3.90 \(\int \frac{4+x^2+3 x^4+5 x^6}{x^8 (2+3 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=76 \[ \frac{x \left (3 x^2+19\right )}{32 \left (x^4+3 x^2+2\right )}-\frac{23}{12 x^3}+\frac{11}{20 x^5}-\frac{1}{7 x^7}+\frac{137}{16 x}+\frac{25}{2} \tan ^{-1}(x)-\frac{123 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}} \]

[Out]

-1/(7*x^7) + 11/(20*x^5) - 23/(12*x^3) + 137/(16*x) + (x*(19 + 3*x^2))/(32*(2 + 3*x^2 + x^4)) + (25*ArcTan[x])
/2 - (123*ArcTan[x/Sqrt[2]])/(32*Sqrt[2])

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Rubi [A]  time = 0.0999244, antiderivative size = 76, normalized size of antiderivative = 1., 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 = {1669, 1664, 203} \[ \frac{x \left (3 x^2+19\right )}{32 \left (x^4+3 x^2+2\right )}-\frac{23}{12 x^3}+\frac{11}{20 x^5}-\frac{1}{7 x^7}+\frac{137}{16 x}+\frac{25}{2} \tan ^{-1}(x)-\frac{123 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(4 + x^2 + 3*x^4 + 5*x^6)/(x^8*(2 + 3*x^2 + x^4)^2),x]

[Out]

-1/(7*x^7) + 11/(20*x^5) - 23/(12*x^3) + 137/(16*x) + (x*(19 + 3*x^2))/(32*(2 + 3*x^2 + x^4)) + (25*ArcTan[x])
/2 - (123*ArcTan[x/Sqrt[2]])/(32*Sqrt[2])

Rule 1669

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]

Rule 1664

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 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])

Rubi steps

\begin{align*} \int \frac{4+x^2+3 x^4+5 x^6}{x^8 \left (2+3 x^2+x^4\right )^2} \, dx &=\frac{x \left (19+3 x^2\right )}{32 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{-8+10 x^2-17 x^4+\frac{21 x^6}{2}-\frac{39 x^8}{8}-\frac{3 x^{10}}{8}}{x^8 \left (2+3 x^2+x^4\right )} \, dx\\ &=\frac{x \left (19+3 x^2\right )}{32 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-\frac{4}{x^8}+\frac{11}{x^6}-\frac{23}{x^4}+\frac{137}{4 x^2}-\frac{50}{1+x^2}+\frac{123}{8 \left (2+x^2\right )}\right ) \, dx\\ &=-\frac{1}{7 x^7}+\frac{11}{20 x^5}-\frac{23}{12 x^3}+\frac{137}{16 x}+\frac{x \left (19+3 x^2\right )}{32 \left (2+3 x^2+x^4\right )}-\frac{123}{32} \int \frac{1}{2+x^2} \, dx+\frac{25}{2} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{7 x^7}+\frac{11}{20 x^5}-\frac{23}{12 x^3}+\frac{137}{16 x}+\frac{x \left (19+3 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac{25}{2} \tan ^{-1}(x)-\frac{123 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.057174, size = 77, normalized size = 1.01 \[ \frac{3 x^3+19 x}{32 \left (x^4+3 x^2+2\right )}-\frac{23}{12 x^3}+\frac{11}{20 x^5}-\frac{1}{7 x^7}+\frac{137}{16 x}+\frac{25}{2} \tan ^{-1}(x)-\frac{123 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + x^2 + 3*x^4 + 5*x^6)/(x^8*(2 + 3*x^2 + x^4)^2),x]

[Out]

-1/(7*x^7) + 11/(20*x^5) - 23/(12*x^3) + 137/(16*x) + (19*x + 3*x^3)/(32*(2 + 3*x^2 + x^4)) + (25*ArcTan[x])/2
- (123*ArcTan[x/Sqrt[2]])/(32*Sqrt[2])

