3.97 \(\int \frac{1}{\left (1-(1+x)^2\right )^3} \, dx\)

Optimal. Leaf size=45 \[ \frac{3 (x+1)}{8 \left (1-(x+1)^2\right )}+\frac{x+1}{4 \left (1-(x+1)^2\right )^2}+\frac{3}{8} \tanh ^{-1}(x+1) \]

[Out]

(1 + x)/(4*(1 - (1 + x)^2)^2) + (3*(1 + x))/(8*(1 - (1 + x)^2)) + (3*ArcTanh[1 +
x])/8

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Rubi [A]  time = 0.0301542, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273 \[ \frac{3 (x+1)}{8 \left (1-(x+1)^2\right )}+\frac{x+1}{4 \left (1-(x+1)^2\right )^2}+\frac{3}{8} \tanh ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]  Int[(1 - (1 + x)^2)^(-3),x]

[Out]

(1 + x)/(4*(1 - (1 + x)^2)^2) + (3*(1 + x))/(8*(1 - (1 + x)^2)) + (3*ArcTanh[1 +
x])/8

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Rubi in Sympy [A]  time = 1.66059, size = 34, normalized size = 0.76 \[ \frac{3 \left (x + 1\right )}{8 \left (- \left (x + 1\right )^{2} + 1\right )} + \frac{x + 1}{4 \left (- \left (x + 1\right )^{2} + 1\right )^{2}} + \frac{3 \operatorname{atanh}{\left (x + 1 \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/(1-(1+x)**2)**3,x)

[Out]

3*(x + 1)/(8*(-(x + 1)**2 + 1)) + (x + 1)/(4*(-(x + 1)**2 + 1)**2) + 3*atanh(x +
1)/8

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Mathematica [A]  time = 0.0289405, size = 37, normalized size = 0.82 \[ \frac{1}{16} \left (\frac{1}{x^2}-\frac{3}{x}-\frac{3}{x+2}-\frac{1}{(x+2)^2}-3 \log (x)+3 \log (x+2)\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[(1 - (1 + x)^2)^(-3),x]

[Out]

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

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Maple [A]  time = 0.013, size = 36, normalized size = 0.8 \[ -{\frac{1}{16\, \left ( 2+x \right ) ^{2}}}-{\frac{3}{32+16\,x}}+{\frac{3\,\ln \left ( 2+x \right ) }{16}}+{\frac{1}{16\,{x}^{2}}}-{\frac{3}{16\,x}}-{\frac{3\,\ln \left ( x \right ) }{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/(1-(1+x)^2)^3,x)

[Out]

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

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Maxima [A]  time = 0.806942, size = 59, normalized size = 1.31 \[ -\frac{3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} + \frac{3}{16} \, \log \left (x + 2\right ) - \frac{3}{16} \, \log \left (x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/8*(3*x^3 + 9*x^2 + 4*x - 2)/(x^4 + 4*x^3 + 4*x^2) + 3/16*log(x + 2) - 3/16*lo
g(x)

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Fricas [A]  time = 0.265981, size = 96, normalized size = 2.13 \[ -\frac{6 \, x^{3} + 18 \, x^{2} - 3 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x + 2\right ) + 3 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x\right ) + 8 \, x - 4}{16 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/16*(6*x^3 + 18*x^2 - 3*(x^4 + 4*x^3 + 4*x^2)*log(x + 2) + 3*(x^4 + 4*x^3 + 4*
x^2)*log(x) + 8*x - 4)/(x^4 + 4*x^3 + 4*x^2)

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Sympy [A]  time = 0.339256, size = 44, normalized size = 0.98 \[ - \frac{3 \log{\left (x \right )}}{16} + \frac{3 \log{\left (x + 2 \right )}}{16} - \frac{3 x^{3} + 9 x^{2} + 4 x - 2}{8 x^{4} + 32 x^{3} + 32 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(1-(1+x)**2)**3,x)

[Out]

-3*log(x)/16 + 3*log(x + 2)/16 - (3*x**3 + 9*x**2 + 4*x - 2)/(8*x**4 + 32*x**3 +
32*x**2)

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GIAC/XCAS [A]  time = 0.261994, size = 53, normalized size = 1.18 \[ -\frac{3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \,{\left (x^{2} + 2 \, x\right )}^{2}} + \frac{3}{16} \,{\rm ln}\left ({\left | x + 2 \right |}\right ) - \frac{3}{16} \,{\rm ln}\left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/8*(3*x^3 + 9*x^2 + 4*x - 2)/(x^2 + 2*x)^2 + 3/16*ln(abs(x + 2)) - 3/16*ln(abs
(x))