\(\int \frac {1}{(x+\sqrt {-3-2 x+x^2})^3} \, dx\) [758]

Optimal result

Integrand size = 16, antiderivative size = 101 \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=-\frac {2}{1-x-\sqrt {-3-2 x+x^2}}+\frac {3}{4 \left (x+\sqrt {-3-2 x+x^2}\right )^2}+\frac {4}{x+\sqrt {-3-2 x+x^2}}+6 \log \left (1-x-\sqrt {-3-2 x+x^2}\right )-6 \log \left (x+\sqrt {-3-2 x+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2141, 907} \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=-\frac {2}{-\sqrt {x^2-2 x-3}-x+1}+\frac {4}{\sqrt {x^2-2 x-3}+x}+\frac {3}{4 \left (\sqrt {x^2-2 x-3}+x\right )^2}+6 \log \left (-\sqrt {x^2-2 x-3}-x+1\right )-6 \log \left (\sqrt {x^2-2 x-3}+x\right ) \]

[In]

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

[Out]

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2141

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[(g + h*x^n)^p*((d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2
*e*x)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && EqQ[e^2 -
c*f^2, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {-3-2 x+x^2}{x^3 (-2+2 x)^2} \, dx,x,x+\sqrt {-3-2 x+x^2}\right ) \\ & = 2 \text {Subst}\left (\int \left (-\frac {1}{(-1+x)^2}+\frac {3}{-1+x}-\frac {3}{4 x^3}-\frac {2}{x^2}-\frac {3}{x}\right ) \, dx,x,x+\sqrt {-3-2 x+x^2}\right ) \\ & = -\frac {2}{1-x-\sqrt {-3-2 x+x^2}}+\frac {3}{4 \left (x+\sqrt {-3-2 x+x^2}\right )^2}+\frac {4}{x+\sqrt {-3-2 x+x^2}}+6 \log \left (1-x-\sqrt {-3-2 x+x^2}\right )-6 \log \left (x+\sqrt {-3-2 x+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=-\frac {189+108 x-48 x^2-16 x^3+4 \sqrt {-3-2 x+x^2} \left (33+31 x+4 x^2\right )+96 (3+2 x)^2 \text {arctanh}\left (\frac {1+x}{2+2 x+\sqrt {-3-2 x+x^2}}\right )}{8 (3+2 x)^2} \]

[In]

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

[Out]

-1/8*(189 + 108*x - 48*x^2 - 16*x^3 + 4*Sqrt[-3 - 2*x + x^2]*(33 + 31*x + 4*x^2) + 96*(3 + 2*x)^2*ArcTanh[(1 +
x)/(2 + 2*x + Sqrt[-3 - 2*x + x^2])])/(3 + 2*x)^2

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68

method	result	size
trager	\(\frac {\left (4 x^{2}+33 x +36\right ) x}{2 \left (2 x +3\right )^{2}}-\frac {\left (4 x^{2}+31 x +33\right ) \sqrt {x^{2}-2 x -3}}{2 \left (2 x +3\right )^{2}}-6 \ln \left (x +3-\sqrt {x^{2}-2 x -3}\right )\)	\(69\)
default	\(-\frac {9}{2 x +3}-3 \ln \left (2 x +3\right )+\frac {x}{2}+\frac {27}{8 \left (2 x +3\right )^{2}}-\frac {\left (\left (x +\frac {3}{2}\right )^{2}-5 x -\frac {21}{4}\right )^{\frac {3}{2}}}{2 \left (x +\frac {3}{2}\right )}-\sqrt {4 \left (x +\frac {3}{2}\right )^{2}-20 x -21}+3 \,\operatorname {arctanh}\left (\frac {-2-\frac {10 x}{3}}{\sqrt {4 \left (x +\frac {3}{2}\right )^{2}-20 x -21}}\right )+\frac {\left (2 x -2\right ) \sqrt {\left (x +\frac {3}{2}\right )^{2}-5 x -\frac {21}{4}}}{4}+3 \ln \left (-1+x +\sqrt {\left (x +\frac {3}{2}\right )^{2}-5 x -\frac {21}{4}}\right )+\frac {\left (\left (x +\frac {3}{2}\right )^{2}-5 x -\frac {21}{4}\right )^{\frac {3}{2}}}{4 \left (x +\frac {3}{2}\right )^{2}}\)	\(146\)

