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

Optimal. Leaf size=72 \[ -\frac {x \left (12 x^2+11\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac {x \left (217 x^2+335\right )}{16 \left (x^4+3 x^2+2\right )}-\frac {257}{8} \tan ^{-1}(x)+\frac {731 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}} \]

[Out]

-1/4*x*(12*x^2+11)/(x^4+3*x^2+2)^2+1/16*x*(217*x^2+335)/(x^4+3*x^2+2)-257/8*arctan(x)+731/32*arctan(1/2*x*2^(1
/2))*2^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1678, 1178, 1166, 203} \[ -\frac {x \left (12 x^2+11\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac {x \left (217 x^2+335\right )}{16 \left (x^4+3 x^2+2\right )}-\frac {257}{8} \tan ^{-1}(x)+\frac {731 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(11 + 12*x^2))/(4*(2 + 3*x^2 + x^4)^2) + (x*(335 + 217*x^2))/(16*(2 + 3*x^2 + x^4)) - (257*ArcTan[x])/8 +
(731*ArcTan[x/Sqrt[2]])/(16*Sqrt[2])

Rule 203

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

Rule 1166

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

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(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[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {4+x^2+3 x^4+5 x^6}{\left (2+3 x^2+x^4\right )^3} \, dx &=-\frac {x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {1}{8} \int \frac {-38+80 x^2}{\left (2+3 x^2+x^4\right )^2} \, dx\\ &=-\frac {x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}+\frac {1}{32} \int \frac {-594+434 x^2}{2+3 x^2+x^4} \, dx\\ &=-\frac {x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {257}{8} \int \frac {1}{1+x^2} \, dx+\frac {731}{16} \int \frac {1}{2+x^2} \, dx\\ &=-\frac {x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac {x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {257}{8} \tan ^{-1}(x)+\frac {731 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 56, normalized size = 0.78 \[ \frac {1}{32} \left (\frac {2 x \left (217 x^6+986 x^4+1391 x^2+626\right )}{\left (x^4+3 x^2+2\right )^2}-1028 \tan ^{-1}(x)+731 \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*x*(626 + 1391*x^2 + 986*x^4 + 217*x^6))/(2 + 3*x^2 + x^4)^2 - 1028*ArcTan[x] + 731*Sqrt[2]*ArcTan[x/Sqrt[2
]])/32

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fricas [A]  time = 0.90, size = 99, normalized size = 1.38 \[ \frac {434 \, x^{7} + 1972 \, x^{5} + 2782 \, x^{3} + 731 \, \sqrt {2} {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - 1028 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \relax (x) + 1252 \, x}{32 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(434*x^7 + 1972*x^5 + 2782*x^3 + 731*sqrt(2)*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(1/2*sqrt(2)*x) -
1028*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(x) + 1252*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)

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giac [A]  time = 0.33, size = 50, normalized size = 0.69 \[ \frac {731}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {217 \, x^{7} + 986 \, x^{5} + 1391 \, x^{3} + 626 \, x}{16 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} - \frac {257}{8} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

731/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(217*x^7 + 986*x^5 + 1391*x^3 + 626*x)/(x^4 + 3*x^2 + 2)^2 - 257/8
*arctan(x)

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maple [A]  time = 0.01, size = 53, normalized size = 0.74 \[ -\frac {257 \arctan \relax (x )}{8}+\frac {731 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{32}-\frac {-\frac {31}{8} x^{3}-\frac {33}{8} x}{\left (x^{2}+1\right )^{2}}+\frac {\frac {155}{16} x^{3}+\frac {181}{8} x}{\left (x^{2}+2\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-31/8*x^3-33/8*x)/(x^2+1)^2-257/8*arctan(x)+(155/16*x^3+181/8*x)/(x^2+2)^2+731/32*2^(1/2)*arctan(1/2*2^(1/2)
*x)

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maxima [A]  time = 1.67, size = 60, normalized size = 0.83 \[ \frac {731}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {217 \, x^{7} + 986 \, x^{5} + 1391 \, x^{3} + 626 \, x}{16 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} - \frac {257}{8} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

731/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(217*x^7 + 986*x^5 + 1391*x^3 + 626*x)/(x^8 + 6*x^6 + 13*x^4 + 12*
x^2 + 4) - 257/8*arctan(x)

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mupad [B]  time = 0.07, size = 59, normalized size = 0.82 \[ \frac {731\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{32}-\frac {257\,\mathrm {atan}\relax (x)}{8}+\frac {\frac {217\,x^7}{16}+\frac {493\,x^5}{8}+\frac {1391\,x^3}{16}+\frac {313\,x}{8}}{x^8+6\,x^6+13\,x^4+12\,x^2+4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(731*2^(1/2)*atan((2^(1/2)*x)/2))/32 - (257*atan(x))/8 + ((313*x)/8 + (1391*x^3)/16 + (493*x^5)/8 + (217*x^7)/
16)/(12*x^2 + 13*x^4 + 6*x^6 + x^8 + 4)

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sympy [A]  time = 0.25, size = 65, normalized size = 0.90 \[ \frac {217 x^{7} + 986 x^{5} + 1391 x^{3} + 626 x}{16 x^{8} + 96 x^{6} + 208 x^{4} + 192 x^{2} + 64} - \frac {257 \operatorname {atan}{\relax (x )}}{8} + \frac {731 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(217*x**7 + 986*x**5 + 1391*x**3 + 626*x)/(16*x**8 + 96*x**6 + 208*x**4 + 192*x**2 + 64) - 257*atan(x)/8 + 731
*sqrt(2)*atan(sqrt(2)*x/2)/32

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