3.265 \(\int \frac{-2+3 x}{(1+x)^3 \sqrt{2 x-x^2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{2 \sqrt{2 x-x^2}}{3 (x+1)}-\frac{5 \sqrt{2 x-x^2}}{6 (x+1)^2}+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3} \sqrt{2 x-x^2}}\right )}{2 \sqrt{3}} \]

[Out]

(-5*Sqrt[2*x - x^2])/(6*(1 + x)^2) - (2*Sqrt[2*x - x^2])/(3*(1 + x)) + ArcTan[(1 - 2*x)/(Sqrt[3]*Sqrt[2*x - x^
2])]/(2*Sqrt[3])

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Rubi [A]  time = 0.044347, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {834, 806, 724, 204} \[ -\frac{2 \sqrt{2 x-x^2}}{3 (x+1)}-\frac{5 \sqrt{2 x-x^2}}{6 (x+1)^2}+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3} \sqrt{2 x-x^2}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[2*x - x^2])/(6*(1 + x)^2) - (2*Sqrt[2*x - x^2])/(3*(1 + x)) + ArcTan[(1 - 2*x)/(Sqrt[3]*Sqrt[2*x - x^
2])]/(2*Sqrt[3])

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{-2+3 x}{(1+x)^3 \sqrt{2 x-x^2}} \, dx &=-\frac{5 \sqrt{2 x-x^2}}{6 (1+x)^2}+\frac{1}{6} \int \frac{-7+5 x}{(1+x)^2 \sqrt{2 x-x^2}} \, dx\\ &=-\frac{5 \sqrt{2 x-x^2}}{6 (1+x)^2}-\frac{2 \sqrt{2 x-x^2}}{3 (1+x)}-\frac{1}{2} \int \frac{1}{(1+x) \sqrt{2 x-x^2}} \, dx\\ &=-\frac{5 \sqrt{2 x-x^2}}{6 (1+x)^2}-\frac{2 \sqrt{2 x-x^2}}{3 (1+x)}+\operatorname{Subst}\left (\int \frac{1}{-12-x^2} \, dx,x,\frac{-2+4 x}{\sqrt{2 x-x^2}}\right )\\ &=-\frac{5 \sqrt{2 x-x^2}}{6 (1+x)^2}-\frac{2 \sqrt{2 x-x^2}}{3 (1+x)}-\frac{\tan ^{-1}\left (\frac{-2+4 x}{2 \sqrt{3} \sqrt{2 x-x^2}}\right )}{2 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0558168, size = 72, normalized size = 0.91 \[ \frac{x \left (4 x^2+x-18\right )-2 \sqrt{3} \sqrt{x-2} \sqrt{x} (x+1)^2 \tanh ^{-1}\left (\frac{\sqrt{\frac{x-2}{x}}}{\sqrt{3}}\right )}{6 \sqrt{-(x-2) x} (x+1)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-18 + x + 4*x^2) - 2*Sqrt[3]*Sqrt[-2 + x]*Sqrt[x]*(1 + x)^2*ArcTanh[Sqrt[(-2 + x)/x]/Sqrt[3]])/(6*Sqrt[-((
-2 + x)*x)]*(1 + x)^2)

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Maple [A]  time = 0.013, size = 74, normalized size = 0.9 \begin{align*} -{\frac{5}{6\, \left ( 1+x \right ) ^{2}}\sqrt{- \left ( 1+x \right ) ^{2}+1+4\,x}}-{\frac{2}{3+3\,x}\sqrt{- \left ( 1+x \right ) ^{2}+1+4\,x}}-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( -2+4\,x \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{- \left ( 1+x \right ) ^{2}+1+4\,x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/6/(1+x)^2*(-(1+x)^2+1+4*x)^(1/2)-2/3/(1+x)*(-(1+x)^2+1+4*x)^(1/2)-1/6*3^(1/2)*arctan(1/6*(-2+4*x)*3^(1/2)/(
-(1+x)^2+1+4*x)^(1/2))

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Maxima [A]  time = 1.48565, size = 89, normalized size = 1.13 \begin{align*} -\frac{1}{6} \, \sqrt{3} \arcsin \left (\frac{2 \, x}{{\left | x + 1 \right |}} - \frac{1}{{\left | x + 1 \right |}}\right ) - \frac{5 \, \sqrt{-x^{2} + 2 \, x}}{6 \,{\left (x^{2} + 2 \, x + 1\right )}} - \frac{2 \, \sqrt{-x^{2} + 2 \, x}}{3 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*arcsin(2*x/abs(x + 1) - 1/abs(x + 1)) - 5/6*sqrt(-x^2 + 2*x)/(x^2 + 2*x + 1) - 2/3*sqrt(-x^2 + 2*
x)/(x + 1)

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Fricas [A]  time = 2.08249, size = 158, normalized size = 2. \begin{align*} \frac{2 \, \sqrt{3}{\left (x^{2} + 2 \, x + 1\right )} \arctan \left (\frac{\sqrt{3} \sqrt{-x^{2} + 2 \, x}}{3 \, x}\right ) - \sqrt{-x^{2} + 2 \, x}{\left (4 \, x + 9\right )}}{6 \,{\left (x^{2} + 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(3)*(x^2 + 2*x + 1)*arctan(1/3*sqrt(3)*sqrt(-x^2 + 2*x)/x) - sqrt(-x^2 + 2*x)*(4*x + 9))/(x^2 + 2*x
+ 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 x - 2}{\sqrt{- x \left (x - 2\right )} \left (x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.07121, size = 198, normalized size = 2.51 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (\frac{2 \,{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}}{x - 1} - 1\right )}\right ) + \frac{\frac{34 \,{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}}{x - 1} - \frac{39 \,{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}^{2}}{{\left (x - 1\right )}^{2}} + \frac{18 \,{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}^{3}}{{\left (x - 1\right )}^{3}} - 26}{24 \,{\left (\frac{\sqrt{-x^{2} + 2 \, x} - 1}{x - 1} - \frac{{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}^{2}}{{\left (x - 1\right )}^{2}} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(sqrt(-x^2 + 2*x) - 1)/(x - 1) - 1)) + 1/24*(34*(sqrt(-x^2 + 2*x) - 1)/(x -
1) - 39*(sqrt(-x^2 + 2*x) - 1)^2/(x - 1)^2 + 18*(sqrt(-x^2 + 2*x) - 1)^3/(x - 1)^3 - 26)/((sqrt(-x^2 + 2*x) -
1)/(x - 1) - (sqrt(-x^2 + 2*x) - 1)^2/(x - 1)^2 - 1)^2