\(\int \frac {4+x^2+3 x^4+5 x^6}{x^4 (2+3 x^2+x^4)^3} \, dx\) [98]

Optimal result

Integrand size = 31, antiderivative size = 86 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {1}{6 x^3}+\frac {17}{8 x}-\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}-\frac {113 \arctan (x)}{8}+\frac {1611 \arctan \left (\frac {x}{\sqrt {2}}\right )}{64 \sqrt {2}} \]

[Out]

-1/6/x^3+17/8/x-1/16*x*(9*x^2+5)/(x^4+3*x^2+2)^2+1/64*x*(571*x^2+951)/(x^4+3*x^2+2)-113/8*arctan(x)+1611/128*a
rctan(1/2*x*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1683, 1678, 209} \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=-\frac {113 \arctan (x)}{8}+\frac {1611 \arctan \left (\frac {x}{\sqrt {2}}\right )}{64 \sqrt {2}}-\frac {1}{6 x^3}-\frac {x \left (9 x^2+5\right )}{16 \left (x^4+3 x^2+2\right )^2}+\frac {x \left (571 x^2+951\right )}{64 \left (x^4+3 x^2+2\right )}+\frac {17}{8 x} \]

[In]

Int[(4 + x^2 + 3*x^4 + 5*x^6)/(x^4*(2 + 3*x^2 + x^4)^3),x]

[Out]

-1/6*1/x^3 + 17/(8*x) - (x*(5 + 9*x^2))/(16*(2 + 3*x^2 + x^4)^2) + (x*(951 + 571*x^2))/(64*(2 + 3*x^2 + x^4))
- (113*ArcTan[x])/8 + (1611*ArcTan[x/Sqrt[2]])/(64*Sqrt[2])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1678

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x
)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]

Rule 1683

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainde
r[x^m*Pq, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 2]}, S
imp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))
), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[x^m*(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[(2*a*(p + 1)*(b^2
- 4*a*c)*PolynomialQuotient[x^m*Pq, a + b*x^2 + c*x^4, x])/x^m + (b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e)
/x^m + c*(4*p + 7)*(b*d - 2*a*e)*x^(2 - m), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && GtQ[Expon[
Pq, x^2], 1] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && ILtQ[m/2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}-\frac {1}{8} \int \frac {-16+20 x^2-\frac {73 x^4}{2}+\frac {45 x^6}{2}}{x^4 \left (2+3 x^2+x^4\right )^2} \, dx \\ & = -\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}+\frac {1}{32} \int \frac {32-88 x^2-\frac {573 x^4}{2}+\frac {571 x^6}{2}}{x^4 \left (2+3 x^2+x^4\right )} \, dx \\ & = -\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}+\frac {1}{32} \int \left (\frac {16}{x^4}-\frac {68}{x^2}-\frac {452}{1+x^2}+\frac {1611}{2 \left (2+x^2\right )}\right ) \, dx \\ & = -\frac {1}{6 x^3}+\frac {17}{8 x}-\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}-\frac {113}{8} \int \frac {1}{1+x^2} \, dx+\frac {1611}{64} \int \frac {1}{2+x^2} \, dx \\ & = -\frac {1}{6 x^3}+\frac {17}{8 x}-\frac {x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}-\frac {113}{8} \tan ^{-1}(x)+\frac {1611 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{64 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {1}{384} \left (-\frac {64}{x^3}+\frac {816}{x}-\frac {24 x \left (5+9 x^2\right )}{\left (2+3 x^2+x^4\right )^2}+\frac {6 x \left (951+571 x^2\right )}{2+3 x^2+x^4}-5424 \arctan (x)+4833 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )\right ) \]

[In]

Integrate[(4 + x^2 + 3*x^4 + 5*x^6)/(x^4*(2 + 3*x^2 + x^4)^3),x]

[Out]

(-64/x^3 + 816/x - (24*x*(5 + 9*x^2))/(2 + 3*x^2 + x^4)^2 + (6*x*(951 + 571*x^2))/(2 + 3*x^2 + x^4) - 5424*Arc
Tan[x] + 4833*Sqrt[2]*ArcTan[x/Sqrt[2]])/384

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\frac {707}{64} x^{10}+\frac {1301}{24} x^{8}+\frac {5663}{64} x^{6}+\frac {5063}{96} x^{4}+\frac {13}{2} x^{2}-\frac {2}{3}}{x^{3} \left (x^{4}+3 x^{2}+2\right )^{2}}+\frac {1611 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{128}-\frac {113 \arctan \left (x \right )}{8}\)	\(61\)
default	\(\frac {\frac {259}{8} x^{3}+\frac {285}{4} x}{8 \left (x^{2}+2\right )^{2}}+\frac {1611 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{128}-\frac {1}{6 x^{3}}+\frac {17}{8 x}-\frac {-\frac {39}{8} x^{3}-\frac {41}{8} x}{\left (x^{2}+1\right )^{2}}-\frac {113 \arctan \left (x \right )}{8}\)	\(64\)

