3.366 \(\int \frac{-32+36 x-42 x^2+21 x^3-10 x^4+3 x^5}{x (1+x^2) (4+x^2)^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{x^2+4}+\log \left (x^2+4\right )-2 \log (x)+\frac{1}{2} \tan ^{-1}\left (\frac{x}{2}\right )+2 \tan ^{-1}(x) \]

[Out]

(4 + x^2)^(-1) + ArcTan[x/2]/2 + 2*ArcTan[x] - 2*Log[x] + Log[4 + x^2]

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Rubi [A]  time = 0.254368, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {6725, 203, 261, 635, 260} \[ \frac{1}{x^2+4}+\log \left (x^2+4\right )-2 \log (x)+\frac{1}{2} \tan ^{-1}\left (\frac{x}{2}\right )+2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(-32 + 36*x - 42*x^2 + 21*x^3 - 10*x^4 + 3*x^5)/(x*(1 + x^2)*(4 + x^2)^2),x]

[Out]

(4 + x^2)^(-1) + ArcTan[x/2]/2 + 2*ArcTan[x] - 2*Log[x] + Log[4 + x^2]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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])

Rule 261

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

Rule 635

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{-32+36 x-42 x^2+21 x^3-10 x^4+3 x^5}{x \left (1+x^2\right ) \left (4+x^2\right )^2} \, dx &=\int \left (-\frac{2}{x}+\frac{2}{1+x^2}-\frac{2 x}{\left (4+x^2\right )^2}+\frac{1+2 x}{4+x^2}\right ) \, dx\\ &=-2 \log (x)+2 \int \frac{1}{1+x^2} \, dx-2 \int \frac{x}{\left (4+x^2\right )^2} \, dx+\int \frac{1+2 x}{4+x^2} \, dx\\ &=\frac{1}{4+x^2}+2 \tan ^{-1}(x)-2 \log (x)+2 \int \frac{x}{4+x^2} \, dx+\int \frac{1}{4+x^2} \, dx\\ &=\frac{1}{4+x^2}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{2}\right )+2 \tan ^{-1}(x)-2 \log (x)+\log \left (4+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.017828, size = 32, normalized size = 1. \[ \frac{1}{x^2+4}+\log \left (x^2+4\right )-2 \log (x)+\frac{1}{2} \tan ^{-1}\left (\frac{x}{2}\right )+2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-32 + 36*x - 42*x^2 + 21*x^3 - 10*x^4 + 3*x^5)/(x*(1 + x^2)*(4 + x^2)^2),x]

[Out]

(4 + x^2)^(-1) + ArcTan[x/2]/2 + 2*ArcTan[x] - 2*Log[x] + Log[4 + x^2]

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Maple [A]  time = 0.009, size = 29, normalized size = 0.9 \begin{align*} \left ({x}^{2}+4 \right ) ^{-1}+{\frac{1}{2}\arctan \left ({\frac{x}{2}} \right ) }+2\,\arctan \left ( x \right ) -2\,\ln \left ( x \right ) +\ln \left ({x}^{2}+4 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^5-10*x^4+21*x^3-42*x^2+36*x-32)/x/(x^2+1)/(x^2+4)^2,x)

[Out]

1/(x^2+4)+1/2*arctan(1/2*x)+2*arctan(x)-2*ln(x)+ln(x^2+4)

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Maxima [A]  time = 1.63389, size = 38, normalized size = 1.19 \begin{align*} \frac{1}{x^{2} + 4} + \frac{1}{2} \, \arctan \left (\frac{1}{2} \, x\right ) + 2 \, \arctan \left (x\right ) + \log \left (x^{2} + 4\right ) - 2 \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^5-10*x^4+21*x^3-42*x^2+36*x-32)/x/(x^2+1)/(x^2+4)^2,x, algorithm="maxima")

[Out]

1/(x^2 + 4) + 1/2*arctan(1/2*x) + 2*arctan(x) + log(x^2 + 4) - 2*log(x)

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Fricas [A]  time = 1.51326, size = 158, normalized size = 4.94 \begin{align*} \frac{{\left (x^{2} + 4\right )} \arctan \left (\frac{1}{2} \, x\right ) + 4 \,{\left (x^{2} + 4\right )} \arctan \left (x\right ) + 2 \,{\left (x^{2} + 4\right )} \log \left (x^{2} + 4\right ) - 4 \,{\left (x^{2} + 4\right )} \log \left (x\right ) + 2}{2 \,{\left (x^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^5-10*x^4+21*x^3-42*x^2+36*x-32)/x/(x^2+1)/(x^2+4)^2,x, algorithm="fricas")

[Out]

1/2*((x^2 + 4)*arctan(1/2*x) + 4*(x^2 + 4)*arctan(x) + 2*(x^2 + 4)*log(x^2 + 4) - 4*(x^2 + 4)*log(x) + 2)/(x^2
+ 4)

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Sympy [A]  time = 0.221786, size = 29, normalized size = 0.91 \begin{align*} - 2 \log{\left (x \right )} + \log{\left (x^{2} + 4 \right )} + \frac{\operatorname{atan}{\left (\frac{x}{2} \right )}}{2} + 2 \operatorname{atan}{\left (x \right )} + \frac{1}{x^{2} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**5-10*x**4+21*x**3-42*x**2+36*x-32)/x/(x**2+1)/(x**2+4)**2,x)

[Out]

-2*log(x) + log(x**2 + 4) + atan(x/2)/2 + 2*atan(x) + 1/(x**2 + 4)

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Giac [A]  time = 1.15844, size = 39, normalized size = 1.22 \begin{align*} \frac{1}{x^{2} + 4} + \frac{1}{2} \, \arctan \left (\frac{1}{2} \, x\right ) + 2 \, \arctan \left (x\right ) + \log \left (x^{2} + 4\right ) - 2 \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^5-10*x^4+21*x^3-42*x^2+36*x-32)/x/(x^2+1)/(x^2+4)^2,x, algorithm="giac")

[Out]

1/(x^2 + 4) + 1/2*arctan(1/2*x) + 2*arctan(x) + log(x^2 + 4) - 2*log(abs(x))