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

Optimal. Leaf size=45 \[ -\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{2 \sqrt {2}} \]

[Out]

-5/4*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-11/23*(5+3*x)/(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 45, 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 = {1660, 12, 619, 215} \[ -\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(5 + 3*x))/(23*Sqrt[3 - x + 2*x^2]) - (5*ArcSinh[(1 - 4*x)/Sqrt[23]])/(2*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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
]

Rubi steps

\begin {align*} \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx &=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}+\frac {2}{23} \int \frac {115}{4 \sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}+\frac {5}{2} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{2 \sqrt {46}}\\ &=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 45, normalized size = 1.00 \[ \frac {5 \sinh ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{2 \sqrt {2}}-\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(5 + 3*x))/(23*Sqrt[3 - x + 2*x^2]) + (5*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(2*Sqrt[2])

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fricas [B]  time = 0.80, size = 82, normalized size = 1.82 \[ \frac {115 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) - 88 \, \sqrt {2 \, x^{2} - x + 3} {\left (3 \, x + 5\right )}}{184 \, {\left (2 \, x^{2} - x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/184*(115*sqrt(2)*(2*x^2 - x + 3)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) - 88*sqr
t(2*x^2 - x + 3)*(3*x + 5))/(2*x^2 - x + 3)

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giac [A]  time = 0.23, size = 53, normalized size = 1.18 \[ -\frac {5}{4} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) - \frac {11 \, {\left (3 \, x + 5\right )}}{23 \, \sqrt {2 \, x^{2} - x + 3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) - 11/23*(3*x + 5)/sqrt(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 64, normalized size = 1.42 \[ -\frac {5 x}{2 \sqrt {2 x^{2}-x +3}}+\frac {5 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{4}-\frac {17}{8 \sqrt {2 x^{2}-x +3}}+\frac {\frac {49 x}{46}-\frac {49}{184}}{\sqrt {2 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2/(2*x^2-x+3)^(1/2)*x-17/8/(2*x^2-x+3)^(1/2)+49/184*(4*x-1)/(2*x^2-x+3)^(1/2)+5/4*2^(1/2)*arcsinh(4/23*23^(
1/2)*(x-1/4))

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maxima [A]  time = 0.95, size = 46, normalized size = 1.02 \[ \frac {5}{4} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {33 \, x}{23 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {55}{23 \, \sqrt {2 \, x^{2} - x + 3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 33/23*x/sqrt(2*x^2 - x + 3) - 55/23/sqrt(2*x^2 - x + 3)

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mupad [B]  time = 0.23, size = 87, normalized size = 1.93 \[ \frac {5\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{4}+\frac {3\,\left (2\,x-12\right )}{23\,\sqrt {2\,x^2-x+3}}-\frac {10\,\left (\frac {11\,x}{2}+\frac {3}{2}\right )}{23\,\sqrt {2\,x^2-x+3}}+\frac {16\,x-4}{23\,\sqrt {2\,x^2-x+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/4 + (3*(2*x - 12))/(23*(2*x^2 - x + 3)^(1/2))
- (10*((11*x)/2 + 3/2))/(23*(2*x^2 - x + 3)^(1/2)) + (16*x - 4)/(23*(2*x^2 - x + 3)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {5 x^{2} + 3 x + 2}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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