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

Optimal. Leaf size=120 \[ \frac {2}{15} \sqrt {3 x^2-x+2} (2 x+1)^4+\frac {19}{60} \sqrt {3 x^2-x+2} (2 x+1)^3+\frac {44}{135} \sqrt {3 x^2-x+2} (2 x+1)^2-\frac {(6298 x+24897) \sqrt {3 x^2-x+2}}{3240}+\frac {9211 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{1296 \sqrt {3}} \]

[Out]

9211/3888*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)+44/135*(1+2*x)^2*(3*x^2-x+2)^(1/2)+19/60*(1+2*x)^3*(3*x^2-x+2
)^(1/2)+2/15*(1+2*x)^4*(3*x^2-x+2)^(1/2)-1/3240*(24897+6298*x)*(3*x^2-x+2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1653, 832, 779, 619, 215} \[ \frac {2}{15} \sqrt {3 x^2-x+2} (2 x+1)^4+\frac {19}{60} \sqrt {3 x^2-x+2} (2 x+1)^3+\frac {44}{135} \sqrt {3 x^2-x+2} (2 x+1)^2-\frac {(6298 x+24897) \sqrt {3 x^2-x+2}}{3240}+\frac {9211 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{1296 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(44*(1 + 2*x)^2*Sqrt[2 - x + 3*x^2])/135 + (19*(1 + 2*x)^3*Sqrt[2 - x + 3*x^2])/60 + (2*(1 + 2*x)^4*Sqrt[2 - x
+ 3*x^2])/15 - ((24897 + 6298*x)*Sqrt[2 - x + 3*x^2])/3240 + (9211*ArcSinh[(1 - 6*x)/Sqrt[23]])/(1296*Sqrt[3]
)

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]

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 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 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 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]))

Rubi steps

\begin {align*} \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx &=\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}+\frac {1}{60} \int \frac {(1+2 x)^3 (-64+228 x)}{\sqrt {2-x+3 x^2}} \, dx\\ &=\frac {19}{60} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}+\frac {1}{720} \int \frac {(1+2 x)^2 (-3390+2112 x)}{\sqrt {2-x+3 x^2}} \, dx\\ &=\frac {44}{135} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {19}{60} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}+\frac {\int \frac {(-46350-37788 x) (1+2 x)}{\sqrt {2-x+3 x^2}} \, dx}{6480}\\ &=\frac {44}{135} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {19}{60} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}-\frac {(24897+6298 x) \sqrt {2-x+3 x^2}}{3240}-\frac {9211 \int \frac {1}{\sqrt {2-x+3 x^2}} \, dx}{1296}\\ &=\frac {44}{135} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {19}{60} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}-\frac {(24897+6298 x) \sqrt {2-x+3 x^2}}{3240}-\frac {9211 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+6 x\right )}{1296 \sqrt {69}}\\ &=\frac {44}{135} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {19}{60} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {2}{15} (1+2 x)^4 \sqrt {2-x+3 x^2}-\frac {(24897+6298 x) \sqrt {2-x+3 x^2}}{3240}+\frac {9211 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{1296 \sqrt {3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 60, normalized size = 0.50 \[ \frac {6 \sqrt {3 x^2-x+2} \left (6912 x^4+22032 x^3+26904 x^2+7538 x-22383\right )-46055 \sqrt {3} \sinh ^{-1}\left (\frac {6 x-1}{\sqrt {23}}\right )}{19440} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[2 - x + 3*x^2]*(-22383 + 7538*x + 26904*x^2 + 22032*x^3 + 6912*x^4) - 46055*Sqrt[3]*ArcSinh[(-1 + 6*x)
/Sqrt[23]])/19440

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 73, normalized size = 0.61 \[ \frac {1}{3240} \, {\left (6912 \, x^{4} + 22032 \, x^{3} + 26904 \, x^{2} + 7538 \, x - 22383\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {9211}{7776} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \]

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-x+2)^(1/2),x, algorithm="fricas")

[Out]

1/3240*(6912*x^4 + 22032*x^3 + 26904*x^2 + 7538*x - 22383)*sqrt(3*x^2 - x + 2) + 9211/7776*sqrt(3)*log(4*sqrt(
3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25)

________________________________________________________________________________________

giac [A]  time = 0.24, size = 68, normalized size = 0.57 \[ \frac {1}{3240} \, {\left (2 \, {\left (12 \, {\left (18 \, {\left (16 \, x + 51\right )} x + 1121\right )} x + 3769\right )} x - 22383\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {9211}{3888} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) \]

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-x+2)^(1/2),x, algorithm="giac")

[Out]

1/3240*(2*(12*(18*(16*x + 51)*x + 1121)*x + 3769)*x - 22383)*sqrt(3*x^2 - x + 2) + 9211/3888*sqrt(3)*log(-2*sq
rt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 96, normalized size = 0.80 \[ \frac {32 \sqrt {3 x^{2}-x +2}\, x^{4}}{15}+\frac {34 \sqrt {3 x^{2}-x +2}\, x^{3}}{5}+\frac {1121 \sqrt {3 x^{2}-x +2}\, x^{2}}{135}+\frac {3769 \sqrt {3 x^{2}-x +2}\, x}{1620}-\frac {9211 \sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{3888}-\frac {829 \sqrt {3 x^{2}-x +2}}{120} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*x^4*(3*x^2-x+2)^(1/2)+34/5*x^3*(3*x^2-x+2)^(1/2)+1121/135*x^2*(3*x^2-x+2)^(1/2)+3769/1620*x*(3*x^2-x+2)^
(1/2)-829/120*(3*x^2-x+2)^(1/2)-9211/3888*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 97, normalized size = 0.81 \[ \frac {32}{15} \, \sqrt {3 \, x^{2} - x + 2} x^{4} + \frac {34}{5} \, \sqrt {3 \, x^{2} - x + 2} x^{3} + \frac {1121}{135} \, \sqrt {3 \, x^{2} - x + 2} x^{2} + \frac {3769}{1620} \, \sqrt {3 \, x^{2} - x + 2} x - \frac {9211}{3888} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) - \frac {829}{120} \, \sqrt {3 \, x^{2} - x + 2} \]

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-x+2)^(1/2),x, algorithm="maxima")

[Out]

32/15*sqrt(3*x^2 - x + 2)*x^4 + 34/5*sqrt(3*x^2 - x + 2)*x^3 + 1121/135*sqrt(3*x^2 - x + 2)*x^2 + 3769/1620*sq
rt(3*x^2 - x + 2)*x - 9211/3888*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) - 829/120*sqrt(3*x^2 - x + 2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (2\,x+1\right )}^3\,\left (4\,x^2+3\,x+1\right )}{\sqrt {3\,x^2-x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 x + 1\right )^{3} \left (4 x^{2} + 3 x + 1\right )}{\sqrt {3 x^{2} - x + 2}}\, dx \]

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-x+2)**(1/2),x)

[Out]

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

________________________________________________________________________________________