\(\int \frac {1}{(-3-2 x+x^2)^3} \, dx\) [163]

Optimal result

Integrand size = 10, antiderivative size = 61 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {1-x}{16 \left (3+2 x-x^2\right )^2}+\frac {3 (1-x)}{128 \left (3+2 x-x^2\right )}+\frac {3}{512} \log (3-x)-\frac {3}{512} \log (1+x) \]

[Out]

1/16*(1-x)/(-x^2+2*x+3)^2+3/128*(1-x)/(-x^2+2*x+3)+3/512*ln(3-x)-3/512*ln(1+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {628, 630, 31} \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {3 (1-x)}{128 \left (-x^2+2 x+3\right )}+\frac {1-x}{16 \left (-x^2+2 x+3\right )^2}+\frac {3}{512} \log (3-x)-\frac {3}{512} \log (x+1) \]

[In]

Int[(-3 - 2*x + x^2)^(-3),x]

[Out]

(1 - x)/(16*(3 + 2*x - x^2)^2) + (3*(1 - x))/(128*(3 + 2*x - x^2)) + (3*Log[3 - x])/512 - (3*Log[1 + x])/512

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = \frac {1-x}{16 \left (3+2 x-x^2\right )^2}-\frac {3}{16} \int \frac {1}{\left (-3-2 x+x^2\right )^2} \, dx \\ & = \frac {1-x}{16 \left (3+2 x-x^2\right )^2}+\frac {3 (1-x)}{128 \left (3+2 x-x^2\right )}+\frac {3}{128} \int \frac {1}{-3-2 x+x^2} \, dx \\ & = \frac {1-x}{16 \left (3+2 x-x^2\right )^2}+\frac {3 (1-x)}{128 \left (3+2 x-x^2\right )}+\frac {3}{512} \int \frac {1}{-3+x} \, dx-\frac {3}{512} \int \frac {1}{1+x} \, dx \\ & = \frac {1-x}{16 \left (3+2 x-x^2\right )^2}+\frac {3 (1-x)}{128 \left (3+2 x-x^2\right )}+\frac {3}{512} \log (3-x)-\frac {3}{512} \log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {1}{512} \left (\frac {4 \left (17-11 x-9 x^2+3 x^3\right )}{\left (-3-2 x+x^2\right )^2}+3 \log (3-x)-3 \log (1+x)\right ) \]

[In]

Integrate[(-3 - 2*x + x^2)^(-3),x]

[Out]

((4*(17 - 11*x - 9*x^2 + 3*x^3))/(-3 - 2*x + x^2)^2 + 3*Log[3 - x] - 3*Log[1 + x])/512

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66

method	result	size
norman	\(\frac {-\frac {11}{128} x -\frac {9}{128} x^{2}+\frac {3}{128} x^{3}+\frac {17}{128}}{\left (x^{2}-2 x -3\right )^{2}}+\frac {3 \ln \left (-3+x \right )}{512}-\frac {3 \ln \left (1+x \right )}{512}\)	\(40\)
risch	\(\frac {-\frac {11}{128} x -\frac {9}{128} x^{2}+\frac {3}{128} x^{3}+\frac {17}{128}}{\left (x^{2}-2 x -3\right )^{2}}+\frac {3 \ln \left (-3+x \right )}{512}-\frac {3 \ln \left (1+x \right )}{512}\)	\(40\)
default	\(\frac {1}{128 \left (1+x \right )^{2}}+\frac {3}{256 \left (1+x \right )}-\frac {3 \ln \left (1+x \right )}{512}-\frac {1}{128 \left (-3+x \right )^{2}}+\frac {3}{256 \left (-3+x \right )}+\frac {3 \ln \left (-3+x \right )}{512}\)	\(42\)
parallelrisch	\(-\frac {3 \ln \left (1+x \right ) x^{4}-3 \ln \left (-3+x \right ) x^{4}-68-12 \ln \left (1+x \right ) x^{3}+12 \ln \left (-3+x \right ) x^{3}-6 \ln \left (1+x \right ) x^{2}+6 \ln \left (-3+x \right ) x^{2}-12 x^{3}+36 \ln \left (1+x \right ) x -36 \ln \left (-3+x \right ) x +36 x^{2}+27 \ln \left (1+x \right )-27 \ln \left (-3+x \right )+44 x}{512 \left (x^{2}-2 x -3\right )^{2}}\)	\(108\)