[In]

int((5*x^6+3*x^4+x^2+4)/x^4/(x^4+3*x^2+2)^3,x,method=_RETURNVERBOSE)

[Out]

(707/64*x^10+1301/24*x^8+5663/64*x^6+5063/96*x^4+13/2*x^2-2/3)/x^3/(x^4+3*x^2+2)^2+1611/128*arctan(1/2*x*2^(1/
2))*2^(1/2)-113/8*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {4242 \, x^{10} + 20816 \, x^{8} + 33978 \, x^{6} + 20252 \, x^{4} + 4833 \, \sqrt {2} {\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 2496 \, x^{2} - 5424 \, {\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )} \arctan \left (x\right ) - 256}{384 \, {\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )}} \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^4/(x^4+3*x^2+2)^3,x, algorithm="fricas")

[Out]

1/384*(4242*x^10 + 20816*x^8 + 33978*x^6 + 20252*x^4 + 4833*sqrt(2)*(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3)*a
rctan(1/2*sqrt(2)*x) + 2496*x^2 - 5424*(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3)*arctan(x) - 256)/(x^11 + 6*x^9
+ 13*x^7 + 12*x^5 + 4*x^3)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=- \frac {113 \operatorname {atan}{\left (x \right )}}{8} + \frac {1611 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{128} + \frac {2121 x^{10} + 10408 x^{8} + 16989 x^{6} + 10126 x^{4} + 1248 x^{2} - 128}{192 x^{11} + 1152 x^{9} + 2496 x^{7} + 2304 x^{5} + 768 x^{3}} \]

[In]

integrate((5*x**6+3*x**4+x**2+4)/x**4/(x**4+3*x**2+2)**3,x)

[Out]

-113*atan(x)/8 + 1611*sqrt(2)*atan(sqrt(2)*x/2)/128 + (2121*x**10 + 10408*x**8 + 16989*x**6 + 10126*x**4 + 124
8*x**2 - 128)/(192*x**11 + 1152*x**9 + 2496*x**7 + 2304*x**5 + 768*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {1611}{128} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {2121 \, x^{10} + 10408 \, x^{8} + 16989 \, x^{6} + 10126 \, x^{4} + 1248 \, x^{2} - 128}{192 \, {\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )}} - \frac {113}{8} \, \arctan \left (x\right ) \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^4/(x^4+3*x^2+2)^3,x, algorithm="maxima")

[Out]

1611/128*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/192*(2121*x^10 + 10408*x^8 + 16989*x^6 + 10126*x^4 + 1248*x^2 - 128
)/(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3) - 113/8*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {1611}{128} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {571 \, x^{7} + 2664 \, x^{5} + 3959 \, x^{3} + 1882 \, x}{64 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} + \frac {51 \, x^{2} - 4}{24 \, x^{3}} - \frac {113}{8} \, \arctan \left (x\right ) \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^4/(x^4+3*x^2+2)^3,x, algorithm="giac")

[Out]

1611/128*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/64*(571*x^7 + 2664*x^5 + 3959*x^3 + 1882*x)/(x^4 + 3*x^2 + 2)^2 + 1
/24*(51*x^2 - 4)/x^3 - 113/8*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^4 \left (2+3 x^2+x^4\right )^3} \, dx=\frac {\frac {707\,x^{10}}{64}+\frac {1301\,x^8}{24}+\frac {5663\,x^6}{64}+\frac {5063\,x^4}{96}+\frac {13\,x^2}{2}-\frac {2}{3}}{x^{11}+6\,x^9+13\,x^7+12\,x^5+4\,x^3}-\frac {113\,\mathrm {atan}\left (x\right )}{8}+\frac {1611\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{128} \]

[In]

int((x^2 + 3*x^4 + 5*x^6 + 4)/(x^4*(3*x^2 + x^4 + 2)^3),x)

[Out]

((13*x^2)/2 + (5063*x^4)/96 + (5663*x^6)/64 + (1301*x^8)/24 + (707*x^10)/64 - 2/3)/(4*x^3 + 12*x^5 + 13*x^7 +
6*x^9 + x^11) - (113*atan(x))/8 + (1611*2^(1/2)*atan((2^(1/2)*x)/2))/128