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

Optimal. Leaf size=118 \[ \frac{1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3+\frac{1}{5} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2+\frac{1}{810} (306 x+25) \left (3 x^2-x+2\right )^{3/2}+\frac{235 (1-6 x) \sqrt{3 x^2-x+2}}{1296}+\frac{5405 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{2592 \sqrt{3}} \]

[Out]

(235*(1 - 6*x)*Sqrt[2 - x + 3*x^2])/1296 + ((1 + 2*x)^2*(2 - x + 3*x^2)^(3/2))/5 + ((1 + 2*x)^3*(2 - x + 3*x^2
)^(3/2))/9 + ((25 + 306*x)*(2 - x + 3*x^2)^(3/2))/810 + (5405*ArcSinh[(1 - 6*x)/Sqrt[23]])/(2592*Sqrt[3])

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Rubi [A]  time = 0.111741, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1653, 832, 779, 612, 619, 215} \[ \frac{1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3+\frac{1}{5} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2+\frac{1}{810} (306 x+25) \left (3 x^2-x+2\right )^{3/2}+\frac{235 (1-6 x) \sqrt{3 x^2-x+2}}{1296}+\frac{5405 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{2592 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(235*(1 - 6*x)*Sqrt[2 - x + 3*x^2])/1296 + ((1 + 2*x)^2*(2 - x + 3*x^2)^(3/2))/5 + ((1 + 2*x)^3*(2 - x + 3*x^2
)^(3/2))/9 + ((25 + 306*x)*(2 - x + 3*x^2)^(3/2))/810 + (5405*ArcSinh[(1 - 6*x)/Sqrt[23]])/(2592*Sqrt[3])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (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 + b*x + 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 + b*x + 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)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + 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 + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + 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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

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 (1+2 x)^2 \sqrt{2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx &=\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{72} \int (1+2 x)^2 (-12+216 x) \sqrt{2-x+3 x^2} \, dx\\ &=\frac{1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{\int (1+2 x) (-1584+2448 x) \sqrt{2-x+3 x^2} \, dx}{1080}\\ &=\frac{1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{810} (25+306 x) \left (2-x+3 x^2\right )^{3/2}-\frac{235}{108} \int \sqrt{2-x+3 x^2} \, dx\\ &=\frac{235 (1-6 x) \sqrt{2-x+3 x^2}}{1296}+\frac{1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{810} (25+306 x) \left (2-x+3 x^2\right )^{3/2}-\frac{5405 \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx}{2592}\\ &=\frac{235 (1-6 x) \sqrt{2-x+3 x^2}}{1296}+\frac{1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{810} (25+306 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (235 \sqrt{\frac{23}{3}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{2592}\\ &=\frac{235 (1-6 x) \sqrt{2-x+3 x^2}}{1296}+\frac{1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac{1}{810} (25+306 x) \left (2-x+3 x^2\right )^{3/2}+\frac{5405 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{2592 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0387511, size = 65, normalized size = 0.55 \[ \frac{6 \sqrt{3 x^2-x+2} \left (17280 x^5+35712 x^4+33552 x^3+22344 x^2+14638 x+5607\right )-27025 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{38880} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[2 - x + 3*x^2]*(5607 + 14638*x + 22344*x^2 + 33552*x^3 + 35712*x^4 + 17280*x^5) - 27025*Sqrt[3]*ArcSin
h[(-1 + 6*x)/Sqrt[23]])/38880

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Maple [A]  time = 0.055, size = 98, normalized size = 0.8 \begin{align*}{\frac{8\,{x}^{3}}{9} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{32\,{x}^{2}}{15} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{83\,x}{45} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{277}{810} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}-{\frac{-235+1410\,x}{1296}\sqrt{3\,{x}^{2}-x+2}}-{\frac{5405\,\sqrt{3}}{7776}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) } \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-x+2)^(1/2),x)

[Out]

8/9*x^3*(3*x^2-x+2)^(3/2)+32/15*x^2*(3*x^2-x+2)^(3/2)+83/45*x*(3*x^2-x+2)^(3/2)+277/810*(3*x^2-x+2)^(3/2)-235/
1296*(-1+6*x)*(3*x^2-x+2)^(1/2)-5405/7776*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))

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Maxima [A]  time = 1.4895, size = 147, normalized size = 1.25 \begin{align*} \frac{8}{9} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x^{3} + \frac{32}{15} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x^{2} + \frac{83}{45} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x + \frac{277}{810} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} - \frac{235}{216} \, \sqrt{3 \, x^{2} - x + 2} x - \frac{5405}{7776} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (6 \, x - 1\right )}\right ) + \frac{235}{1296} \, \sqrt{3 \, x^{2} - x + 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-x+2)^(1/2),x, algorithm="maxima")

[Out]

8/9*(3*x^2 - x + 2)^(3/2)*x^3 + 32/15*(3*x^2 - x + 2)^(3/2)*x^2 + 83/45*(3*x^2 - x + 2)^(3/2)*x + 277/810*(3*x
^2 - x + 2)^(3/2) - 235/216*sqrt(3*x^2 - x + 2)*x - 5405/7776*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) + 235/1
296*sqrt(3*x^2 - x + 2)

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Fricas [A]  time = 1.55612, size = 243, normalized size = 2.06 \begin{align*} \frac{1}{6480} \,{\left (17280 \, x^{5} + 35712 \, x^{4} + 33552 \, x^{3} + 22344 \, x^{2} + 14638 \, x + 5607\right )} \sqrt{3 \, x^{2} - x + 2} + \frac{5405}{15552} \, \sqrt{3} \log \left (4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\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-x+2)^(1/2),x, algorithm="fricas")

[Out]

1/6480*(17280*x^5 + 35712*x^4 + 33552*x^3 + 22344*x^2 + 14638*x + 5607)*sqrt(3*x^2 - x + 2) + 5405/15552*sqrt(
3)*log(4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (2 x + 1\right )^{2} \sqrt{3 x^{2} - x + 2} \left (4 x^{2} + 3 x + 1\right )\, dx \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-x+2)**(1/2),x)

[Out]

Integral((2*x + 1)**2*sqrt(3*x**2 - x + 2)*(4*x**2 + 3*x + 1), x)

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Giac [A]  time = 1.16029, size = 99, normalized size = 0.84 \begin{align*} \frac{1}{6480} \,{\left (2 \,{\left (12 \,{\left (6 \,{\left (8 \,{\left (15 \, x + 31\right )} x + 233\right )} x + 931\right )} x + 7319\right )} x + 5607\right )} \sqrt{3 \, x^{2} - x + 2} + \frac{5405}{7776} \, \sqrt{3} \log \left (-2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )} + 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-x+2)^(1/2),x, algorithm="giac")

[Out]

1/6480*(2*(12*(6*(8*(15*x + 31)*x + 233)*x + 931)*x + 7319)*x + 5607)*sqrt(3*x^2 - x + 2) + 5405/7776*sqrt(3)*
log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1)