3.10.26 \(\int \frac {1}{(1+x) \sqrt {2 x+x^2}} \, dx\) [926]

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

[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 = {702, 209} \begin {gather*} \text {ArcTan}\left (\sqrt {x^2+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 + x)*Sqrt[2*x + x^2]),x]

[Out]

ArcTan[Sqrt[2*x + x^2]]

Rule 209

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

Rule 702

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 \text {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] Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(12)=24\).
time = 0.04, size = 41, normalized size = 3.42 \begin {gather*} -\frac {2 \sqrt {x} \sqrt {2+x} \tan ^{-1}\left (1+x-\sqrt {x} \sqrt {2+x}\right )}{\sqrt {x (2+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.24, size = 13, normalized size = 1.08

method	result	size
default	\(-\arctan \left (\frac {1}{\sqrt {\left (1+x \right )^{2}-1}}\right )\)	\(13\)
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{2}+2 x}}{1+x}\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 9, normalized size = 0.75 \begin {gather*} -\arcsin \left (\frac {1}{{\left | x + 1 \right |}}\right ) \end {gather*}

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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Fricas [A]
time = 0.34, size = 17, normalized size = 1.42 \begin {gather*} 2 \, \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \end {gather*}

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

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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Giac [A]
time = 3.17, size = 17, normalized size = 1.42 \begin {gather*} 2 \, \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \end {gather*}

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

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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