3.431 \(\int \frac{4+4 x}{x^2 (1+x^2)} \, dx\)

Optimal. Leaf size=22 \[ -2 \log \left (x^2+1\right )-\frac{4}{x}+4 \log (x)-4 \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0189116, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {801, 635, 203, 260} \[ -2 \log \left (x^2+1\right )-\frac{4}{x}+4 \log (x)-4 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(4 + 4*x)/(x^2*(1 + x^2)),x]

[Out]

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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 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{4+4 x}{x^2 \left (1+x^2\right )} \, dx &=\int \left (\frac{4}{x^2}+\frac{4}{x}-\frac{4 (1+x)}{1+x^2}\right ) \, dx\\ &=-\frac{4}{x}+4 \log (x)-4 \int \frac{1+x}{1+x^2} \, dx\\ &=-\frac{4}{x}+4 \log (x)-4 \int \frac{1}{1+x^2} \, dx-4 \int \frac{x}{1+x^2} \, dx\\ &=-\frac{4}{x}-4 \tan ^{-1}(x)+4 \log (x)-2 \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0054922, size = 24, normalized size = 1.09 \[ 4 \left (-\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}+\log (x)-\tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 4*x)/(x^2*(1 + x^2)),x]

[Out]

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

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Maple [A]  time = 0.006, size = 23, normalized size = 1.1 \begin{align*} -4\,{x}^{-1}-4\,\arctan \left ( x \right ) +4\,\ln \left ( x \right ) -2\,\ln \left ({x}^{2}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4+4*x)/x^2/(x^2+1),x)

[Out]

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

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Maxima [A]  time = 1.48236, size = 30, normalized size = 1.36 \begin{align*} -\frac{4}{x} - 4 \, \arctan \left (x\right ) - 2 \, \log \left (x^{2} + 1\right ) + 4 \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+4*x)/x^2/(x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.999919, size = 76, normalized size = 3.45 \begin{align*} -\frac{2 \,{\left (2 \, x \arctan \left (x\right ) + x \log \left (x^{2} + 1\right ) - 2 \, x \log \left (x\right ) + 2\right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+4*x)/x^2/(x^2+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.117294, size = 20, normalized size = 0.91 \begin{align*} 4 \log{\left (x \right )} - 2 \log{\left (x^{2} + 1 \right )} - 4 \operatorname{atan}{\left (x \right )} - \frac{4}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+4*x)/x**2/(x**2+1),x)

[Out]

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

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Giac [A]  time = 1.1062, size = 31, normalized size = 1.41 \begin{align*} -\frac{4}{x} - 4 \, \arctan \left (x\right ) - 2 \, \log \left (x^{2} + 1\right ) + 4 \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+4*x)/x^2/(x^2+1),x, algorithm="giac")

[Out]

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