### 3.107 $$\int \frac{4+x^2+3 x^4+5 x^6}{x^5 (3+2 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=80 $\frac{25 \left (5 x^2+7\right )}{216 \left (x^4+2 x^2+3\right )}+\frac{13}{54 x^2}-\frac{1}{9 x^4}-\frac{13}{108} \log \left (x^4+2 x^2+3\right )+\frac{125 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{216 \sqrt{2}}+\frac{13 \log (x)}{27}$

[Out]

-1/(9*x^4) + 13/(54*x^2) + (25*(7 + 5*x^2))/(216*(3 + 2*x^2 + x^4)) + (125*ArcTan[(1 + x^2)/Sqrt[2]])/(216*Sqr
t[2]) + (13*Log[x])/27 - (13*Log[3 + 2*x^2 + x^4])/108

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Rubi [A]  time = 0.136748, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.226, Rules used = {1663, 1646, 1628, 634, 618, 204, 628} $\frac{25 \left (5 x^2+7\right )}{216 \left (x^4+2 x^2+3\right )}+\frac{13}{54 x^2}-\frac{1}{9 x^4}-\frac{13}{108} \log \left (x^4+2 x^2+3\right )+\frac{125 \tan ^{-1}\left (\frac{x^2+1}{\sqrt{2}}\right )}{216 \sqrt{2}}+\frac{13 \log (x)}{27}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(9*x^4) + 13/(54*x^2) + (25*(7 + 5*x^2))/(216*(3 + 2*x^2 + x^4)) + (125*ArcTan[(1 + x^2)/Sqrt[2]])/(216*Sqr
t[2]) + (13*Log[x])/27 - (13*Log[3 + 2*x^2 + x^4])/108

Rule 1663

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]

Rule 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*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[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1628

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

Rule 634

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

Rule 618

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

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{4+x^2+3 x^4+5 x^6}{x^5 \left (3+2 x^2+x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{4+x+3 x^2+5 x^3}{x^3 \left (3+2 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{\frac{32}{3}-\frac{40 x}{9}+\frac{200 x^2}{27}+\frac{250 x^3}{27}}{x^3 \left (3+2 x+x^2\right )} \, dx,x,x^2\right )\\ &=\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \left (\frac{32}{9 x^3}-\frac{104}{27 x^2}+\frac{104}{27 x}-\frac{2 (-73+52 x)}{27 \left (3+2 x+x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{1}{9 x^4}+\frac{13}{54 x^2}+\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{13 \log (x)}{27}-\frac{1}{216} \operatorname{Subst}\left (\int \frac{-73+52 x}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{9 x^4}+\frac{13}{54 x^2}+\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{13 \log (x)}{27}-\frac{13}{108} \operatorname{Subst}\left (\int \frac{2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )+\frac{125}{216} \operatorname{Subst}\left (\int \frac{1}{3+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{9 x^4}+\frac{13}{54 x^2}+\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{13 \log (x)}{27}-\frac{13}{108} \log \left (3+2 x^2+x^4\right )-\frac{125}{108} \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=-\frac{1}{9 x^4}+\frac{13}{54 x^2}+\frac{25 \left (7+5 x^2\right )}{216 \left (3+2 x^2+x^4\right )}+\frac{125 \tan ^{-1}\left (\frac{1+x^2}{\sqrt{2}}\right )}{216 \sqrt{2}}+\frac{13 \log (x)}{27}-\frac{13}{108} \log \left (3+2 x^2+x^4\right )\\ \end{align*}

Mathematica [C]  time = 0.0613122, size = 105, normalized size = 1.31 $\frac{1}{864} \left (\frac{100 \left (5 x^2+7\right )}{x^4+2 x^2+3}+\frac{208}{x^2}-\frac{96}{x^4}-\sqrt{2} \left (52 \sqrt{2}+125 i\right ) \log \left (x^2-i \sqrt{2}+1\right )+\sqrt{2} \left (-52 \sqrt{2}+125 i\right ) \log \left (x^2+i \sqrt{2}+1\right )+416 \log (x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-96/x^4 + 208/x^2 + (100*(7 + 5*x^2))/(3 + 2*x^2 + x^4) + 416*Log[x] - Sqrt[2]*(125*I + 52*Sqrt[2])*Log[1 - I
*Sqrt[2] + x^2] + Sqrt[2]*(125*I - 52*Sqrt[2])*Log[1 + I*Sqrt[2] + x^2])/864

