3.90 \(\int \frac {x}{\sqrt {5+2 x+x^2}} \, dx\)

Optimal. Leaf size=23 \[ \sqrt {x^2+2 x+5}-\sinh ^{-1}\left (\frac {x+1}{2}\right ) \]

[Out]

-arcsinh(1/2+1/2*x)+(x^2+2*x+5)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {640, 619, 215} \[ \sqrt {x^2+2 x+5}-\sinh ^{-1}\left (\frac {x+1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[5 + 2*x + x^2],x]

[Out]

Sqrt[5 + 2*x + x^2] - ArcSinh[(1 + x)/2]

Rule 215

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 640

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

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {5+2 x+x^2}} \, dx &=\sqrt {5+2 x+x^2}-\int \frac {1}{\sqrt {5+2 x+x^2}} \, dx\\ &=\sqrt {5+2 x+x^2}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{16}}} \, dx,x,2+2 x\right )\\ &=\sqrt {5+2 x+x^2}-\sinh ^{-1}\left (\frac {1+x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.09 \[ \sqrt {x^2+2 x+5}-\sinh ^{-1}\left (\frac {1}{4} (2 x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[5 + 2*x + x^2],x]

[Out]

Sqrt[5 + 2*x + x^2] - ArcSinh[(2 + 2*x)/4]

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fricas [A]  time = 0.41, size = 27, normalized size = 1.17 \[ \sqrt {x^{2} + 2 \, x + 5} + \log \left (-x + \sqrt {x^{2} + 2 \, x + 5} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+2*x+5)^(1/2),x, algorithm="fricas")

[Out]

sqrt(x^2 + 2*x + 5) + log(-x + sqrt(x^2 + 2*x + 5) - 1)

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giac [A]  time = 1.02, size = 27, normalized size = 1.17 \[ \sqrt {x^{2} + 2 \, x + 5} + \log \left (-x + \sqrt {x^{2} + 2 \, x + 5} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+2*x+5)^(1/2),x, algorithm="giac")

[Out]

sqrt(x^2 + 2*x + 5) + log(-x + sqrt(x^2 + 2*x + 5) - 1)

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maple [A]  time = 0.00, size = 20, normalized size = 0.87 \[ -\arcsinh \left (\frac {x}{2}+\frac {1}{2}\right )+\sqrt {x^{2}+2 x +5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+2*x+5)^(1/2),x)

[Out]

-arcsinh(1/2*x+1/2)+(x^2+2*x+5)^(1/2)

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maxima [A]  time = 1.27, size = 19, normalized size = 0.83 \[ \sqrt {x^{2} + 2 \, x + 5} - \operatorname {arsinh}\left (\frac {1}{2} \, x + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+2*x+5)^(1/2),x, algorithm="maxima")

[Out]

sqrt(x^2 + 2*x + 5) - arcsinh(1/2*x + 1/2)

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mupad [B]  time = 0.08, size = 27, normalized size = 1.17 \[ \sqrt {x^2+2\,x+5}-\ln \left (x+\sqrt {x^2+2\,x+5}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(2*x + x^2 + 5)^(1/2),x)

[Out]

(2*x + x^2 + 5)^(1/2) - log(x + (2*x + x^2 + 5)^(1/2) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {x^{2} + 2 x + 5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**2+2*x+5)**(1/2),x)

[Out]

Integral(x/sqrt(x**2 + 2*x + 5), x)

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