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

Optimal. Leaf size=82 \[ \frac{1}{6} \sqrt{3 x^2+2} (2 x+1)^3+\frac{5}{18} \sqrt{3 x^2+2} (2 x+1)^2-\frac{1}{27} (3 x+61) \sqrt{3 x^2+2}-\sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right ) \]

[Out]

(5*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/18 + ((1 + 2*x)^3*Sqrt[2 + 3*x^2])/6 - ((61 + 3*x)*Sqrt[2 + 3*x^2])/27 - Sqrt[
3]*ArcSinh[Sqrt[3/2]*x]

________________________________________________________________________________________

Rubi [A]  time = 0.0884072, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1654, 833, 780, 215} \[ \frac{1}{6} \sqrt{3 x^2+2} (2 x+1)^3+\frac{5}{18} \sqrt{3 x^2+2} (2 x+1)^2-\frac{1}{27} (3 x+61) \sqrt{3 x^2+2}-\sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/18 + ((1 + 2*x)^3*Sqrt[2 + 3*x^2])/6 - ((61 + 3*x)*Sqrt[2 + 3*x^2])/27 - Sqrt[
3]*ArcSinh[Sqrt[3/2]*x]

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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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)^2 \left (1+3 x+4 x^2\right )}{\sqrt{2+3 x^2}} \, dx &=\frac{1}{6} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{1}{48} \int \frac{(1+2 x)^2 (-48+120 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{5}{18} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{1}{6} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{1}{432} \int \frac{(-1392-144 x) (1+2 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{5}{18} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{1}{6} (1+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{27} (61+3 x) \sqrt{2+3 x^2}-3 \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{5}{18} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{1}{6} (1+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{27} (61+3 x) \sqrt{2+3 x^2}-\sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\\ \end{align*}

Mathematica [A]  time = 0.0426069, size = 48, normalized size = 0.59 \[ \frac{1}{27} \sqrt{3 x^2+2} \left (36 x^3+84 x^2+54 x-49\right )-\sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + 3*x^2]*(-49 + 54*x + 84*x^2 + 36*x^3))/27 - Sqrt[3]*ArcSinh[Sqrt[3/2]*x]

________________________________________________________________________________________

Maple [A]  time = 0.055, size = 65, normalized size = 0.8 \begin{align*}{\frac{4\,{x}^{3}}{3}\sqrt{3\,{x}^{2}+2}}+2\,x\sqrt{3\,{x}^{2}+2}-{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) \sqrt{3}+{\frac{28\,{x}^{2}}{9}\sqrt{3\,{x}^{2}+2}}-{\frac{49}{27}\sqrt{3\,{x}^{2}+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*x^3*(3*x^2+2)^(1/2)+2*x*(3*x^2+2)^(1/2)-arcsinh(1/2*x*6^(1/2))*3^(1/2)+28/9*x^2*(3*x^2+2)^(1/2)-49/27*(3*x
^2+2)^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.49621, size = 86, normalized size = 1.05 \begin{align*} \frac{4}{3} \, \sqrt{3 \, x^{2} + 2} x^{3} + \frac{28}{9} \, \sqrt{3 \, x^{2} + 2} x^{2} + 2 \, \sqrt{3 \, x^{2} + 2} x - \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) - \frac{49}{27} \, \sqrt{3 \, x^{2} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*sqrt(3*x^2 + 2)*x^3 + 28/9*sqrt(3*x^2 + 2)*x^2 + 2*sqrt(3*x^2 + 2)*x - sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 49
/27*sqrt(3*x^2 + 2)

________________________________________________________________________________________

Fricas [A]  time = 1.52905, size = 147, normalized size = 1.79 \begin{align*} \frac{1}{27} \,{\left (36 \, x^{3} + 84 \, x^{2} + 54 \, x - 49\right )} \sqrt{3 \, x^{2} + 2} + \frac{1}{2} \, \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)^2*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/27*(36*x^3 + 84*x^2 + 54*x - 49)*sqrt(3*x^2 + 2) + 1/2*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)

________________________________________________________________________________________

Sympy [A]  time = 1.15238, size = 75, normalized size = 0.91 \begin{align*} \frac{4 x^{3} \sqrt{3 x^{2} + 2}}{3} + \frac{28 x^{2} \sqrt{3 x^{2} + 2}}{9} + 2 x \sqrt{3 x^{2} + 2} - \frac{49 \sqrt{3 x^{2} + 2}}{27} - \sqrt{3} \operatorname{asinh}{\left (\frac{\sqrt{6} x}{2} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x**3*sqrt(3*x**2 + 2)/3 + 28*x**2*sqrt(3*x**2 + 2)/9 + 2*x*sqrt(3*x**2 + 2) - 49*sqrt(3*x**2 + 2)/27 - sqrt(
3)*asinh(sqrt(6)*x/2)

________________________________________________________________________________________

Giac [A]  time = 1.1942, size = 65, normalized size = 0.79 \begin{align*} \frac{1}{27} \,{\left (6 \,{\left (2 \,{\left (3 \, x + 7\right )} x + 9\right )} x - 49\right )} \sqrt{3 \, x^{2} + 2} + \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)^2*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

1/27*(6*(2*(3*x + 7)*x + 9)*x - 49)*sqrt(3*x^2 + 2) + sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))