3.167 \(\int \frac{x^8}{(4+x^2)^4} \, dx\)

Optimal. Leaf size=58 \[ -\frac{x^7}{6 \left (x^2+4\right )^3}-\frac{7 x^5}{24 \left (x^2+4\right )^2}-\frac{35 x^3}{48 \left (x^2+4\right )}+\frac{35 x}{16}-\frac{35}{8} \tan ^{-1}\left (\frac{x}{2}\right ) \]

[Out]

(35*x)/16 - x^7/(6*(4 + x^2)^3) - (7*x^5)/(24*(4 + x^2)^2) - (35*x^3)/(48*(4 + x^2)) - (35*ArcTan[x/2])/8

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Rubi [A]  time = 0.0166097, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {288, 321, 203} \[ -\frac{x^7}{6 \left (x^2+4\right )^3}-\frac{7 x^5}{24 \left (x^2+4\right )^2}-\frac{35 x^3}{48 \left (x^2+4\right )}+\frac{35 x}{16}-\frac{35}{8} \tan ^{-1}\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^8/(4 + x^2)^4,x]

[Out]

(35*x)/16 - x^7/(6*(4 + x^2)^3) - (7*x^5)/(24*(4 + x^2)^2) - (35*x^3)/(48*(4 + x^2)) - (35*ArcTan[x/2])/8

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^8}{\left (4+x^2\right )^4} \, dx &=-\frac{x^7}{6 \left (4+x^2\right )^3}+\frac{7}{6} \int \frac{x^6}{\left (4+x^2\right )^3} \, dx\\ &=-\frac{x^7}{6 \left (4+x^2\right )^3}-\frac{7 x^5}{24 \left (4+x^2\right )^2}+\frac{35}{24} \int \frac{x^4}{\left (4+x^2\right )^2} \, dx\\ &=-\frac{x^7}{6 \left (4+x^2\right )^3}-\frac{7 x^5}{24 \left (4+x^2\right )^2}-\frac{35 x^3}{48 \left (4+x^2\right )}+\frac{35}{16} \int \frac{x^2}{4+x^2} \, dx\\ &=\frac{35 x}{16}-\frac{x^7}{6 \left (4+x^2\right )^3}-\frac{7 x^5}{24 \left (4+x^2\right )^2}-\frac{35 x^3}{48 \left (4+x^2\right )}-\frac{35}{4} \int \frac{1}{4+x^2} \, dx\\ &=\frac{35 x}{16}-\frac{x^7}{6 \left (4+x^2\right )^3}-\frac{7 x^5}{24 \left (4+x^2\right )^2}-\frac{35 x^3}{48 \left (4+x^2\right )}-\frac{35}{8} \tan ^{-1}\left (\frac{x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0190208, size = 40, normalized size = 0.69 \[ \frac{x \left (12 x^6+231 x^4+1120 x^2+1680\right )}{12 \left (x^2+4\right )^3}-\frac{35}{8} \tan ^{-1}\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(4 + x^2)^4,x]

[Out]

(x*(1680 + 1120*x^2 + 231*x^4 + 12*x^6))/(12*(4 + x^2)^3) - (35*ArcTan[x/2])/8

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Maple [A]  time = 0.007, size = 32, normalized size = 0.6 \begin{align*} x-16\,{\frac{1}{ \left ({x}^{2}+4 \right ) ^{3}} \left ( -{\frac{29\,{x}^{5}}{64}}-{\frac{17\,{x}^{3}}{6}}-{\frac{19\,x}{4}} \right ) }-{\frac{35}{8}\arctan \left ({\frac{x}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^2+4)^4,x)

[Out]

x-16*(-29/64*x^5-17/6*x^3-19/4*x)/(x^2+4)^3-35/8*arctan(1/2*x)

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Maxima [A]  time = 1.40538, size = 55, normalized size = 0.95 \begin{align*} x + \frac{87 \, x^{5} + 544 \, x^{3} + 912 \, x}{12 \,{\left (x^{6} + 12 \, x^{4} + 48 \, x^{2} + 64\right )}} - \frac{35}{8} \, \arctan \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^2+4)^4,x, algorithm="maxima")

[Out]

x + 1/12*(87*x^5 + 544*x^3 + 912*x)/(x^6 + 12*x^4 + 48*x^2 + 64) - 35/8*arctan(1/2*x)

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Fricas [A]  time = 1.87502, size = 166, normalized size = 2.86 \begin{align*} \frac{24 \, x^{7} + 462 \, x^{5} + 2240 \, x^{3} - 105 \,{\left (x^{6} + 12 \, x^{4} + 48 \, x^{2} + 64\right )} \arctan \left (\frac{1}{2} \, x\right ) + 3360 \, x}{24 \,{\left (x^{6} + 12 \, x^{4} + 48 \, x^{2} + 64\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^2+4)^4,x, algorithm="fricas")

[Out]

1/24*(24*x^7 + 462*x^5 + 2240*x^3 - 105*(x^6 + 12*x^4 + 48*x^2 + 64)*arctan(1/2*x) + 3360*x)/(x^6 + 12*x^4 + 4
8*x^2 + 64)

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Sympy [A]  time = 0.140521, size = 39, normalized size = 0.67 \begin{align*} x + \frac{87 x^{5} + 544 x^{3} + 912 x}{12 x^{6} + 144 x^{4} + 576 x^{2} + 768} - \frac{35 \operatorname{atan}{\left (\frac{x}{2} \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(x**2+4)**4,x)

[Out]

x + (87*x**5 + 544*x**3 + 912*x)/(12*x**6 + 144*x**4 + 576*x**2 + 768) - 35*atan(x/2)/8

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Giac [A]  time = 1.05332, size = 42, normalized size = 0.72 \begin{align*} x + \frac{87 \, x^{5} + 544 \, x^{3} + 912 \, x}{12 \,{\left (x^{2} + 4\right )}^{3}} - \frac{35}{8} \, \arctan \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^2+4)^4,x, algorithm="giac")

[Out]

x + 1/12*(87*x^5 + 544*x^3 + 912*x)/(x^2 + 4)^3 - 35/8*arctan(1/2*x)