3.1.100 \(\int \frac {x^2}{(1+x^2)^2} \, dx\) [100]

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

[Out]

-1/2*x/(x^2+1)+1/2*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 = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {294, 209} \begin {gather*} \frac {\text {ArcTan}(x)}{2}-\frac {x}{2 \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + x^2)^2,x]

[Out]

-1/2*x/(1 + x^2) + 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 294

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]

Rubi steps

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

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + x^2)^2,x]

[Out]

-1/2*x/(1 + x^2) + ArcTan[x]/2

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Maple [A]
time = 0.08, size = 16, normalized size = 0.84

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.08, size = 15, normalized size = 0.79 \begin {gather*} -\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2 + 1) + 1/2*arctan(x)

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Fricas [A]
time = 0.99, size = 21, normalized size = 1.11 \begin {gather*} \frac {{\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(x^2/(x^2+1)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.03, size = 12, normalized size = 0.63 \begin {gather*} - \frac {x}{2 x^{2} + 2} + \frac {\operatorname {atan}{\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(2*x**2 + 2) + atan(x)/2

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Giac [A]
time = 0.44, size = 15, normalized size = 0.79 \begin {gather*} -\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2 + 1) + 1/2*arctan(x)

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Mupad [B]
time = 0.03, size = 17, normalized size = 0.89 \begin {gather*} \frac {\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(x^2/(x^2 + 1)^2,x)

[Out]

atan(x)/2 - x/(2*(x^2 + 1))

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