3.1.82 \(\int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx\) [82]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {1675, 654, 633, 221} \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 221

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 633

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 654

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 1675

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 \text {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 = 0.16, size = 55, normalized size = 0.93 \begin {gather*} \frac {1}{64} \left (4 (39+20 x) \sqrt {3-x+2 x^2}+17 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\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]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/64

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Maple [A]
time = 0.10, size = 45, normalized size = 0.76

method	result	size
risch	\(\frac {\left (20 x +39\right ) \sqrt {2 x^{2}-x +3}}{16}-\frac {17 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}\)	\(35\)
default	\(\frac {5 x \sqrt {2 x^{2}-x +3}}{4}+\frac {39 \sqrt {2 x^{2}-x +3}}{16}-\frac {17 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}\)	\(45\)
trager	\(\left (\frac {5 x}{4}+\frac {39}{16}\right ) \sqrt {2 x^{2}-x +3}-\frac {17 \RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \RootOf \left (\textit {\_Z}^{2}-2\right ) x -\RootOf \left (\textit {\_Z}^{2}-2\right )+4 \sqrt {2 x^{2}-x +3}\right )}{64}\)	\(61\)

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,method=_RETURNVERBOSE)

[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 = 0.51, 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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Fricas [A]
time = 1.77, 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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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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Giac [A]
time = 4.93, 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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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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