### 3.198 $$\int \frac{-1+x}{2+2 x+x^2} \, dx$$

Optimal. Leaf size=20 $\frac{1}{2} \log \left (x^2+2 x+2\right )-2 \tan ^{-1}(x+1)$

[Out]

-2*ArcTan[1 + x] + Log[2 + 2*x + x^2]/2

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Rubi [A]  time = 0.0090319, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {634, 617, 204, 628} $\frac{1}{2} \log \left (x^2+2 x+2\right )-2 \tan ^{-1}(x+1)$

Antiderivative was successfully veriﬁed.

[In]

Int[(-1 + x)/(2 + 2*x + x^2),x]

[Out]

-2*ArcTan[1 + x] + Log[2 + 2*x + x^2]/2

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.005253, size = 20, normalized size = 1. $\frac{1}{2} \log \left (x^2+2 x+2\right )-2 \tan ^{-1}(x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-1 + x)/(2 + 2*x + x^2),x]

[Out]

-2*ArcTan[1 + x] + Log[2 + 2*x + x^2]/2

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Maple [A]  time = 0.001, size = 19, normalized size = 1. \begin{align*} -2\,\arctan \left ( 1+x \right ) +{\frac{\ln \left ({x}^{2}+2\,x+2 \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.4061, size = 24, normalized size = 1.2 \begin{align*} -2 \, \arctan \left (x + 1\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73515, size = 58, normalized size = 2.9 \begin{align*} -2 \, \arctan \left (x + 1\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.0911, size = 17, normalized size = 0.85 \begin{align*} \frac{\log{\left (x^{2} + 2 x + 2 \right )}}{2} - 2 \operatorname{atan}{\left (x + 1 \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 + 2*x + 2)/2 - 2*atan(x + 1)

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Giac [A]  time = 1.06146, size = 24, normalized size = 1.2 \begin{align*} -2 \, \arctan \left (x + 1\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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