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

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

[Out]

ArcTan[2*Sqrt[x + x^2]]

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Rubi [A]  time = 0.0081645, antiderivative size = 12, normalized size of antiderivative = 1., 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} \[ \tan ^{-1}\left (2 \sqrt{x^2+x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + 2*x)*Sqrt[x + x^2]),x]

[Out]

ArcTan[2*Sqrt[x + x^2]]

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{(1+2 x) \sqrt{x+x^2}} \, dx &=4 \operatorname{Subst}\left (\int \frac{1}{2+8 x^2} \, dx,x,\sqrt{x+x^2}\right )\\ &=\tan ^{-1}\left (2 \sqrt{x+x^2}\right )\\ \end{align*}

Mathematica [B]  time = 0.0120436, size = 37, normalized size = 3.08 \[ \frac{2 \sqrt{x} \sqrt{x+1} \tan ^{-1}\left (\frac{\sqrt{x}}{\sqrt{x+1}}\right )}{\sqrt{x (x+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 15, normalized size = 1.3 \begin{align*} -\arctan \left ({\frac{1}{\sqrt{4\, \left ( x+1/2 \right ) ^{2}-1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(1/(4*(x+1/2)^2-1)^(1/2))

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Maxima [A]  time = 1.68146, size = 15, normalized size = 1.25 \begin{align*} -\arcsin \left (\frac{1}{{\left | 2 \, x + 1 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(1/abs(2*x + 1))

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Fricas [A]  time = 1.48214, size = 51, normalized size = 4.25 \begin{align*} 2 \, \arctan \left (-2 \, x + 2 \, \sqrt{x^{2} + x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-2*x + 2*sqrt(x^2 + x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1142, size = 23, normalized size = 1.92 \begin{align*} 2 \, \arctan \left (-2 \, x + 2 \, \sqrt{x^{2} + x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-2*x + 2*sqrt(x^2 + x) - 1)