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	\(-\frac {-8 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right )+8 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right )-5+64 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{4}-32 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{8}+24 x^{8}+32 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{2}+64 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{6}+16 x^{6}+32 i \ln \left (x^{2}+\frac {1}{2}-\frac {i}{2}\right ) x^{8}-64 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{4}-64 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{6}-32 i \ln \left (x^{2}+\frac {1}{2}+\frac {i}{2}\right ) x^{2}-12 x^{2}}{16 \left (2 x^{4}+2 x^{2}+1\right )^{2}}\)	\(168\)

[In]

int((x^5+x)/(2*x^4+2*x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 16 \, {\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (2 \, x^{2} + 1\right ) + 11}{16 \, {\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )}} \]

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="fricas")

[Out]

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)*arctan(2*x^2 + 1) + 11)/(4*x^8 + 8*x^6
+ 8*x^4 + 4*x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 x^{6} + 48 x^{4} + 36 x^{2} + 11}{64 x^{8} + 128 x^{6} + 128 x^{4} + 64 x^{2} + 16} + \operatorname {atan}{\left (2 x^{2} + 1 \right )} \]

[In]

integrate((x**5+x)/(2*x**4+2*x**2+1)**3,x)

[Out]

(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)

Maxima [F]

\[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\int { \frac {x^{5} + x}{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{3}} \,d x } \]

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.74 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\frac {32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 11}{16 \, {\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{2}} + \arctan \left (2 \, x^{2} + 1\right ) \]

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx=\mathrm {atan}\left (2\,x^2+1\right )+\frac {\frac {x^6}{2}+\frac {3\,x^4}{4}+\frac {9\,x^2}{16}+\frac {11}{64}}{x^8+2\,x^6+2\,x^4+x^2+\frac {1}{4}} \]

[In]

int((x + x^5)/(2*x^2 + 2*x^4 + 1)^3,x)

[Out]

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)