\(\int \frac {1}{(1+x^2)^3} \, dx\) [274]

Optimal result

Integrand size = 7, antiderivative size = 31 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {x}{4 \left (1+x^2\right )^2}+\frac {3 x}{8 \left (1+x^2\right )}+\frac {3 \arctan (x)}{8} \]

[Out]

1/4*x/(x^2+1)^2+3*x/(8*x^2+8)+3/8*arctan(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {205, 209} \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3 \arctan (x)}{8}+\frac {3 x}{8 \left (x^2+1\right )}+\frac {x}{4 \left (x^2+1\right )^2} \]

[In]

Int[(1 + x^2)^(-3),x]

[Out]

x/(4*(1 + x^2)^2) + (3*x)/(8*(1 + x^2)) + (3*ArcTan[x])/8

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}{4 \left (1+x^2\right )^2}+\frac {3}{4} \int \frac {1}{\left (1+x^2\right )^2} \, dx \\ & = \frac {x}{4 \left (1+x^2\right )^2}+\frac {3 x}{8 \left (1+x^2\right )}+\frac {3}{8} \int \frac {1}{1+x^2} \, dx \\ & = \frac {x}{4 \left (1+x^2\right )^2}+\frac {3 x}{8 \left (1+x^2\right )}+\frac {3 \arctan (x)}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {1}{8} \left (\frac {x \left (5+3 x^2\right )}{\left (1+x^2\right )^2}+3 \arctan (x)\right ) \]

[In]

Integrate[(1 + x^2)^(-3),x]

[Out]

((x*(5 + 3*x^2))/(1 + x^2)^2 + 3*ArcTan[x])/8

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74

method	result	size
meijerg	\(\frac {x \left (3 x^{2}+5\right )}{8 \left (x^{2}+1\right )^{2}}+\frac {3 \arctan \left (x \right )}{8}\)	\(23\)
risch	\(\frac {\frac {3}{8} x^{3}+\frac {5}{8} x}{\left (x^{2}+1\right )^{2}}+\frac {3 \arctan \left (x \right )}{8}\)	\(23\)
default	\(\frac {x}{4 \left (x^{2}+1\right )^{2}}+\frac {3 x}{8 \left (x^{2}+1\right )}+\frac {3 \arctan \left (x \right )}{8}\)	\(26\)
parallelrisch	\(-\frac {3 i \ln \left (x -i\right ) x^{4}-3 i \ln \left (i+x \right ) x^{4}+6 i \ln \left (x -i\right ) x^{2}-6 i \ln \left (i+x \right ) x^{2}-6 x^{3}+3 i \ln \left (x -i\right )-3 i \ln \left (i+x \right )-10 x}{16 \left (x^{2}+1\right )^{2}}\)	\(79\)

[In]

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

[Out]

1/8*x*(3*x^2+5)/(x^2+1)^2+3/8*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) + 5 \, x}{8 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

[In]

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

[Out]

1/8*(3*x^3 + 3*(x^4 + 2*x^2 + 1)*arctan(x) + 5*x)/(x^4 + 2*x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3 x^{3} + 5 x}{8 x^{4} + 16 x^{2} + 8} + \frac {3 \operatorname {atan}{\left (x \right )}}{8} \]

[In]

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

[Out]

(3*x**3 + 5*x)/(8*x**4 + 16*x**2 + 8) + 3*atan(x)/8

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + 5 \, x}{8 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} + \frac {3}{8} \, \arctan \left (x\right ) \]

[In]

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

[Out]

1/8*(3*x^3 + 5*x)/(x^4 + 2*x^2 + 1) + 3/8*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + 5 \, x}{8 \, {\left (x^{2} + 1\right )}^{2}} + \frac {3}{8} \, \arctan \left (x\right ) \]

[In]

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

[Out]

1/8*(3*x^3 + 5*x)/(x^2 + 1)^2 + 3/8*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (1+x^2\right )^3} \, dx=\frac {3\,\mathrm {atan}\left (x\right )}{8}+x\,\left (\frac {1}{4\,{\left (x^2+1\right )}^2}+\frac {3}{8\,x^2+8}\right ) \]

[In]

int(1/(x^2 + 1)^3,x)

[Out]

(3*atan(x))/8 + x*(1/(4*(x^2 + 1)^2) + 3/(8*x^2 + 8))