3.82 \(\int \frac{2+3 x+5 x^2}{\sqrt{3-x+2 x^2}} \, dx\)

Optimal. Leaf size=59 \[ \frac{5}{4} \sqrt{2 x^2-x+3} x+\frac{39}{16} \sqrt{2 x^2-x+3}+\frac{17 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}} \]

[Out]

(39*Sqrt[3 - x + 2*x^2])/16 + (5*x*Sqrt[3 - x + 2*x^2])/4 + (17*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2])

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Rubi [A]  time = 0.032932, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1661, 640, 619, 215} \[ \frac{5}{4} \sqrt{2 x^2-x+3} x+\frac{39}{16} \sqrt{2 x^2-x+3}+\frac{17 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x + 5*x^2)/Sqrt[3 - x + 2*x^2],x]

[Out]

(39*Sqrt[3 - x + 2*x^2])/16 + (5*x*Sqrt[3 - x + 2*x^2])/4 + (17*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2])

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{2+3 x+5 x^2}{\sqrt{3-x+2 x^2}} \, dx &=\frac{5}{4} x \sqrt{3-x+2 x^2}+\frac{1}{4} \int \frac{-7+\frac{39 x}{2}}{\sqrt{3-x+2 x^2}} \, dx\\ &=\frac{39}{16} \sqrt{3-x+2 x^2}+\frac{5}{4} x \sqrt{3-x+2 x^2}-\frac{17}{32} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx\\ &=\frac{39}{16} \sqrt{3-x+2 x^2}+\frac{5}{4} x \sqrt{3-x+2 x^2}-\frac{17 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{32 \sqrt{46}}\\ &=\frac{39}{16} \sqrt{3-x+2 x^2}+\frac{5}{4} x \sqrt{3-x+2 x^2}+\frac{17 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0382121, size = 45, normalized size = 0.76 \[ \frac{1}{64} \left (4 \sqrt{2 x^2-x+3} (20 x+39)+17 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x + 5*x^2)/Sqrt[3 - x + 2*x^2],x]

[Out]

(4*(39 + 20*x)*Sqrt[3 - x + 2*x^2] + 17*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]])/64

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Maple [A]  time = 0.051, size = 45, normalized size = 0.8 \begin{align*}{\frac{5\,x}{4}\sqrt{2\,{x}^{2}-x+3}}+{\frac{39}{16}\sqrt{2\,{x}^{2}-x+3}}-{\frac{17\,\sqrt{2}}{64}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x)

[Out]

5/4*x*(2*x^2-x+3)^(1/2)+39/16*(2*x^2-x+3)^(1/2)-17/64*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

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Maxima [A]  time = 1.53221, size = 62, normalized size = 1.05 \begin{align*} \frac{5}{4} \, \sqrt{2 \, x^{2} - x + 3} x - \frac{17}{64} \, \sqrt{2} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{39}{16} \, \sqrt{2 \, x^{2} - x + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

5/4*sqrt(2*x^2 - x + 3)*x - 17/64*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 39/16*sqrt(2*x^2 - x + 3)

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Fricas [A]  time = 1.32083, size = 163, normalized size = 2.76 \begin{align*} \frac{1}{16} \, \sqrt{2 \, x^{2} - x + 3}{\left (20 \, x + 39\right )} + \frac{17}{128} \, \sqrt{2} \log \left (4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/16*sqrt(2*x^2 - x + 3)*(20*x + 39) + 17/128*sqrt(2)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 1
6*x - 25)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 x^{2} + 3 x + 2}{\sqrt{2 x^{2} - x + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*x+2)/(2*x**2-x+3)**(1/2),x)

[Out]

Integral((5*x**2 + 3*x + 2)/sqrt(2*x**2 - x + 3), x)

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Giac [A]  time = 1.15619, size = 72, normalized size = 1.22 \begin{align*} \frac{1}{16} \, \sqrt{2 \, x^{2} - x + 3}{\left (20 \, x + 39\right )} + \frac{17}{64} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

1/16*sqrt(2*x^2 - x + 3)*(20*x + 39) + 17/64*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)