### 3.120 $$\int \frac{(1+2 x) (1+3 x+4 x^2)}{\sqrt{2+3 x^2}} \, dx$$

Optimal. Leaf size=62 $\frac{2}{9} \sqrt{3 x^2+2} (2 x+1)^2+\frac{7}{27} (3 x+1) \sqrt{3 x^2+2}-\frac{7 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}$

[Out]

(2*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/9 + (7*(1 + 3*x)*Sqrt[2 + 3*x^2])/27 - (7*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.0516871, antiderivative size = 62, 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 = {1654, 780, 215} $\frac{2}{9} \sqrt{3 x^2+2} (2 x+1)^2+\frac{7}{27} (3 x+1) \sqrt{3 x^2+2}-\frac{7 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/9 + (7*(1 + 3*x)*Sqrt[2 + 3*x^2])/27 - (7*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

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 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 )}{\sqrt{2+3 x^2}} \, dx &=\frac{2}{9} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{1}{36} \int \frac{(1+2 x) (-28+84 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{2}{9} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{7}{27} (1+3 x) \sqrt{2+3 x^2}-\frac{7}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{2}{9} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{7}{27} (1+3 x) \sqrt{2+3 x^2}-\frac{7 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0283797, size = 44, normalized size = 0.71 $\frac{1}{27} \left (\sqrt{3 x^2+2} \left (24 x^2+45 x+13\right )-21 \sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[2 + 3*x^2]*(13 + 45*x + 24*x^2) - 21*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])/27

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Maple [A]  time = 0.049, size = 51, normalized size = 0.8 \begin{align*}{\frac{8\,{x}^{2}}{9}\sqrt{3\,{x}^{2}+2}}+{\frac{13}{27}\sqrt{3\,{x}^{2}+2}}+{\frac{5\,x}{3}\sqrt{3\,{x}^{2}+2}}-{\frac{7\,\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)^(1/2),x)

[Out]

8/9*x^2*(3*x^2+2)^(1/2)+13/27*(3*x^2+2)^(1/2)+5/3*x*(3*x^2+2)^(1/2)-7/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)

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Maxima [A]  time = 1.46932, size = 68, normalized size = 1.1 \begin{align*} \frac{8}{9} \, \sqrt{3 \, x^{2} + 2} x^{2} + \frac{5}{3} \, \sqrt{3 \, x^{2} + 2} x - \frac{7}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{13}{27} \, \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)^(1/2),x, algorithm="maxima")

[Out]

8/9*sqrt(3*x^2 + 2)*x^2 + 5/3*sqrt(3*x^2 + 2)*x - 7/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 13/27*sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.5361, size = 136, normalized size = 2.19 \begin{align*} \frac{1}{27} \,{\left (24 \, x^{2} + 45 \, x + 13\right )} \sqrt{3 \, x^{2} + 2} + \frac{7}{18} \, \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((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/27*(24*x^2 + 45*x + 13)*sqrt(3*x^2 + 2) + 7/18*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)

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Sympy [A]  time = 0.569814, size = 63, normalized size = 1.02 \begin{align*} \frac{8 x^{2} \sqrt{3 x^{2} + 2}}{9} + \frac{5 x \sqrt{3 x^{2} + 2}}{3} + \frac{13 \sqrt{3 x^{2} + 2}}{27} - \frac{7 \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((1+2*x)*(4*x**2+3*x+1)/(3*x**2+2)**(1/2),x)

[Out]

8*x**2*sqrt(3*x**2 + 2)/9 + 5*x*sqrt(3*x**2 + 2)/3 + 13*sqrt(3*x**2 + 2)/27 - 7*sqrt(3)*asinh(sqrt(6)*x/2)/9

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Giac [A]  time = 1.21317, size = 59, normalized size = 0.95 \begin{align*} \frac{1}{27} \,{\left (3 \,{\left (8 \, x + 15\right )} x + 13\right )} \sqrt{3 \, x^{2} + 2} + \frac{7}{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((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

1/27*(3*(8*x + 15)*x + 13)*sqrt(3*x^2 + 2) + 7/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))