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

Optimal. Leaf size=39 \[ -\frac {(x+1)^2}{4 \left (x^2+1\right )^2}-\frac {1-x}{4 \left (x^2+1\right )}+\frac {1}{4} \tan ^{-1}(x) \]

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {819, 639, 203} \begin {gather*} -\frac {(x+1)^2}{4 \left (x^2+1\right )^2}-\frac {1-x}{4 \left (x^2+1\right )}+\frac {1}{4} \tan ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(1 + x)^2)/(1 + x^2)^3,x]

[Out]

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

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])

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.72 \begin {gather*} \frac {1}{4} \left (\frac {x^3-2 x^2-x-2}{\left (x^2+1\right )^2}+\tan ^{-1}(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 + x)^2)/(1 + x^2)^3,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (1+x)^2}{\left (1+x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x*(1 + x)^2)/(1 + x^2)^3,x]

[Out]

IntegrateAlgebraic[(x*(1 + x)^2)/(1 + x^2)^3, x]

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fricas [A]  time = 0.41, size = 40, normalized size = 1.03 \begin {gather*} \frac {x^{3} - 2 \, x^{2} + {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \relax (x) - x - 2}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 27, normalized size = 0.69 \begin {gather*} \frac {x^{3} - 2 \, x^{2} - x - 2}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {1}{4} \, \arctan \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 29, normalized size = 0.74 \begin {gather*} \frac {\arctan \relax (x )}{4}+\frac {\frac {1}{4} x^{3}-\frac {1}{2} x^{2}-\frac {1}{4} x -\frac {1}{2}}{\left (x^{2}+1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.18, size = 32, normalized size = 0.82 \begin {gather*} \frac {x^{3} - 2 \, x^{2} - x - 2}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} + \frac {1}{4} \, \arctan \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 29, normalized size = 0.74 \begin {gather*} \frac {\mathrm {atan}\relax (x)}{4}-\frac {-\frac {x^3}{4}+\frac {x^2}{2}+\frac {x}{4}+\frac {1}{2}}{{\left (x^2+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x + 1)^2)/(x^2 + 1)^3,x)

[Out]

atan(x)/4 - (x/4 + x^2/2 - x^3/4 + 1/2)/(x^2 + 1)^2

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sympy [A]  time = 0.13, size = 27, normalized size = 0.69 \begin {gather*} \frac {\operatorname {atan}{\relax (x )}}{4} + \frac {x^{3} - 2 x^{2} - x - 2}{4 x^{4} + 8 x^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(x)/4 + (x**3 - 2*x**2 - x - 2)/(4*x**4 + 8*x**2 + 4)

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