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Maple [A]  time = 0.012, size = 68, normalized size = 0.9 \begin{align*} -{\frac{1}{54\,{x}^{4}+108\,{x}^{2}+162} \left ( -{\frac{125\,{x}^{2}}{4}}-{\frac{175}{4}} \right ) }-{\frac{13\,\ln \left ({x}^{4}+2\,{x}^{2}+3 \right ) }{108}}+{\frac{125\,\sqrt{2}}{432}\arctan \left ({\frac{ \left ( 2\,{x}^{2}+2 \right ) \sqrt{2}}{4}} \right ) }-{\frac{1}{9\,{x}^{4}}}+{\frac{13}{54\,{x}^{2}}}+{\frac{13\,\ln \left ( x \right ) }{27}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*(-125/4*x^2-175/4)/(x^4+2*x^2+3)-13/108*ln(x^4+2*x^2+3)+125/432*2^(1/2)*arctan(1/4*(2*x^2+2)*2^(1/2))-1/
9/x^4+13/54/x^2+13/27*ln(x)

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Maxima [A]  time = 1.4745, size = 96, normalized size = 1.2 \begin{align*} \frac{125}{432} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + \frac{59 \, x^{6} + 85 \, x^{4} + 36 \, x^{2} - 24}{72 \,{\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )}} - \frac{13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac{13}{54} \, \log \left (x^{2}\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^5/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

125/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/72*(59*x^6 + 85*x^4 + 36*x^2 - 24)/(x^8 + 2*x^6 + 3*x^4) - 1
3/108*log(x^4 + 2*x^2 + 3) + 13/54*log(x^2)

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Fricas [A]  time = 1.49019, size = 289, normalized size = 3.61 \begin{align*} \frac{354 \, x^{6} + 510 \, x^{4} + 125 \, \sqrt{2}{\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + 216 \, x^{2} - 52 \,{\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) + 208 \,{\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\right )} \log \left (x\right ) - 144}{432 \,{\left (x^{8} + 2 \, x^{6} + 3 \, x^{4}\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^5/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

1/432*(354*x^6 + 510*x^4 + 125*sqrt(2)*(x^8 + 2*x^6 + 3*x^4)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 216*x^2 - 52*(x^8
+ 2*x^6 + 3*x^4)*log(x^4 + 2*x^2 + 3) + 208*(x^8 + 2*x^6 + 3*x^4)*log(x) - 144)/(x^8 + 2*x^6 + 3*x^4)

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Sympy [A]  time = 0.214274, size = 80, normalized size = 1. \begin{align*} \frac{13 \log{\left (x \right )}}{27} - \frac{13 \log{\left (x^{4} + 2 x^{2} + 3 \right )}}{108} + \frac{125 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x^{2}}{2} + \frac{\sqrt{2}}{2} \right )}}{432} + \frac{59 x^{6} + 85 x^{4} + 36 x^{2} - 24}{72 x^{8} + 144 x^{6} + 216 x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13*log(x)/27 - 13*log(x**4 + 2*x**2 + 3)/108 + 125*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)/2)/432 + (59*x**6 + 8
5*x**4 + 36*x**2 - 24)/(72*x**8 + 144*x**6 + 216*x**4)

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Giac [A]  time = 1.08324, size = 107, normalized size = 1.34 \begin{align*} \frac{125}{432} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x^{2} + 1\right )}\right ) + \frac{26 \, x^{4} + 177 \, x^{2} + 253}{216 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac{39 \, x^{4} - 26 \, x^{2} + 12}{108 \, x^{4}} - \frac{13}{108} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) + \frac{13}{54} \, \log \left (x^{2}\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^5/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

125/432*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/216*(26*x^4 + 177*x^2 + 253)/(x^4 + 2*x^2 + 3) - 1/108*(39*x
^4 - 26*x^2 + 12)/x^4 - 13/108*log(x^4 + 2*x^2 + 3) + 13/54*log(x^2)