[In]

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

[Out]

1/2*(4*x^2+33*x+36)*x/(2*x+3)^2-1/2*(4*x^2+31*x+33)/(2*x+3)^2*(x^2-2*x-3)^(1/2)-6*ln(x+3-(x^2-2*x-3)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=\frac {8 \, x^{3} - 10 \, x^{2} - 12 \, {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (x^{2} - \sqrt {x^{2} - 2 \, x - 3} {\left (x + 1\right )} - 3\right ) - 12 \, {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (2 \, x + 3\right ) + 12 \, {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-x + \sqrt {x^{2} - 2 \, x - 3}\right ) - 2 \, {\left (4 \, x^{2} + 31 \, x + 33\right )} \sqrt {x^{2} - 2 \, x - 3} - 156 \, x - 171}{4 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} \]

[In]

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

[Out]

1/4*(8*x^3 - 10*x^2 - 12*(4*x^2 + 12*x + 9)*log(x^2 - sqrt(x^2 - 2*x - 3)*(x + 1) - 3) - 12*(4*x^2 + 12*x + 9)
*log(2*x + 3) + 12*(4*x^2 + 12*x + 9)*log(-x + sqrt(x^2 - 2*x - 3)) - 2*(4*x^2 + 31*x + 33)*sqrt(x^2 - 2*x - 3
) - 156*x - 171)/(4*x^2 + 12*x + 9)

Sympy [F]

\[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=\int \frac {1}{\left (x + \sqrt {x^{2} - 2 x - 3}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=\int { \frac {1}{{\left (x + \sqrt {x^{2} - 2 \, x - 3}\right )}^{3}} \,d x } \]

[In]

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

[Out]

integrate((x + sqrt(x^2 - 2*x - 3))^(-3), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (85) = 170\).

Time = 0.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.82 \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - 2 \, x - 3} - \frac {104 \, {\left (x - \sqrt {x^{2} - 2 \, x - 3}\right )}^{3} + 315 \, {\left (x - \sqrt {x^{2} - 2 \, x - 3}\right )}^{2} + 162 \, x - 162 \, \sqrt {x^{2} - 2 \, x - 3} + 27}{8 \, {\left ({\left (x - \sqrt {x^{2} - 2 \, x - 3}\right )}^{2} + 3 \, x - 3 \, \sqrt {x^{2} - 2 \, x - 3}\right )}^{2}} - \frac {9 \, {\left (16 \, x + 21\right )}}{8 \, {\left (2 \, x + 3\right )}^{2}} - 3 \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - 3 \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} + 1 \right |}\right ) + 3 \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} \right |}\right ) - 3 \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} - 3 \right |}\right ) \]

[In]

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

[Out]

1/2*x - 1/2*sqrt(x^2 - 2*x - 3) - 1/8*(104*(x - sqrt(x^2 - 2*x - 3))^3 + 315*(x - sqrt(x^2 - 2*x - 3))^2 + 162
*x - 162*sqrt(x^2 - 2*x - 3) + 27)/((x - sqrt(x^2 - 2*x - 3))^2 + 3*x - 3*sqrt(x^2 - 2*x - 3))^2 - 9/8*(16*x +
21)/(2*x + 3)^2 - 3*log(abs(2*x + 3)) - 3*log(abs(-x + sqrt(x^2 - 2*x - 3) + 1)) + 3*log(abs(-x + sqrt(x^2 -
2*x - 3))) - 3*log(abs(-x + sqrt(x^2 - 2*x - 3) - 3))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (x+\sqrt {-3-2 x+x^2}\right )^3} \, dx=\int \frac {1}{{\left (x+\sqrt {x^2-2\,x-3}\right )}^3} \,d x \]

[In]

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

[Out]

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