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

Optimal. Leaf size=26 \[ \sqrt{x^2+2 x}-\tan ^{-1}\left (\sqrt{x^2+2 x}\right ) \]

[Out]

Sqrt[2*x + x^2] - ArcTan[Sqrt[2*x + x^2]]

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Rubi [A]  time = 0.0143339, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {685, 688, 203} \[ \sqrt{x^2+2 x}-\tan ^{-1}\left (\sqrt{x^2+2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2*x + x^2]/(1 + x),x]

[Out]

Sqrt[2*x + x^2] - ArcTan[Sqrt[2*x + x^2]]

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
 b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

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

Mathematica [A]  time = 0.0285005, size = 38, normalized size = 1.46 \[ \sqrt{x (x+2)} \left (1-\frac{2 \tan ^{-1}\left (\sqrt{\frac{x}{x+2}}\right )}{\sqrt{x} \sqrt{x+2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[2*x + x^2]/(1 + x),x]

[Out]

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

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Maple [A]  time = 0.06, size = 21, normalized size = 0.8 \begin{align*} \sqrt{ \left ( 1+x \right ) ^{2}-1}+\arctan \left ({\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}-1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72709, size = 23, normalized size = 0.88 \begin{align*} \sqrt{x^{2} + 2 \, x} + \arcsin \left (\frac{1}{{\left | x + 1 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 2*x) + arcsin(1/abs(x + 1))

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Fricas [A]  time = 2.09229, size = 73, normalized size = 2.81 \begin{align*} \sqrt{x^{2} + 2 \, x} - 2 \, \arctan \left (-x + \sqrt{x^{2} + 2 \, x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 2*x) - 2*arctan(-x + sqrt(x^2 + 2*x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(x + 2))/(x + 1), x)

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Giac [A]  time = 1.3209, size = 36, normalized size = 1.38 \begin{align*} \sqrt{x^{2} + 2 \, x} - 2 \, \arctan \left (-x + \sqrt{x^{2} + 2 \, x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 2*x) - 2*arctan(-x + sqrt(x^2 + 2*x) - 1)