3.1359 \(\int (5-x) (3+2 x) \sqrt{2+3 x^2} \, dx\)

Optimal. Leaf size=56 \[ \frac{1}{18} (14-3 x) \left (3 x^2+2\right )^{3/2}+\frac{23}{3} x \sqrt{3 x^2+2}+\frac{46 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

[Out]

(23*x*Sqrt[2 + 3*x^2])/3 + ((14 - 3*x)*(2 + 3*x^2)^(3/2))/18 + (46*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.0141275, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {780, 195, 215} \[ \frac{1}{18} (14-3 x) \left (3 x^2+2\right )^{3/2}+\frac{23}{3} x \sqrt{3 x^2+2}+\frac{46 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 - x)*(3 + 2*x)*Sqrt[2 + 3*x^2],x]

[Out]

(23*x*Sqrt[2 + 3*x^2])/3 + ((14 - 3*x)*(2 + 3*x^2)^(3/2))/18 + (46*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 (5-x) (3+2 x) \sqrt{2+3 x^2} \, dx &=\frac{1}{18} (14-3 x) \left (2+3 x^2\right )^{3/2}+\frac{46}{3} \int \sqrt{2+3 x^2} \, dx\\ &=\frac{23}{3} x \sqrt{2+3 x^2}+\frac{1}{18} (14-3 x) \left (2+3 x^2\right )^{3/2}+\frac{46}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{23}{3} x \sqrt{2+3 x^2}+\frac{1}{18} (14-3 x) \left (2+3 x^2\right )^{3/2}+\frac{46 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0398892, size = 50, normalized size = 0.89 \[ \frac{1}{18} \left (92 \sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )-\sqrt{3 x^2+2} \left (9 x^3-42 x^2-132 x-28\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)*(3 + 2*x)*Sqrt[2 + 3*x^2],x]

[Out]

(-(Sqrt[2 + 3*x^2]*(-28 - 132*x - 42*x^2 + 9*x^3)) + 92*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])/18

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Maple [A]  time = 0.004, size = 49, normalized size = 0.9 \begin{align*} -{\frac{x}{6} \left ( 3\,{x}^{2}+2 \right ) ^{{\frac{3}{2}}}}+{\frac{23\,x}{3}\sqrt{3\,{x}^{2}+2}}+{\frac{46\,\sqrt{3}}{9}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }+{\frac{7}{9} \left ( 3\,{x}^{2}+2 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)*(3+2*x)*(3*x^2+2)^(1/2),x)

[Out]

-1/6*x*(3*x^2+2)^(3/2)+23/3*x*(3*x^2+2)^(1/2)+46/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)+7/9*(3*x^2+2)^(3/2)

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Maxima [A]  time = 1.46978, size = 65, normalized size = 1.16 \begin{align*} -\frac{1}{6} \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}} x + \frac{7}{9} \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}} + \frac{23}{3} \, \sqrt{3 \, x^{2} + 2} x + \frac{46}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)*(3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*(3*x^2 + 2)^(3/2)*x + 7/9*(3*x^2 + 2)^(3/2) + 23/3*sqrt(3*x^2 + 2)*x + 46/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x
)

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Fricas [A]  time = 2.69668, size = 151, normalized size = 2.7 \begin{align*} -\frac{1}{18} \,{\left (9 \, x^{3} - 42 \, x^{2} - 132 \, x - 28\right )} \sqrt{3 \, x^{2} + 2} + \frac{23}{9} \, \sqrt{3} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)*(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

-1/18*(9*x^3 - 42*x^2 - 132*x - 28)*sqrt(3*x^2 + 2) + 23/9*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)

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Sympy [A]  time = 9.36279, size = 94, normalized size = 1.68 \begin{align*} - \frac{3 x^{5}}{2 \sqrt{3 x^{2} + 2}} - \frac{3 x^{3}}{2 \sqrt{3 x^{2} + 2}} + \frac{15 x \sqrt{3 x^{2} + 2}}{2} - \frac{x}{3 \sqrt{3 x^{2} + 2}} + \frac{7 \left (3 x^{2} + 2\right )^{\frac{3}{2}}}{9} + \frac{46 \sqrt{3} \operatorname{asinh}{\left (\frac{\sqrt{6} x}{2} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)*(3*x**2+2)**(1/2),x)

[Out]

-3*x**5/(2*sqrt(3*x**2 + 2)) - 3*x**3/(2*sqrt(3*x**2 + 2)) + 15*x*sqrt(3*x**2 + 2)/2 - x/(3*sqrt(3*x**2 + 2))
+ 7*(3*x**2 + 2)**(3/2)/9 + 46*sqrt(3)*asinh(sqrt(6)*x/2)/9

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Giac [A]  time = 1.20328, size = 65, normalized size = 1.16 \begin{align*} -\frac{1}{18} \,{\left (3 \,{\left ({\left (3 \, x - 14\right )} x - 44\right )} x - 28\right )} \sqrt{3 \, x^{2} + 2} - \frac{46}{9} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)*(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

-1/18*(3*((3*x - 14)*x - 44)*x - 28)*sqrt(3*x^2 + 2) - 46/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))