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

Optimal. Leaf size=70 \[ \frac {2}{9} \sqrt {3 x^2-x+2} (2 x+1)^2+\frac {1}{54} (62 x+69) \sqrt {3 x^2-x+2}+\frac {251 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{108 \sqrt {3}} \]

[Out]

251/324*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)+2/9*(1+2*x)^2*(3*x^2-x+2)^(1/2)+1/54*(69+62*x)*(3*x^2-x+2)^(1/2
)

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Rubi [A]  time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1653, 779, 619, 215} \[ \frac {2}{9} \sqrt {3 x^2-x+2} (2 x+1)^2+\frac {1}{54} (62 x+69) \sqrt {3 x^2-x+2}+\frac {251 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{108 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 + 2*x)^2*Sqrt[2 - x + 3*x^2])/9 + ((69 + 62*x)*Sqrt[2 - x + 3*x^2])/54 + (251*ArcSinh[(1 - 6*x)/Sqrt[23]
])/(108*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 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) \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx &=\frac {2}{9} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {1}{36} \int \frac {(1+2 x) (-24+124 x)}{\sqrt {2-x+3 x^2}} \, dx\\ &=\frac {2}{9} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {1}{54} (69+62 x) \sqrt {2-x+3 x^2}-\frac {251}{108} \int \frac {1}{\sqrt {2-x+3 x^2}} \, dx\\ &=\frac {2}{9} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {1}{54} (69+62 x) \sqrt {2-x+3 x^2}-\frac {251 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+6 x\right )}{108 \sqrt {69}}\\ &=\frac {2}{9} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {1}{54} (69+62 x) \sqrt {2-x+3 x^2}+\frac {251 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{108 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.71 \[ \frac {1}{324} \left (6 \sqrt {3 x^2-x+2} \left (48 x^2+110 x+81\right )-251 \sqrt {3} \sinh ^{-1}\left (\frac {6 x-1}{\sqrt {23}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[2 - x + 3*x^2]*(81 + 110*x + 48*x^2) - 251*Sqrt[3]*ArcSinh[(-1 + 6*x)/Sqrt[23]])/324

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fricas [A]  time = 0.76, size = 63, normalized size = 0.90 \[ \frac {1}{54} \, {\left (48 \, x^{2} + 110 \, x + 81\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {251}{648} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(48*x^2 + 110*x + 81)*sqrt(3*x^2 - x + 2) + 251/648*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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giac [A]  time = 0.26, size = 58, normalized size = 0.83 \[ \frac {1}{54} \, {\left (2 \, {\left (24 \, x + 55\right )} x + 81\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {251}{324} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(2*(24*x + 55)*x + 81)*sqrt(3*x^2 - x + 2) + 251/324*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x +
2)) + 1)

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maple [A]  time = 0.01, size = 62, normalized size = 0.89 \[ \frac {8 \sqrt {3 x^{2}-x +2}\, x^{2}}{9}+\frac {55 \sqrt {3 x^{2}-x +2}\, x}{27}-\frac {251 \sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{324}+\frac {3 \sqrt {3 x^{2}-x +2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/9*(3*x^2-x+2)^(1/2)*x^2+55/27*(3*x^2-x+2)^(1/2)*x+3/2*(3*x^2-x+2)^(1/2)-251/324*3^(1/2)*arcsinh(6/23*23^(1/2
)*(x-1/6))

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maxima [A]  time = 0.96, size = 63, normalized size = 0.90 \[ \frac {8}{9} \, \sqrt {3 \, x^{2} - x + 2} x^{2} + \frac {55}{27} \, \sqrt {3 \, x^{2} - x + 2} x - \frac {251}{324} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) + \frac {3}{2} \, \sqrt {3 \, x^{2} - x + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/9*sqrt(3*x^2 - x + 2)*x^2 + 55/27*sqrt(3*x^2 - x + 2)*x - 251/324*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) +
3/2*sqrt(3*x^2 - x + 2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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