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

Optimal. Leaf size=12 \[ \tan ^{-1}\left (\sqrt {x^2+2 x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {688, 203} \[ \tan ^{-1}\left (\sqrt {x^2+2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + x)*Sqrt[2*x + x^2]),x]

[Out]

ArcTan[Sqrt[2*x + x^2]]

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 688

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

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 37, normalized size = 3.08 \[ \frac {2 \sqrt {x} \sqrt {x+2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt {x+2}}\right )}{\sqrt {x (x+2)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[2 + x]*ArcTan[Sqrt[x]/Sqrt[2 + x]])/Sqrt[x*(2 + x)]

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fricas [A]  time = 0.60, size = 17, normalized size = 1.42 \[ 2 \, \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-x + sqrt(x^2 + 2*x) - 1)

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giac [A]  time = 0.46, size = 17, normalized size = 1.42 \[ 2 \, \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-x + sqrt(x^2 + 2*x) - 1)

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maple [A]  time = 0.00, size = 13, normalized size = 1.08 \[ -\arctan \left (\frac {1}{\sqrt {\left (x +1\right )^{2}-1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 9, normalized size = 0.75 \[ -\arcsin \left (\frac {1}{{\left | x + 1 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(1/abs(x + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.08 \[ \int \frac {1}{\sqrt {x^2+2\,x}\,\left (x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x + x^2)^(1/2)*(x + 1)),x)

[Out]

int(1/((2*x + x^2)^(1/2)*(x + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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