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

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

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1661, 640, 619, 215} \begin {gather*} \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}} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.76 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.28, size = 60, normalized size = 1.02 \begin {gather*} \frac {1}{16} \sqrt {2 x^2-x+3} (20 x+39)+\frac {17 \log \left (2 \sqrt {2} \sqrt {2 x^2-x+3}-4 x+1\right )}{32 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((39 + 20*x)*Sqrt[3 - x + 2*x^2])/16 + (17*Log[1 - 4*x + 2*Sqrt[2]*Sqrt[3 - x + 2*x^2]])/(32*Sqrt[2])

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fricas [A]  time = 0.41, size = 58, normalized size = 0.98 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.53, size = 53, normalized size = 0.90 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 45, normalized size = 0.76 \begin {gather*} \frac {5 \sqrt {2 x^{2}-x +3}\, x}{4}-\frac {17 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}+\frac {39 \sqrt {2 x^{2}-x +3}}{16} \end {gather*}

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*(2*x^2-x+3)^(1/2)*x+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 = 0.96, size = 46, normalized size = 0.78 \begin {gather*} \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 {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {5\,x^2+3\,x+2}{\sqrt {2\,x^2-x+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 x^{2} + 3 x + 2}{\sqrt {2 x^{2} - x + 3}}\, dx \end {gather*}

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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