3.287 \(\int \frac {1}{1+\sqrt {2+2 x+x^2}} \, dx\)

Optimal. Leaf size=29 \[ -\frac {\sqrt {x^2+2 x+2}}{x+1}+\frac {1}{x+1}+\sinh ^{-1}(x+1) \]

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Rubi [A]  time = 0.04, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 684, 619, 215} \[ -\frac {\sqrt {x^2+2 x+2}}{x+1}+\frac {1}{x+1}+\sinh ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sqrt[2 + 2*x + x^2])^(-1),x]

[Out]

(1 + x)^(-1) - Sqrt[2 + 2*x + x^2]/(1 + x) + ArcSinh[1 + x]

Rule 215

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 684

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[(b*p)/(d*e*(m + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1
), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.03 \[ \frac {-\sqrt {x^2+2 x+2}+(x+1) \sinh ^{-1}(x+1)+1}{x+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sqrt[2 + 2*x + x^2])^(-1),x]

[Out]

(1 - Sqrt[2 + 2*x + x^2] + (1 + x)*ArcSinh[1 + x])/(1 + x)

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IntegrateAlgebraic [A]  time = 0.29, size = 46, normalized size = 1.59 \[ \frac {\sqrt {x^2+2 x+2}}{-x-1}-\log \left (\sqrt {x^2+2 x+2}-x-1\right )+\frac {1}{x+1} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + Sqrt[2 + 2*x + x^2])^(-1),x]

[Out]

(1 + x)^(-1) + Sqrt[2 + 2*x + x^2]/(-1 - x) - Log[-1 - x + Sqrt[2 + 2*x + x^2]]

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fricas [A]  time = 0.68, size = 39, normalized size = 1.34 \[ -\frac {{\left (x + 1\right )} \log \left (-x + \sqrt {x^{2} + 2 \, x + 2} - 1\right ) + x + \sqrt {x^{2} + 2 \, x + 2}}{x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.66, size = 60, normalized size = 2.07 \[ \frac {2}{{\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )}^{2} + 2 \, x - 2 \, \sqrt {x^{2} + 2 \, x + 2}} + \frac {1}{x + 1} - \log \left (-x + \sqrt {x^{2} + 2 \, x + 2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 40, normalized size = 1.38

method	result	size
default	\(-\frac {\left (\left (1+x \right )^{2}+1\right )^{\frac {3}{2}}}{1+x}+\left (1+x \right ) \sqrt {\left (1+x \right )^{2}+1}+\arcsinh \left (1+x \right )+\frac {1}{1+x}\)	\(40\)
trager	\(-\frac {x}{1+x}-\frac {\sqrt {x^{2}+2 x +2}}{1+x}-\ln \left (\sqrt {x^{2}+2 x +2}-1-x \right )\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(1+x)*((1+x)^2+1)^(3/2)+(1+x)*((1+x)^2+1)^(1/2)+arcsinh(1+x)+1/(1+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{2} + 2 \, x + 2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \frac {1}{x+1}+\int \frac {\sqrt {x^2+2\,x+2}}{{\left (x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x + x^2 + 2)^(1/2) + 1),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{2} + 2 x + 2} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(x**2 + 2*x + 2) + 1), x)

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