### 3.126 $$\int \frac{(1+2 x) (1+3 x+4 x^2)}{(2+3 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=55 $\frac{2-51 x}{18 \sqrt{3 x^2+2}}+\frac{8}{9} \sqrt{3 x^2+2}+\frac{10 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}$

[Out]

(2 - 51*x)/(18*Sqrt[2 + 3*x^2]) + (8*Sqrt[2 + 3*x^2])/9 + (10*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.0444929, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {1814, 641, 215} $\frac{2-51 x}{18 \sqrt{3 x^2+2}}+\frac{8}{9} \sqrt{3 x^2+2}+\frac{10 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]

[Out]

(2 - 51*x)/(18*Sqrt[2 + 3*x^2]) + (8*Sqrt[2 + 3*x^2])/9 + (10*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 641

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

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 \frac{(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx &=\frac{2-51 x}{18 \sqrt{2+3 x^2}}-\frac{1}{2} \int \frac{-\frac{20}{3}-\frac{16 x}{3}}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{2-51 x}{18 \sqrt{2+3 x^2}}+\frac{8}{9} \sqrt{2+3 x^2}+\frac{10}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{2-51 x}{18 \sqrt{2+3 x^2}}+\frac{8}{9} \sqrt{2+3 x^2}+\frac{10 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0304869, size = 48, normalized size = 0.87 $\frac{48 x^2+20 \sqrt{9 x^2+6} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )-51 x+34}{18 \sqrt{3 x^2+2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]

[Out]

(34 - 51*x + 48*x^2 + 20*Sqrt[6 + 9*x^2]*ArcSinh[Sqrt[3/2]*x])/(18*Sqrt[2 + 3*x^2])

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Maple [A]  time = 0.052, size = 51, normalized size = 0.9 \begin{align*}{\frac{8\,{x}^{2}}{3}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{17}{9}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}-{\frac{17\,x}{6}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{10\,\sqrt{3}}{9}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x)

[Out]

8/3*x^2/(3*x^2+2)^(1/2)+17/9/(3*x^2+2)^(1/2)-17/6*x/(3*x^2+2)^(1/2)+10/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)

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Maxima [A]  time = 1.47274, size = 68, normalized size = 1.24 \begin{align*} \frac{8 \, x^{2}}{3 \, \sqrt{3 \, x^{2} + 2}} + \frac{10}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) - \frac{17 \, x}{6 \, \sqrt{3 \, x^{2} + 2}} + \frac{17}{9 \, \sqrt{3 \, x^{2} + 2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="maxima")

[Out]

8/3*x^2/sqrt(3*x^2 + 2) + 10/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 17/6*x/sqrt(3*x^2 + 2) + 17/9/sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.85049, size = 170, normalized size = 3.09 \begin{align*} \frac{10 \, \sqrt{3}{\left (3 \, x^{2} + 2\right )} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) +{\left (48 \, x^{2} - 51 \, x + 34\right )} \sqrt{3 \, x^{2} + 2}}{18 \,{\left (3 \, x^{2} + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="fricas")

[Out]

1/18*(10*sqrt(3)*(3*x^2 + 2)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + (48*x^2 - 51*x + 34)*sqrt(3*x^2 + 2
))/(3*x^2 + 2)

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Sympy [B]  time = 12.906, size = 114, normalized size = 2.07 \begin{align*} \frac{30 \sqrt{3} x^{2} \operatorname{asinh}{\left (\frac{\sqrt{6} x}{2} \right )}}{27 x^{2} + 18} + \frac{8 x^{2}}{3 \sqrt{3 x^{2} + 2}} - \frac{30 x \sqrt{3 x^{2} + 2}}{27 x^{2} + 18} + \frac{x}{2 \sqrt{3 x^{2} + 2}} + \frac{20 \sqrt{3} \operatorname{asinh}{\left (\frac{\sqrt{6} x}{2} \right )}}{27 x^{2} + 18} + \frac{17}{9 \sqrt{3 x^{2} + 2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x**2+3*x+1)/(3*x**2+2)**(3/2),x)

[Out]

30*sqrt(3)*x**2*asinh(sqrt(6)*x/2)/(27*x**2 + 18) + 8*x**2/(3*sqrt(3*x**2 + 2)) - 30*x*sqrt(3*x**2 + 2)/(27*x*
*2 + 18) + x/(2*sqrt(3*x**2 + 2)) + 20*sqrt(3)*asinh(sqrt(6)*x/2)/(27*x**2 + 18) + 17/(9*sqrt(3*x**2 + 2))

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Giac [A]  time = 1.20877, size = 59, normalized size = 1.07 \begin{align*} -\frac{10}{9} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) + \frac{3 \,{\left (16 \, x - 17\right )} x + 34}{18 \, \sqrt{3 \, x^{2} + 2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="giac")

[Out]

-10/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/18*(3*(16*x - 17)*x + 34)/sqrt(3*x^2 + 2)