3.3.65 \(\int \frac {-2+3 x}{(1+x)^3 \sqrt {2 x-x^2}} \, dx\) [265]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 = {848, 820, 738, 210} \begin {gather*} \frac {\text {ArcTan}\left (\frac {1-2 x}{\sqrt {3} \sqrt {2 x-x^2}}\right )}{2 \sqrt {3}}-\frac {2 \sqrt {2 x-x^2}}{3 (x+1)}-\frac {5 \sqrt {2 x-x^2}}{6 (x+1)^2} \end {gather*}

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 210

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

Rule 738

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 820

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)/(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[S
implify[m + 2*p + 3], 0]

Rule 848

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])

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)}+\text {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*}

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Mathematica [A]
time = 0.22, size = 78, normalized size = 0.99 \begin {gather*} \frac {x \left (-18+x+4 x^2\right )-2 \sqrt {3} \sqrt {-2+x} \sqrt {x} (1+x)^2 \tanh ^{-1}\left (\frac {1-\sqrt {-2+x} \sqrt {x}+x}{\sqrt {3}}\right )}{6 \sqrt {-((-2+x) x)} (1+x)^2} \end {gather*}

Antiderivative was successfully verified.

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

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Maple [A]
time = 0.11, size = 74, normalized size = 0.94

method	result	size
risch	\(\frac {x \left (-2+x \right ) \left (4 x +9\right )}{6 \left (1+x \right )^{2} \sqrt {-x \left (-2+x \right )}}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-2+4 x \right ) \sqrt {3}}{6 \sqrt {-\left (1+x \right )^{2}+1+4 x}}\right )}{6}\)	\(56\)
trager	\(-\frac {\left (4 x +9\right ) \sqrt {-x^{2}+2 x}}{6 \left (1+x \right )^{2}}-\frac {\RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x +3 \sqrt {-x^{2}+2 x}+\RootOf \left (\textit {\_Z}^{2}+3\right )}{1+x}\right )}{6}\)	\(69\)
default	\(-\frac {5 \sqrt {-\left (1+x \right )^{2}+1+4 x}}{6 \left (1+x \right )^{2}}-\frac {2 \sqrt {-\left (1+x \right )^{2}+1+4 x}}{3 \left (1+x \right )}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-2+4 x \right ) \sqrt {3}}{6 \sqrt {-\left (1+x \right )^{2}+1+4 x}}\right )}{6}\)	\(74\)

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,method=_RETURNVERBOSE)

[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 = 3.73, size = 66, normalized size = 0.84 \begin {gather*} -\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 {gather*}

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 = 0.43, size = 64, normalized size = 0.81 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 x - 2}{\sqrt {- x \left (x - 2\right )} \left (x + 1\right )^{3}}\, dx \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (64) = 128\).
time = 1.00, size = 147, normalized size = 1.86 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {3\,x-2}{\sqrt {2\,x-x^2}\,{\left (x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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