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

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Rubi [A]  time = 0.0145362, antiderivative size = 50, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0182636, 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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Maple [A]  time = 0.004, size = 42, normalized size = 0.8 \begin{align*}{\frac{1}{3} \left ({x}^{2}+2\,x+4 \right ) ^{{\frac{3}{2}}}}-{\frac{2\,x+2}{4}\sqrt{{x}^{2}+2\,x+4}}-{\frac{3}{2}{\it Arcsinh} \left ({\frac{ \left ( 1+x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

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*(1+x)*3^(1/2))

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Maxima [A]  time = 1.42293, size = 66, normalized size = 1.32 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.94938, size = 109, normalized size = 2.18 \begin{align*} \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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{x^{2} + 2 x + 4}\, dx \end{align*}

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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Giac [A]  time = 1.05913, size = 54, normalized size = 1.08 \begin{align*} \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 ) \end{align*}

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)