3.340 \(\int x \sqrt {4+2 x+x^2} \, dx\)

Optimal. Leaf size=50 \[ \frac {1}{3} \left (x^2+2 x+4\right )^{3/2}-\frac {1}{2} (x+1) \sqrt {x^2+2 x+4}-\frac {3}{2} \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right ) \]

[Out]

1/3*(x^2+2*x+4)^(3/2)-3/2*arcsinh(1/3*(1+x)*3^(1/2))-1/2*(1+x)*(x^2+2*x+4)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {640, 612, 619, 215} \[ \frac {1}{3} \left (x^2+2 x+4\right )^{3/2}-\frac {1}{2} (x+1) \sqrt {x^2+2 x+4}-\frac {3}{2} \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[4 + 2*x + x^2],x]

[Out]

-((1 + x)*Sqrt[4 + 2*x + x^2])/2 + (4 + 2*x + x^2)^(3/2)/3 - (3*ArcSinh[(1 + x)/Sqrt[3]])/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 612

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

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 x \sqrt {4+2 x+x^2} \, dx &=\frac {1}{3} \left (4+2 x+x^2\right )^{3/2}-\int \sqrt {4+2 x+x^2} \, dx\\ &=-\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {1}{3} \left (4+2 x+x^2\right )^{3/2}-\frac {3}{2} \int \frac {1}{\sqrt {4+2 x+x^2}} \, dx\\ &=-\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {1}{3} \left (4+2 x+x^2\right )^{3/2}-\frac {1}{4} \sqrt {3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{12}}} \, dx,x,2+2 x\right )\\ &=-\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {1}{3} \left (4+2 x+x^2\right )^{3/2}-\frac {3}{2} \sinh ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.76 \[ \frac {1}{6} \left (\sqrt {x^2+2 x+4} \left (2 x^2+x+5\right )-9 \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[4 + 2*x + x^2],x]

[Out]

(Sqrt[4 + 2*x + x^2]*(5 + x + 2*x^2) - 9*ArcSinh[(1 + x)/Sqrt[3]])/6

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fricas [A]  time = 0.40, size = 39, normalized size = 0.78 \[ \frac {1}{6} \, {\left (2 \, x^{2} + x + 5\right )} \sqrt {x^{2} + 2 \, x + 4} + \frac {3}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+2*x+4)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*x^2 + x + 5)*sqrt(x^2 + 2*x + 4) + 3/2*log(-x + sqrt(x^2 + 2*x + 4) - 1)

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giac [A]  time = 0.95, size = 40, normalized size = 0.80 \[ \frac {1}{6} \, {\left ({\left (2 \, x + 1\right )} x + 5\right )} \sqrt {x^{2} + 2 \, x + 4} + \frac {3}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+2*x+4)^(1/2),x, algorithm="giac")

[Out]

1/6*((2*x + 1)*x + 5)*sqrt(x^2 + 2*x + 4) + 3/2*log(-x + sqrt(x^2 + 2*x + 4) - 1)

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maple [A]  time = 0.01, size = 42, normalized size = 0.84 \[ -\frac {3 \arcsinh \left (\frac {\left (x +1\right ) \sqrt {3}}{3}\right )}{2}+\frac {\left (x^{2}+2 x +4\right )^{\frac {3}{2}}}{3}-\frac {\left (2 x +2\right ) \sqrt {x^{2}+2 x +4}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2+2*x+4)^(1/2),x)

[Out]

1/3*(x^2+2*x+4)^(3/2)-1/4*(2*x+2)*(x^2+2*x+4)^(1/2)-3/2*arcsinh(1/3*(x+1)*3^(1/2))

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maxima [A]  time = 1.42, size = 49, normalized size = 0.98 \[ \frac {1}{3} \, {\left (x^{2} + 2 \, x + 4\right )}^{\frac {3}{2}} - \frac {1}{2} \, \sqrt {x^{2} + 2 \, x + 4} x - \frac {1}{2} \, \sqrt {x^{2} + 2 \, x + 4} - \frac {3}{2} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+2*x+4)^(1/2),x, algorithm="maxima")

[Out]

1/3*(x^2 + 2*x + 4)^(3/2) - 1/2*sqrt(x^2 + 2*x + 4)*x - 1/2*sqrt(x^2 + 2*x + 4) - 3/2*arcsinh(1/3*sqrt(3)*(x +
1))

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mupad [B]  time = 0.21, size = 39, normalized size = 0.78 \[ \frac {\sqrt {x^2+2\,x+4}\,\left (8\,x^2+4\,x+20\right )}{24}-\frac {3\,\ln \left (x+\sqrt {x^2+2\,x+4}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*x + x^2 + 4)^(1/2),x)

[Out]

((2*x + x^2 + 4)^(1/2)*(4*x + 8*x^2 + 20))/24 - (3*log(x + (2*x + x^2 + 4)^(1/2) + 1))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {x^{2} + 2 x + 4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**2+2*x+4)**(1/2),x)

[Out]

Integral(x*sqrt(x**2 + 2*x + 4), x)

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