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

Optimal. Leaf size=49 \[ \frac{\left (25 x^2+24\right ) x}{2 \left (x^4+3 x^2+2\right )}+5 x-\frac{15}{2} \tan ^{-1}(x)-\frac{7 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{\sqrt{2}} \]

[Out]

5*x + (x*(24 + 25*x^2))/(2*(2 + 3*x^2 + x^4)) - (15*ArcTan[x])/2 - (7*ArcTan[x/Sqrt[2]])/Sqrt[2]

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Rubi [A]  time = 0.0663581, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1668, 1676, 1166, 203} \[ \frac{\left (25 x^2+24\right ) x}{2 \left (x^4+3 x^2+2\right )}+5 x-\frac{15}{2} \tan ^{-1}(x)-\frac{7 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

5*x + (x*(24 + 25*x^2))/(2*(2 + 3*x^2 + x^4)) - (15*ArcTan[x])/2 - (7*ArcTan[x/Sqrt[2]])/Sqrt[2]

Rule 1668

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[(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] + 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] && GtQ[Expon[Pq, x^2], 1] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[m/2, 0]

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

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 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{x^2 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx &=\frac{x \left (24+25 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{48-2 x^2-20 x^4}{2+3 x^2+x^4} \, dx\\ &=\frac{x \left (24+25 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-20+\frac{2 \left (44+29 x^2\right )}{2+3 x^2+x^4}\right ) \, dx\\ &=5 x+\frac{x \left (24+25 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \int \frac{44+29 x^2}{2+3 x^2+x^4} \, dx\\ &=5 x+\frac{x \left (24+25 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-7 \int \frac{1}{2+x^2} \, dx-\frac{15}{2} \int \frac{1}{1+x^2} \, dx\\ &=5 x+\frac{x \left (24+25 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{15}{2} \tan ^{-1}(x)-\frac{7 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.038853, size = 50, normalized size = 1.02 \[ \frac{25 x^3+24 x}{2 \left (x^4+3 x^2+2\right )}+5 x-\frac{15}{2} \tan ^{-1}(x)-\frac{7 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

5*x + (24*x + 25*x^3)/(2*(2 + 3*x^2 + x^4)) - (15*ArcTan[x])/2 - (7*ArcTan[x/Sqrt[2]])/Sqrt[2]

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Maple [A]  time = 0.01, size = 41, normalized size = 0.8 \begin{align*} 5\,x+13\,{\frac{x}{{x}^{2}+2}}-{\frac{7\,\sqrt{2}}{2}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{x}{2\,{x}^{2}+2}}-{\frac{15\,\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x+13*x/(x^2+2)-7/2*arctan(1/2*x*2^(1/2))*2^(1/2)-1/2*x/(x^2+1)-15/2*arctan(x)

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Maxima [A]  time = 1.48706, size = 58, normalized size = 1.18 \begin{align*} -\frac{7}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 5 \, x + \frac{25 \, x^{3} + 24 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{15}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x + 1/2*(25*x^3 + 24*x)/(x^4 + 3*x^2 + 2) - 15/2*arctan(x)

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Fricas [A]  time = 2.13975, size = 180, normalized size = 3.67 \begin{align*} \frac{10 \, x^{5} + 55 \, x^{3} - 7 \, \sqrt{2}{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 15 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (x\right ) + 44 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*x^5 + 55*x^3 - 7*sqrt(2)*(x^4 + 3*x^2 + 2)*arctan(1/2*sqrt(2)*x) - 15*(x^4 + 3*x^2 + 2)*arctan(x) + 44
*x)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.185744, size = 48, normalized size = 0.98 \begin{align*} 5 x + \frac{25 x^{3} + 24 x}{2 x^{4} + 6 x^{2} + 4} - \frac{15 \operatorname{atan}{\left (x \right )}}{2} - \frac{7 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x + (25*x**3 + 24*x)/(2*x**4 + 6*x**2 + 4) - 15*atan(x)/2 - 7*sqrt(2)*atan(sqrt(2)*x/2)/2

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Giac [A]  time = 1.10117, size = 58, normalized size = 1.18 \begin{align*} -\frac{7}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 5 \, x + \frac{25 \, x^{3} + 24 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{15}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x + 1/2*(25*x^3 + 24*x)/(x^4 + 3*x^2 + 2) - 15/2*arctan(x)