3.4.40 \(\int x \sqrt {4+2 x+x^2} \, dx\) [340]

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

[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 = {654, 626, 633, 221} \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[4 + 2*x + x^2],x]

[Out]

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

Rule 221

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 626

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 633

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 654

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} \text {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 = 0.05, size = 47, normalized size = 0.94 \begin {gather*} \frac {1}{6} \sqrt {4+2 x+x^2} \left (5+x+2 x^2\right )+\frac {3}{2} \log \left (-1-x+\sqrt {4+2 x+x^2}\right ) \end {gather*}

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x*Sqrt[4 + 2*x + x^2],x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [A]
time = 0.16, size = 42, normalized size = 0.84

method	result	size
risch	\(\frac {\left (2 x^{2}+x +5\right ) \sqrt {x^{2}+2 x +4}}{6}-\frac {3 \arcsinh \left (\frac {\left (1+x \right ) \sqrt {3}}{3}\right )}{2}\)	\(33\)
trager	\(\left (\frac {1}{3} x^{2}+\frac {1}{6} x +\frac {5}{6}\right ) \sqrt {x^{2}+2 x +4}-\frac {3 \ln \left (x +1+\sqrt {x^{2}+2 x +4}\right )}{2}\)	\(39\)
default	\(\frac {\left (x^{2}+2 x +4\right )^{\frac {3}{2}}}{3}-\frac {\left (2+2 x \right ) \sqrt {x^{2}+2 x +4}}{4}-\frac {3 \arcsinh \left (\frac {\left (1+x \right ) \sqrt {3}}{3}\right )}{2}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 49, normalized size = 0.98 \begin {gather*} \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 {gather*}

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 = 0.33, size = 39, normalized size = 0.78 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {x^{2} + 2 x + 4}\, dx \end {gather*}

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 = 0.01, size = 50, normalized size = 1.00 \begin {gather*} 2 \left (\left (\frac {x}{6}+\frac 1{12}\right ) x+\frac {5}{12}\right ) \sqrt {x^{2}+2 x+4}+\frac {3}{2} \ln \left (\sqrt {x^{2}+2 x+4}-x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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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Mupad [B]
time = 0.21, size = 39, normalized size = 0.78 \begin {gather*} \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} \end {gather*}

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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