3.111 \(\int \frac{3+2 x^2}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=19 \[ \frac{x}{2 \left (x^2+1\right )}+\frac{5}{2} \tan ^{-1}(x) \]

[Out]

x/(2*(1 + x^2)) + (5*ArcTan[x])/2

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Rubi [A]  time = 0.0043252, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {385, 203} \[ \frac{x}{2 \left (x^2+1\right )}+\frac{5}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x^2)/(1 + x^2)^2,x]

[Out]

x/(2*(1 + x^2)) + (5*ArcTan[x])/2

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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])

Rubi steps

\begin{align*} \int \frac{3+2 x^2}{\left (1+x^2\right )^2} \, dx &=\frac{x}{2 \left (1+x^2\right )}+\frac{5}{2} \int \frac{1}{1+x^2} \, dx\\ &=\frac{x}{2 \left (1+x^2\right )}+\frac{5}{2} \tan ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0071309, size = 19, normalized size = 1. \[ \frac{x}{2 \left (x^2+1\right )}+\frac{5}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*x^2)/(1 + x^2)^2,x]

[Out]

x/(2*(1 + x^2)) + (5*ArcTan[x])/2

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Maple [A]  time = 0.005, size = 16, normalized size = 0.8 \begin{align*}{\frac{x}{2\,{x}^{2}+2}}+{\frac{5\,\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+3)/(x^2+1)^2,x)

[Out]

1/2*x/(x^2+1)+5/2*arctan(x)

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Maxima [A]  time = 1.48689, size = 20, normalized size = 1.05 \begin{align*} \frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{5}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3)/(x^2+1)^2,x, algorithm="maxima")

[Out]

1/2*x/(x^2 + 1) + 5/2*arctan(x)

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Fricas [A]  time = 1.28191, size = 58, normalized size = 3.05 \begin{align*} \frac{5 \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) + x}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3)/(x^2+1)^2,x, algorithm="fricas")

[Out]

1/2*(5*(x^2 + 1)*arctan(x) + x)/(x^2 + 1)

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Sympy [A]  time = 0.099179, size = 14, normalized size = 0.74 \begin{align*} \frac{x}{2 x^{2} + 2} + \frac{5 \operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+3)/(x**2+1)**2,x)

[Out]

x/(2*x**2 + 2) + 5*atan(x)/2

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Giac [A]  time = 1.12801, size = 20, normalized size = 1.05 \begin{align*} \frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{5}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+3)/(x^2+1)^2,x, algorithm="giac")

[Out]

1/2*x/(x^2 + 1) + 5/2*arctan(x)