[In]

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

[Out]

(-11/128*x-9/128*x^2+3/128*x^3+17/128)/(x^2-2*x-3)^2+3/512*ln(-3+x)-3/512*ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {12 \, x^{3} - 36 \, x^{2} - 3 \, {\left (x^{4} - 4 \, x^{3} - 2 \, x^{2} + 12 \, x + 9\right )} \log \left (x + 1\right ) + 3 \, {\left (x^{4} - 4 \, x^{3} - 2 \, x^{2} + 12 \, x + 9\right )} \log \left (x - 3\right ) - 44 \, x + 68}{512 \, {\left (x^{4} - 4 \, x^{3} - 2 \, x^{2} + 12 \, x + 9\right )}} \]

[In]

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

[Out]

1/512*(12*x^3 - 36*x^2 - 3*(x^4 - 4*x^3 - 2*x^2 + 12*x + 9)*log(x + 1) + 3*(x^4 - 4*x^3 - 2*x^2 + 12*x + 9)*lo
g(x - 3) - 44*x + 68)/(x^4 - 4*x^3 - 2*x^2 + 12*x + 9)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {3 x^{3} - 9 x^{2} - 11 x + 17}{128 x^{4} - 512 x^{3} - 256 x^{2} + 1536 x + 1152} + \frac {3 \log {\left (x - 3 \right )}}{512} - \frac {3 \log {\left (x + 1 \right )}}{512} \]

[In]

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

[Out]

(3*x**3 - 9*x**2 - 11*x + 17)/(128*x**4 - 512*x**3 - 256*x**2 + 1536*x + 1152) + 3*log(x - 3)/512 - 3*log(x +
1)/512

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {3 \, x^{3} - 9 \, x^{2} - 11 \, x + 17}{128 \, {\left (x^{4} - 4 \, x^{3} - 2 \, x^{2} + 12 \, x + 9\right )}} - \frac {3}{512} \, \log \left (x + 1\right ) + \frac {3}{512} \, \log \left (x - 3\right ) \]

[In]

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

[Out]

1/128*(3*x^3 - 9*x^2 - 11*x + 17)/(x^4 - 4*x^3 - 2*x^2 + 12*x + 9) - 3/512*log(x + 1) + 3/512*log(x - 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=\frac {3 \, x^{3} - 9 \, x^{2} - 11 \, x + 17}{128 \, {\left (x^{2} - 2 \, x - 3\right )}^{2}} - \frac {3}{512} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {3}{512} \, \log \left ({\left | x - 3 \right |}\right ) \]

[In]

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

[Out]

1/128*(3*x^3 - 9*x^2 - 11*x + 17)/(x^2 - 2*x - 3)^2 - 3/512*log(abs(x + 1)) + 3/512*log(abs(x - 3))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (-3-2 x+x^2\right )^3} \, dx=-\frac {3\,\ln \left (\frac {x+1}{x-3}\right )}{512}-6\,\left (\frac {1}{256\,\left (-x^2+2\,x+3\right )}+\frac {1}{96\,{\left (-x^2+2\,x+3\right )}^2}\right )\,\left (x-1\right ) \]

[In]

int(-1/(2*x - x^2 + 3)^3,x)

[Out]

- (3*log((x + 1)/(x - 3)))/512 - 6*(1/(256*(2*x - x^2 + 3)) + 1/(96*(2*x - x^2 + 3)^2))*(x - 1)