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*(x^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} \[ \tan ^{-1}\left (2 \sqrt {x^2+x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + 2*x)*Sqrt[x + x^2]),x]

[Out]

ArcTan[2*Sqrt[x + x^2]]

Rule 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+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*}

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Mathematica [B]  time = 0.02, 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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fricas [A]  time = 0.65, size = 17, normalized size = 1.42 \[ 2 \, \arctan \left (-2 \, x + 2 \, \sqrt {x^{2} + x} - 1\right ) \]

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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giac [A]  time = 0.62, size = 17, normalized size = 1.42 \[ 2 \, \arctan \left (-2 \, x + 2 \, \sqrt {x^{2} + x} - 1\right ) \]

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)

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maple [A]  time = 0.01, size = 15, normalized size = 1.25 \[ -\arctan \left (\frac {1}{\sqrt {4 \left (x +\frac {1}{2}\right )^{2}-1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.93, size = 11, normalized size = 0.92 \[ -\arcsin \left (\frac {1}{{\left | 2 \, x + 1 \right |}}\right ) \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x + 1)*(x + x^2)^(1/2)),x)

[Out]

int(1/((2*x + 1)*(x + x^2)^(1/2)), x)

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

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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