3.2.98 \(\int \frac {-1+x}{2+2 x+x^2} \, dx\) [198]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {648, 631, 210, 642} \begin {gather*} \frac {1}{2} \log \left (x^2+2 x+2\right )-2 \text {ArcTan}(x+1) \end {gather*}

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 210

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

Rule 631

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 642

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]

Rule 648

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]

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 \text {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*}

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Mathematica [A]
time = 0.00, size = 20, normalized size = 1.00 \begin {gather*} -2 \tan ^{-1}(1+x)+\frac {1}{2} \log \left (2+2 x+x^2\right ) \end {gather*}

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 = 0.08, size = 19, normalized size = 0.95

method	result	size
default	\(-2 \arctan \left (1+x \right )+\frac {\ln \left (x^{2}+2 x +2\right )}{2}\)	\(19\)
risch	\(-2 \arctan \left (1+x \right )+\frac {\ln \left (x^{2}+2 x +2\right )}{2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.75, size = 18, normalized size = 0.90 \begin {gather*} -2 \, \arctan \left (x + 1\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end {gather*}

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 = 1.94, size = 18, normalized size = 0.90 \begin {gather*} -2 \, \arctan \left (x + 1\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end {gather*}

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 = 0.03, size = 17, normalized size = 0.85 \begin {gather*} \frac {\log {\left (x^{2} + 2 x + 2 \right )}}{2} - 2 \operatorname {atan}{\left (x + 1 \right )} \end {gather*}

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 = 0.42, size = 18, normalized size = 0.90 \begin {gather*} -2 \, \arctan \left (x + 1\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 2\right ) \end {gather*}

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)

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Mupad [B]
time = 0.16, size = 18, normalized size = 0.90 \begin {gather*} \frac {\ln \left (x^2+2\,x+2\right )}{2}-2\,\mathrm {atan}\left (x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)/(2*x + x^2 + 2),x)

[Out]

log(2*x + x^2 + 2)/2 - 2*atan(x + 1)

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