\(\int \frac {1}{(1+x^2)^2} \, dx\) [161]

Optimal result

Integrand size = 7, antiderivative size = 19 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {x}{2 \left (1+x^2\right )}+\frac {\arctan (x)}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {205, 209} \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {\arctan (x)}{2}+\frac {x}{2 \left (x^2+1\right )} \]

[In]

Int[(1 + x^2)^(-2),x]

[Out]

x/(2*(1 + x^2)) + ArcTan[x]/2

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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])

Rubi steps \begin{align*} \text {integral}& = \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 {\arctan (x)}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} \left (\frac {x}{1+x^2}+\arctan (x)\right ) \]

[In]

Integrate[(1 + x^2)^(-2),x]

[Out]

(x/(1 + x^2) + ArcTan[x])/2

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {x}{2 x^{2}+2}+\frac {\arctan \left (x \right )}{2}\)	\(16\)
risch	\(\frac {x}{2 x^{2}+2}+\frac {\arctan \left (x \right )}{2}\)	\(16\)
meijerg	\(\frac {x}{2 x^{2}+2}+\frac {\arctan \left (x \right )}{2}\)	\(17\)
parallelrisch	\(-\frac {i \ln \left (x -i\right ) x^{2}-i \ln \left (x +i\right ) x^{2}+i \ln \left (x -i\right )-i \ln \left (x +i\right )-2 x}{4 \left (x^{2}+1\right )}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 1\right )} \arctan \left (x\right ) + x}{2 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {x}{2 x^{2} + 2} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

x/(2*x**2 + 2) + atan(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) \]

[In]

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

[Out]

1/2*x/(x^2 + 1) + 1/2*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) \]

[In]

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

[Out]

1/2*x/(x^2 + 1) + 1/2*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (x\right )}{2}+\frac {x}{2\,\left (x^2+1\right )} \]

[In]

int(1/(x^2 + 1)^2,x)

[Out]

atan(x)/2 + x/(2*(x^2 + 1))