3.2.11 \(\int \frac {3+2 x^2}{(1+x^2)^2} \, dx\) [111]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, 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 = {393, 209} \begin {gather*} \frac {5 \text {ArcTan}(x)}{2}+\frac {x}{2 \left (x^2+1\right )} \end {gather*}

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 393

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])

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*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} \frac {x}{2 \left (1+x^2\right )}+\frac {5}{2} \tan ^{-1}(x) \end {gather*}

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 = 0.07, size = 16, normalized size = 0.84

method	result	size
default	\(\frac {x}{2 x^{2}+2}+\frac {5 \arctan \left (x \right )}{2}\)	\(16\)
risch	\(\frac {x}{2 x^{2}+2}+\frac {5 \arctan \left (x \right )}{2}\)	\(16\)
meijerg	\(-\frac {x}{x^{2}+1}+\frac {5 \arctan \left (x \right )}{2}+\frac {3 x}{2 x^{2}+2}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 15, normalized size = 0.79 \begin {gather*} \frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {5}{2} \, \arctan \left (x\right ) \end {gather*}

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 = 0.89, size = 20, normalized size = 1.05 \begin {gather*} \frac {5 \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) + x}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}

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 = 0.04, size = 14, normalized size = 0.74 \begin {gather*} \frac {x}{2 x^{2} + 2} + \frac {5 \operatorname {atan}{\left (x \right )}}{2} \end {gather*}

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 = 1.13, size = 15, normalized size = 0.79 \begin {gather*} \frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {5}{2} \, \arctan \left (x\right ) \end {gather*}

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)

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Mupad [B]
time = 0.04, size = 16, normalized size = 0.84 \begin {gather*} \frac {5\,\mathrm {atan}\left (x\right )}{2}+\frac {x}{2\,\left (x^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + 3)/(x^2 + 1)^2,x)

[Out]

(5*atan(x))/2 + x/(2*(x^2 + 1))

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