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Maple [A]  time = 0.013, size = 58, normalized size = 0.8 \begin{align*} -{\frac{13\,x}{32\,{x}^{2}+64}}-{\frac{123\,\sqrt{2}}{64}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }+{\frac{x}{2\,{x}^{2}+2}}+{\frac{25\,\arctan \left ( x \right ) }{2}}-{\frac{1}{7\,{x}^{7}}}+{\frac{11}{20\,{x}^{5}}}-{\frac{23}{12\,{x}^{3}}}+{\frac{137}{16\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^6+3*x^4+x^2+4)/x^8/(x^4+3*x^2+2)^2,x)

[Out]

-13/32*x/(x^2+2)-123/64*arctan(1/2*x*2^(1/2))*2^(1/2)+1/2*x/(x^2+1)+25/2*arctan(x)-1/7/x^7+11/20/x^5-23/12/x^3
+137/16/x

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Maxima [A]  time = 1.53116, size = 84, normalized size = 1.11 \begin{align*} -\frac{123}{64} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{29085 \, x^{10} + 81865 \, x^{8} + 40068 \, x^{6} - 7816 \, x^{4} + 2256 \, x^{2} - 960}{3360 \,{\left (x^{11} + 3 \, x^{9} + 2 \, x^{7}\right )}} + \frac{25}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^8/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

-123/64*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/3360*(29085*x^10 + 81865*x^8 + 40068*x^6 - 7816*x^4 + 2256*x^2 - 960
)/(x^11 + 3*x^9 + 2*x^7) + 25/2*arctan(x)

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Fricas [A]  time = 1.56454, size = 271, normalized size = 3.57 \begin{align*} \frac{58170 \, x^{10} + 163730 \, x^{8} + 80136 \, x^{6} - 15632 \, x^{4} - 12915 \, \sqrt{2}{\left (x^{11} + 3 \, x^{9} + 2 \, x^{7}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 4512 \, x^{2} + 84000 \,{\left (x^{11} + 3 \, x^{9} + 2 \, x^{7}\right )} \arctan \left (x\right ) - 1920}{6720 \,{\left (x^{11} + 3 \, x^{9} + 2 \, x^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^8/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/6720*(58170*x^10 + 163730*x^8 + 80136*x^6 - 15632*x^4 - 12915*sqrt(2)*(x^11 + 3*x^9 + 2*x^7)*arctan(1/2*sqrt
(2)*x) + 4512*x^2 + 84000*(x^11 + 3*x^9 + 2*x^7)*arctan(x) - 1920)/(x^11 + 3*x^9 + 2*x^7)

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Sympy [A]  time = 0.263308, size = 66, normalized size = 0.87 \begin{align*} \frac{25 \operatorname{atan}{\left (x \right )}}{2} - \frac{123 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{64} + \frac{29085 x^{10} + 81865 x^{8} + 40068 x^{6} - 7816 x^{4} + 2256 x^{2} - 960}{3360 x^{11} + 10080 x^{9} + 6720 x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**6+3*x**4+x**2+4)/x**8/(x**4+3*x**2+2)**2,x)

[Out]

25*atan(x)/2 - 123*sqrt(2)*atan(sqrt(2)*x/2)/64 + (29085*x**10 + 81865*x**8 + 40068*x**6 - 7816*x**4 + 2256*x*
*2 - 960)/(3360*x**11 + 10080*x**9 + 6720*x**7)

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Giac [A]  time = 1.07734, size = 84, normalized size = 1.11 \begin{align*} -\frac{123}{64} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{3 \, x^{3} + 19 \, x}{32 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + \frac{14385 \, x^{6} - 3220 \, x^{4} + 924 \, x^{2} - 240}{1680 \, x^{7}} + \frac{25}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^8/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

-123/64*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/32*(3*x^3 + 19*x)/(x^4 + 3*x^2 + 2) + 1/1680*(14385*x^6 - 3220*x^4 +
924*x^2 - 240)/x^7 + 25/2*arctan(x)