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

Optimal. Leaf size=106 \[ \frac{2}{15} \sqrt{3 x^2+2} (2 x+1)^4+\frac{13}{60} \sqrt{3 x^2+2} (2 x+1)^3-\frac{19}{540} \sqrt{3 x^2+2} (2 x+1)^2-\frac{1}{810} (2073 x+3937) \sqrt{3 x^2+2}+\frac{5 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

[Out]

(-19*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/540 + (13*(1 + 2*x)^3*Sqrt[2 + 3*x^2])/60 + (2*(1 + 2*x)^4*Sqrt[2 + 3*x^2])/
15 - ((3937 + 2073*x)*Sqrt[2 + 3*x^2])/810 + (5*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.114004, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1654, 833, 780, 215} \[ \frac{2}{15} \sqrt{3 x^2+2} (2 x+1)^4+\frac{13}{60} \sqrt{3 x^2+2} (2 x+1)^3-\frac{19}{540} \sqrt{3 x^2+2} (2 x+1)^2-\frac{1}{810} (2073 x+3937) \sqrt{3 x^2+2}+\frac{5 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-19*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/540 + (13*(1 + 2*x)^3*Sqrt[2 + 3*x^2])/60 + (2*(1 + 2*x)^4*Sqrt[2 + 3*x^2])/
15 - ((3937 + 2073*x)*Sqrt[2 + 3*x^2])/810 + (5*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 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)^3 \left (1+3 x+4 x^2\right )}{\sqrt{2+3 x^2}} \, dx &=\frac{2}{15} (1+2 x)^4 \sqrt{2+3 x^2}+\frac{1}{60} \int \frac{(1+2 x)^3 (-68+156 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{13}{60} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{2}{15} (1+2 x)^4 \sqrt{2+3 x^2}+\frac{1}{720} \int \frac{(-2688-228 x) (1+2 x)^2}{\sqrt{2+3 x^2}} \, dx\\ &=-\frac{19}{540} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{13}{60} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{2}{15} (1+2 x)^4 \sqrt{2+3 x^2}+\frac{\int \frac{(-22368-49752 x) (1+2 x)}{\sqrt{2+3 x^2}} \, dx}{6480}\\ &=-\frac{19}{540} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{13}{60} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{2}{15} (1+2 x)^4 \sqrt{2+3 x^2}-\frac{1}{810} (3937+2073 x) \sqrt{2+3 x^2}+\frac{5}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=-\frac{19}{540} (1+2 x)^2 \sqrt{2+3 x^2}+\frac{13}{60} (1+2 x)^3 \sqrt{2+3 x^2}+\frac{2}{15} (1+2 x)^4 \sqrt{2+3 x^2}-\frac{1}{810} (3937+2073 x) \sqrt{2+3 x^2}+\frac{5 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0677779, size = 54, normalized size = 0.51 \[ \frac{1}{405} \left (\sqrt{3 x^2+2} \left (864 x^4+2430 x^3+2292 x^2-135 x-1841\right )+225 \sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + 3*x^2]*(-1841 - 135*x + 2292*x^2 + 2430*x^3 + 864*x^4) + 225*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])/405

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Maple [A]  time = 0.062, size = 79, normalized size = 0.8 \begin{align*}{\frac{32\,{x}^{4}}{15}\sqrt{3\,{x}^{2}+2}}+{\frac{764\,{x}^{2}}{135}\sqrt{3\,{x}^{2}+2}}-{\frac{1841}{405}\sqrt{3\,{x}^{2}+2}}+6\,{x}^{3}\sqrt{3\,{x}^{2}+2}-{\frac{x}{3}\sqrt{3\,{x}^{2}+2}}+{\frac{5\,\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)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x)

[Out]

32/15*x^4*(3*x^2+2)^(1/2)+764/135*x^2*(3*x^2+2)^(1/2)-1841/405*(3*x^2+2)^(1/2)+6*x^3*(3*x^2+2)^(1/2)-1/3*x*(3*
x^2+2)^(1/2)+5/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)

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Maxima [A]  time = 1.44823, size = 105, normalized size = 0.99 \begin{align*} \frac{32}{15} \, \sqrt{3 \, x^{2} + 2} x^{4} + 6 \, \sqrt{3 \, x^{2} + 2} x^{3} + \frac{764}{135} \, \sqrt{3 \, x^{2} + 2} x^{2} - \frac{1}{3} \, \sqrt{3 \, x^{2} + 2} x + \frac{5}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) - \frac{1841}{405} \, \sqrt{3 \, x^{2} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*sqrt(3*x^2 + 2)*x^4 + 6*sqrt(3*x^2 + 2)*x^3 + 764/135*sqrt(3*x^2 + 2)*x^2 - 1/3*sqrt(3*x^2 + 2)*x + 5/9*
sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 1841/405*sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.5656, size = 174, normalized size = 1.64 \begin{align*} \frac{1}{405} \,{\left (864 \, x^{4} + 2430 \, x^{3} + 2292 \, x^{2} - 135 \, x - 1841\right )} \sqrt{3 \, x^{2} + 2} + \frac{5}{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)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/405*(864*x^4 + 2430*x^3 + 2292*x^2 - 135*x - 1841)*sqrt(3*x^2 + 2) + 5/18*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 +
2)*x - 3*x^2 - 1)

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Sympy [A]  time = 2.18135, size = 94, normalized size = 0.89 \begin{align*} \frac{32 x^{4} \sqrt{3 x^{2} + 2}}{15} + 6 x^{3} \sqrt{3 x^{2} + 2} + \frac{764 x^{2} \sqrt{3 x^{2} + 2}}{135} - \frac{x \sqrt{3 x^{2} + 2}}{3} - \frac{1841 \sqrt{3 x^{2} + 2}}{405} + \frac{5 \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)**3*(4*x**2+3*x+1)/(3*x**2+2)**(1/2),x)

[Out]

32*x**4*sqrt(3*x**2 + 2)/15 + 6*x**3*sqrt(3*x**2 + 2) + 764*x**2*sqrt(3*x**2 + 2)/135 - x*sqrt(3*x**2 + 2)/3 -
1841*sqrt(3*x**2 + 2)/405 + 5*sqrt(3)*asinh(sqrt(6)*x/2)/9

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Giac [A]  time = 1.21552, size = 73, normalized size = 0.69 \begin{align*} \frac{1}{405} \,{\left (3 \,{\left (2 \,{\left (9 \,{\left (16 \, x + 45\right )} x + 382\right )} x - 45\right )} x - 1841\right )} \sqrt{3 \, x^{2} + 2} - \frac{5}{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)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

1/405*(3*(2*(9*(16*x + 45)*x + 382)*x - 45)*x - 1841)*sqrt(3*x^2 + 2) - 5/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^
2 + 2))