3.247 $$\int \frac{(1+2 x)^2 (1+3 x+4 x^2)}{(2-x+3 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=82 $\frac{2 (1249-2273 x)}{621 \sqrt{3 x^2-x+2}}+\frac{8}{9} x \sqrt{3 x^2-x+2}+\frac{112}{27} \sqrt{3 x^2-x+2}-\frac{64 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{9 \sqrt{3}}$

[Out]

(2*(1249 - 2273*x))/(621*Sqrt[2 - x + 3*x^2]) + (112*Sqrt[2 - x + 3*x^2])/27 + (8*x*Sqrt[2 - x + 3*x^2])/9 - (
64*ArcSinh[(1 - 6*x)/Sqrt[23]])/(9*Sqrt[3])

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Rubi [A]  time = 0.108118, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.156, Rules used = {1660, 1661, 640, 619, 215} $\frac{2 (1249-2273 x)}{621 \sqrt{3 x^2-x+2}}+\frac{8}{9} x \sqrt{3 x^2-x+2}+\frac{112}{27} \sqrt{3 x^2-x+2}-\frac{64 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{9 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(1249 - 2273*x))/(621*Sqrt[2 - x + 3*x^2]) + (112*Sqrt[2 - x + 3*x^2])/27 + (8*x*Sqrt[2 - x + 3*x^2])/9 - (
64*ArcSinh[(1 - 6*x)/Sqrt[23]])/(9*Sqrt[3])

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 640

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

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 \frac{(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{3/2}} \, dx &=\frac{2 (1249-2273 x)}{621 \sqrt{2-x+3 x^2}}+\frac{2}{23} \int \frac{\frac{2116}{27}+\frac{1150 x}{9}+\frac{184 x^2}{3}}{\sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2 (1249-2273 x)}{621 \sqrt{2-x+3 x^2}}+\frac{8}{9} x \sqrt{2-x+3 x^2}+\frac{1}{69} \int \frac{\frac{3128}{9}+\frac{2576 x}{3}}{\sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2 (1249-2273 x)}{621 \sqrt{2-x+3 x^2}}+\frac{112}{27} \sqrt{2-x+3 x^2}+\frac{8}{9} x \sqrt{2-x+3 x^2}+\frac{64}{9} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2 (1249-2273 x)}{621 \sqrt{2-x+3 x^2}}+\frac{112}{27} \sqrt{2-x+3 x^2}+\frac{8}{9} x \sqrt{2-x+3 x^2}+\frac{64 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{9 \sqrt{69}}\\ &=\frac{2 (1249-2273 x)}{621 \sqrt{2-x+3 x^2}}+\frac{112}{27} \sqrt{2-x+3 x^2}+\frac{8}{9} x \sqrt{2-x+3 x^2}-\frac{64 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{9 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0318988, size = 61, normalized size = 0.74 $\frac{2 \left (828 x^3+3588 x^2+736 \sqrt{9 x^2-3 x+6} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )-3009 x+3825\right )}{621 \sqrt{3 x^2-x+2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(3825 - 3009*x + 3588*x^2 + 828*x^3 + 736*Sqrt[6 - 3*x + 9*x^2]*ArcSinh[(-1 + 6*x)/Sqrt[23]]))/(621*Sqrt[2
- x + 3*x^2])

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Maple [A]  time = 0.053, size = 98, normalized size = 1.2 \begin{align*}{\frac{8\,{x}^{3}}{3}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{104\,{x}^{2}}{9}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}-{\frac{64\,x}{9}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{107}{9}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}-{\frac{-89+534\,x}{207}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{64\,\sqrt{3}}{27}{\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)^(3/2),x)

[Out]

8/3*x^3/(3*x^2-x+2)^(1/2)+104/9*x^2/(3*x^2-x+2)^(1/2)-64/9*x/(3*x^2-x+2)^(1/2)+107/9/(3*x^2-x+2)^(1/2)-89/207*
(-1+6*x)/(3*x^2-x+2)^(1/2)+64/27*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))

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Maxima [A]  time = 1.55439, size = 108, normalized size = 1.32 \begin{align*} \frac{8 \, x^{3}}{3 \, \sqrt{3 \, x^{2} - x + 2}} + \frac{104 \, x^{2}}{9 \, \sqrt{3 \, x^{2} - x + 2}} + \frac{64}{27} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (6 \, x - 1\right )}\right ) - \frac{2006 \, x}{207 \, \sqrt{3 \, x^{2} - x + 2}} + \frac{850}{69 \, \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)^(3/2),x, algorithm="maxima")

[Out]

8/3*x^3/sqrt(3*x^2 - x + 2) + 104/9*x^2/sqrt(3*x^2 - x + 2) + 64/27*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) -
2006/207*x/sqrt(3*x^2 - x + 2) + 850/69/sqrt(3*x^2 - x + 2)

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Fricas [A]  time = 1.12729, size = 244, normalized size = 2.98 \begin{align*} \frac{2 \,{\left (368 \, \sqrt{3}{\left (3 \, x^{2} - x + 2\right )} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + 3 \,{\left (276 \, x^{3} + 1196 \, x^{2} - 1003 \, x + 1275\right )} \sqrt{3 \, x^{2} - x + 2}\right )}}{621 \,{\left (3 \, x^{2} - 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-x+2)^(3/2),x, algorithm="fricas")

[Out]

2/621*(368*sqrt(3)*(3*x^2 - x + 2)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 3*(276
*x^3 + 1196*x^2 - 1003*x + 1275)*sqrt(3*x^2 - x + 2))/(3*x^2 - x + 2)

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

[Out]

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

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Giac [A]  time = 1.15107, size = 84, normalized size = 1.02 \begin{align*} -\frac{64}{27} \, \sqrt{3} \log \left (-2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )} + 1\right ) + \frac{2 \,{\left ({\left (92 \,{\left (3 \, x + 13\right )} x - 1003\right )} x + 1275\right )}}{207 \, \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)^(3/2),x, algorithm="giac")

[Out]

-64/27*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1) + 2/207*((92*(3*x + 13)*x - 1003)*x + 127
5)/sqrt(3*x^2 - x + 2)