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

Optimal. Leaf size=63 \[ -\frac {2 (367 x+73)}{207 \sqrt {3 x^2-x+2}}+\frac {8}{9} \sqrt {3 x^2-x+2}-\frac {14 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{3 \sqrt {3}} \]

[Out]

-14/9*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)-2/207*(73+367*x)/(3*x^2-x+2)^(1/2)+8/9*(3*x^2-x+2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1660, 640, 619, 215} \[ -\frac {2 (367 x+73)}{207 \sqrt {3 x^2-x+2}}+\frac {8}{9} \sqrt {3 x^2-x+2}-\frac {14 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(73 + 367*x))/(207*Sqrt[2 - x + 3*x^2]) + (8*Sqrt[2 - x + 3*x^2])/9 - (14*ArcSinh[(1 - 6*x)/Sqrt[23]])/(3*
Sqrt[3])

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 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 {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{3/2}} \, dx &=-\frac {2 (73+367 x)}{207 \sqrt {2-x+3 x^2}}+\frac {2}{23} \int \frac {\frac {437}{9}+\frac {92 x}{3}}{\sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {2 (73+367 x)}{207 \sqrt {2-x+3 x^2}}+\frac {8}{9} \sqrt {2-x+3 x^2}+\frac {14}{3} \int \frac {1}{\sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {2 (73+367 x)}{207 \sqrt {2-x+3 x^2}}+\frac {8}{9} \sqrt {2-x+3 x^2}+\frac {14 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+6 x\right )}{3 \sqrt {69}}\\ &=-\frac {2 (73+367 x)}{207 \sqrt {2-x+3 x^2}}+\frac {8}{9} \sqrt {2-x+3 x^2}-\frac {14 \sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 50, normalized size = 0.79 \[ \frac {2 \left (92 x^2-153 x+37\right )}{69 \sqrt {3 x^2-x+2}}+\frac {14 \sinh ^{-1}\left (\frac {6 x-1}{\sqrt {23}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(37 - 153*x + 92*x^2))/(69*Sqrt[2 - x + 3*x^2]) + (14*ArcSinh[(-1 + 6*x)/Sqrt[23]])/(3*Sqrt[3])

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fricas [A]  time = 0.65, size = 87, normalized size = 1.38 \[ \frac {161 \, \sqrt {3} {\left (3 \, x^{2} - x + 2\right )} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + 6 \, {\left (92 \, x^{2} - 153 \, x + 37\right )} \sqrt {3 \, x^{2} - x + 2}}{207 \, {\left (3 \, x^{2} - x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(3/2),x, algorithm="fricas")

[Out]

1/207*(161*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) + 6*(92*
x^2 - 153*x + 37)*sqrt(3*x^2 - x + 2))/(3*x^2 - x + 2)

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giac [A]  time = 0.27, size = 57, normalized size = 0.90 \[ -\frac {14}{9} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) + \frac {2 \, {\left ({\left (92 \, x - 153\right )} x + 37\right )}}{69 \, \sqrt {3 \, x^{2} - x + 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(3/2),x, algorithm="giac")

[Out]

-14/9*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1) + 2/69*((92*x - 153)*x + 37)/sqrt(3*x^2 -
x + 2)

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maple [A]  time = 0.01, size = 81, normalized size = 1.29 \[ \frac {8 x^{2}}{3 \sqrt {3 x^{2}-x +2}}-\frac {14 x}{3 \sqrt {3 x^{2}-x +2}}+\frac {14 \sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{9}+\frac {10}{9 \sqrt {3 x^{2}-x +2}}+\frac {\frac {16 x}{69}-\frac {8}{207}}{\sqrt {3 x^{2}-x +2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3/(3*x^2-x+2)^(1/2)*x^2-14/3/(3*x^2-x+2)^(1/2)*x+10/9/(3*x^2-x+2)^(1/2)+8/207*(6*x-1)/(3*x^2-x+2)^(1/2)+14/9
*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))

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maxima [A]  time = 0.94, size = 63, normalized size = 1.00 \[ \frac {8 \, x^{2}}{3 \, \sqrt {3 \, x^{2} - x + 2}} + \frac {14}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) - \frac {102 \, x}{23 \, \sqrt {3 \, x^{2} - x + 2}} + \frac {74}{69 \, \sqrt {3 \, x^{2} - x + 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(3/2),x, algorithm="maxima")

[Out]

8/3*x^2/sqrt(3*x^2 - x + 2) + 14/9*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) - 102/23*x/sqrt(3*x^2 - x + 2) + 7
4/69/sqrt(3*x^2 - x + 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\left (2\,x+1\right )\,\left (4\,x^2+3\,x+1\right )}{{\left (3\,x^2-x+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 x + 1\right ) \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} - x + 2\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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