3.1.89 \(\int \frac {4+x^2+3 x^4+5 x^6}{x^6 (2+3 x^2+x^4)^2} \, dx\) [89]

Optimal. Leaf size=69 \[ -\frac {1}{5 x^5}+\frac {11}{12 x^3}-\frac {23}{4 x}-\frac {x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {23}{2} \tan ^{-1}(x)+\frac {97 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}} \]

[Out]

-1/5/x^5+11/12/x^3-23/4/x-1/16*x*(-5*x^2+3)/(x^4+3*x^2+2)-23/2*arctan(x)+97/32*arctan(1/2*x*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1683, 1678, 209} \begin {gather*} -\frac {23 \text {ArcTan}(x)}{2}+\frac {97 \text {ArcTan}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}}-\frac {1}{5 x^5}+\frac {11}{12 x^3}-\frac {x \left (3-5 x^2\right )}{16 \left (x^4+3 x^2+2\right )}-\frac {23}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*1/x^5 + 11/(12*x^3) - 23/(4*x) - (x*(3 - 5*x^2))/(16*(2 + 3*x^2 + x^4)) - (23*ArcTan[x])/2 + (97*ArcTan[x
/Sqrt[2]])/(16*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*} \int \frac {4+x^2+3 x^4+5 x^6}{x^6 \left (2+3 x^2+x^4\right )^2} \, dx &=-\frac {x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {1}{4} \int \frac {-8+10 x^2-17 x^4+\frac {39 x^6}{4}-\frac {5 x^8}{4}}{x^6 \left (2+3 x^2+x^4\right )} \, dx\\ &=-\frac {x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {1}{4} \int \left (-\frac {4}{x^6}+\frac {11}{x^4}-\frac {23}{x^2}+\frac {46}{1+x^2}-\frac {97}{4 \left (2+x^2\right )}\right ) \, dx\\ &=-\frac {1}{5 x^5}+\frac {11}{12 x^3}-\frac {23}{4 x}-\frac {x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}+\frac {97}{16} \int \frac {1}{2+x^2} \, dx-\frac {23}{2} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {1}{5 x^5}+\frac {11}{12 x^3}-\frac {23}{4 x}-\frac {x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac {23}{2} \tan ^{-1}(x)+\frac {97 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 61, normalized size = 0.88 \begin {gather*} \frac {1}{480} \left (-\frac {96}{x^5}+\frac {440}{x^3}-\frac {2760}{x}+\frac {30 x \left (-3+5 x^2\right )}{2+3 x^2+x^4}-5520 \tan ^{-1}(x)+1455 \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-96/x^5 + 440/x^3 - 2760/x + (30*x*(-3 + 5*x^2))/(2 + 3*x^2 + x^4) - 5520*ArcTan[x] + 1455*Sqrt[2]*ArcTan[x/S
qrt[2]])/480

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Maple [A]
time = 0.04, size = 53, normalized size = 0.77

method result size
default \(-\frac {x}{2 \left (x^{2}+1\right )}-\frac {23 \arctan \left (x \right )}{2}-\frac {1}{5 x^{5}}+\frac {11}{12 x^{3}}-\frac {23}{4 x}+\frac {13 x}{16 \left (x^{2}+2\right )}+\frac {97 \arctan \left (\frac {\sqrt {2}\, x}{2}\right ) \sqrt {2}}{32}\) \(53\)
risch \(\frac {-\frac {87}{16} x^{8}-\frac {793}{48} x^{6}-\frac {179}{20} x^{4}+\frac {37}{30} x^{2}-\frac {2}{5}}{x^{5} \left (x^{4}+3 x^{2}+2\right )}+\frac {97 \arctan \left (\frac {\sqrt {2}\, x}{2}\right ) \sqrt {2}}{32}-\frac {23 \arctan \left (x \right )}{2}\) \(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2+1)-23/2*arctan(x)-1/5/x^5+11/12/x^3-23/4/x+13/16*x/(x^2+2)+97/32*arctan(1/2*2^(1/2)*x)*2^(1/2)

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Maxima [A]
time = 0.50, size = 57, normalized size = 0.83 \begin {gather*} \frac {97}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {1305 \, x^{8} + 3965 \, x^{6} + 2148 \, x^{4} - 296 \, x^{2} + 96}{240 \, {\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )}} - \frac {23}{2} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

97/32*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/240*(1305*x^8 + 3965*x^6 + 2148*x^4 - 296*x^2 + 96)/(x^9 + 3*x^7 + 2*x
^5) - 23/2*arctan(x)

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Fricas [A]
time = 0.38, size = 84, normalized size = 1.22 \begin {gather*} -\frac {2610 \, x^{8} + 7930 \, x^{6} + 4296 \, x^{4} - 1455 \, \sqrt {2} {\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - 592 \, x^{2} + 5520 \, {\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )} \arctan \left (x\right ) + 192}{480 \, {\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(2610*x^8 + 7930*x^6 + 4296*x^4 - 1455*sqrt(2)*(x^9 + 3*x^7 + 2*x^5)*arctan(1/2*sqrt(2)*x) - 592*x^2 +
5520*(x^9 + 3*x^7 + 2*x^5)*arctan(x) + 192)/(x^9 + 3*x^7 + 2*x^5)

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Sympy [A]
time = 0.10, size = 61, normalized size = 0.88 \begin {gather*} - \frac {23 \operatorname {atan}{\left (x \right )}}{2} + \frac {97 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{32} + \frac {- 1305 x^{8} - 3965 x^{6} - 2148 x^{4} + 296 x^{2} - 96}{240 x^{9} + 720 x^{7} + 480 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-23*atan(x)/2 + 97*sqrt(2)*atan(sqrt(2)*x/2)/32 + (-1305*x**8 - 3965*x**6 - 2148*x**4 + 296*x**2 - 96)/(240*x*
*9 + 720*x**7 + 480*x**5)

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Giac [A]
time = 5.11, size = 57, normalized size = 0.83 \begin {gather*} \frac {97}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {5 \, x^{3} - 3 \, x}{16 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac {345 \, x^{4} - 55 \, x^{2} + 12}{60 \, x^{5}} - \frac {23}{2} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

97/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(5*x^3 - 3*x)/(x^4 + 3*x^2 + 2) - 1/60*(345*x^4 - 55*x^2 + 12)/x^5
- 23/2*arctan(x)

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Mupad [B]
time = 0.92, size = 57, normalized size = 0.83 \begin {gather*} \frac {97\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{32}-\frac {23\,\mathrm {atan}\left (x\right )}{2}-\frac {\frac {87\,x^8}{16}+\frac {793\,x^6}{48}+\frac {179\,x^4}{20}-\frac {37\,x^2}{30}+\frac {2}{5}}{x^9+3\,x^7+2\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(97*2^(1/2)*atan((2^(1/2)*x)/2))/32 - (23*atan(x))/2 - ((179*x^4)/20 - (37*x^2)/30 + (793*x^6)/48 + (87*x^8)/1
6 + 2/5)/(2*x^5 + 3*x^7 + x^9)

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