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

Optimal. Leaf size=75 \[ -\frac{\left (15 x^2+244\right ) x}{8 \left (x^4+3 x^2+2\right )}+\frac{\left (103 x^2+102\right ) x}{4 \left (x^4+3 x^2+2\right )^2}+5 x+\frac{413}{8} \tan ^{-1}(x)-\frac{191 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{2 \sqrt{2}} \]

[Out]

5*x + (x*(102 + 103*x^2))/(4*(2 + 3*x^2 + x^4)^2) - (x*(244 + 15*x^2))/(8*(2 + 3*x^2 + x^4)) + (413*ArcTan[x])
/8 - (191*ArcTan[x/Sqrt[2]])/(2*Sqrt[2])

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Rubi [A]  time = 0.0913924, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {1668, 1678, 1676, 1166, 203} \[ -\frac{\left (15 x^2+244\right ) x}{8 \left (x^4+3 x^2+2\right )}+\frac{\left (103 x^2+102\right ) x}{4 \left (x^4+3 x^2+2\right )^2}+5 x+\frac{413}{8} \tan ^{-1}(x)-\frac{191 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

5*x + (x*(102 + 103*x^2))/(4*(2 + 3*x^2 + x^4)^2) - (x*(244 + 15*x^2))/(8*(2 + 3*x^2 + x^4)) + (413*ArcTan[x])
/8 - (191*ArcTan[x/Sqrt[2]])/(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 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
]

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^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx &=\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{1}{8} \int \frac{204-606 x^2-216 x^4+96 x^6-40 x^8}{\left (2+3 x^2+x^4\right )^2} \, dx\\ &=\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \frac{568-924 x^2+160 x^4}{2+3 x^2+x^4} \, dx\\ &=\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \left (160+\frac{4 \left (62-351 x^2\right )}{2+3 x^2+x^4}\right ) \, dx\\ &=5 x+\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{1}{8} \int \frac{62-351 x^2}{2+3 x^2+x^4} \, dx\\ &=5 x+\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{413}{8} \int \frac{1}{1+x^2} \, dx-\frac{191}{2} \int \frac{1}{2+x^2} \, dx\\ &=5 x+\frac{x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{413}{8} \tan ^{-1}(x)-\frac{191 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{2 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.061187, size = 60, normalized size = 0.8 \[ \frac{1}{8} \left (\frac{x \left (40 x^8+225 x^6+231 x^4-76 x^2-124\right )}{\left (x^4+3 x^2+2\right )^2}+413 \tan ^{-1}(x)-382 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x*(-124 - 76*x^2 + 231*x^4 + 225*x^6 + 40*x^8))/(2 + 3*x^2 + x^4)^2 + 413*ArcTan[x] - 382*Sqrt[2]*ArcTan[x/S
qrt[2]])/8

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Maple [A]  time = 0.011, size = 56, normalized size = 0.8 \begin{align*} 5\,x-16\,{\frac{1}{ \left ({x}^{2}+2 \right ) ^{2}} \left ( -1/32\,{x}^{3}+{\frac{25\,x}{16}} \right ) }-{\frac{191\,\sqrt{2}}{4}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }+{\frac{1}{ \left ({x}^{2}+1 \right ) ^{2}} \left ( -{\frac{19\,{x}^{3}}{8}}-{\frac{21\,x}{8}} \right ) }+{\frac{413\,\arctan \left ( x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x-16*(-1/32*x^3+25/16*x)/(x^2+2)^2-191/4*arctan(1/2*x*2^(1/2))*2^(1/2)+(-19/8*x^3-21/8*x)/(x^2+1)^2+413/8*ar
ctan(x)

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Maxima [A]  time = 1.4955, size = 85, normalized size = 1.13 \begin{align*} -\frac{191}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 5 \, x - \frac{15 \, x^{7} + 289 \, x^{5} + 556 \, x^{3} + 284 \, x}{8 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} + \frac{413}{8} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-191/4*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x - 1/8*(15*x^7 + 289*x^5 + 556*x^3 + 284*x)/(x^8 + 6*x^6 + 13*x^4 +
12*x^2 + 4) + 413/8*arctan(x)

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Fricas [A]  time = 1.62588, size = 285, normalized size = 3.8 \begin{align*} \frac{40 \, x^{9} + 225 \, x^{7} + 231 \, x^{5} - 76 \, x^{3} - 382 \, \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 ) + 413 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (x\right ) - 124 \, x}{8 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(40*x^9 + 225*x^7 + 231*x^5 - 76*x^3 - 382*sqrt(2)*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(1/2*sqrt(2)*
x) + 413*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(x) - 124*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)

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Sympy [A]  time = 0.240573, size = 68, normalized size = 0.91 \begin{align*} 5 x - \frac{15 x^{7} + 289 x^{5} + 556 x^{3} + 284 x}{8 x^{8} + 48 x^{6} + 104 x^{4} + 96 x^{2} + 32} + \frac{413 \operatorname{atan}{\left (x \right )}}{8} - \frac{191 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x - (15*x**7 + 289*x**5 + 556*x**3 + 284*x)/(8*x**8 + 48*x**6 + 104*x**4 + 96*x**2 + 32) + 413*atan(x)/8 - 1
91*sqrt(2)*atan(sqrt(2)*x/2)/4

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Giac [A]  time = 1.08504, size = 72, normalized size = 0.96 \begin{align*} -\frac{191}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 5 \, x - \frac{15 \, x^{7} + 289 \, x^{5} + 556 \, x^{3} + 284 \, x}{8 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} + \frac{413}{8} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-191/4*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x - 1/8*(15*x^7 + 289*x^5 + 556*x^3 + 284*x)/(x^4 + 3*x^2 + 2)^2 + 41
3/8*arctan(x)