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

Optimal. Leaf size=36 \[ \frac {1}{8 (1-x)}-\frac {1}{4 (x+1)}-\frac {1}{8 (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {44, 207} \[ \frac {1}{8 (1-x)}-\frac {1}{4 (x+1)}-\frac {1}{8 (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x)^2*(1 + x)^3),x]

[Out]

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

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

Rubi steps

\begin {align*} \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx &=\int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}+\frac {1}{4 (1+x)^2}-\frac {3}{8 \left (-1+x^2\right )}\right ) \, dx\\ &=\frac {1}{8 (1-x)}-\frac {1}{8 (1+x)^2}-\frac {1}{4 (1+x)}-\frac {3}{8} \int \frac {1}{-1+x^2} \, dx\\ &=\frac {1}{8 (1-x)}-\frac {1}{8 (1+x)^2}-\frac {1}{4 (1+x)}+\frac {3}{8} \tanh ^{-1}(x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 38, normalized size = 1.06 \[ \frac {1}{16} \left (\frac {-6 x^2-6 x+4}{(x-1) (x+1)^2}-3 \log (x-1)+3 \log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((-1 + x)^2*(1 + x)^3),x]

[Out]

((4 - 6*x - 6*x^2)/((-1 + x)*(1 + x)^2) - 3*Log[-1 + x] + 3*Log[1 + x])/16

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.03, size = 31, normalized size = 0.86 \[ \frac {-3 x^2-3 x+2}{8 (x-1) (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.11, size = 59, normalized size = 1.64 \[ -\frac {6 \, x^{2} - 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x - 1\right ) + 6 \, x - 4}{16 \, {\left (x^{3} + x^{2} - x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.06, size = 43, normalized size = 1.19 \[ -\frac {1}{8 \, {\left (x - 1\right )}} + \frac {\frac {12}{x - 1} + 5}{32 \, {\left (\frac {2}{x - 1} + 1\right )}^{2}} + \frac {3}{16} \, \log \left ({\left | -\frac {2}{x - 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/(x - 1) + 1/32*(12/(x - 1) + 5)/(2/(x - 1) + 1)^2 + 3/16*log(abs(-2/(x - 1) - 1))

________________________________________________________________________________________

maple [A]  time = 0.29, size = 35, normalized size = 0.97

method	result	size
default	\(-\frac {1}{8 \left (-1+x \right )}-\frac {3 \ln \left (-1+x \right )}{16}-\frac {1}{8 \left (1+x \right )^{2}}-\frac {1}{4 \left (1+x \right )}+\frac {3 \ln \left (1+x \right )}{16}\)	\(35\)
norman	\(\frac {-\frac {3}{8} x -\frac {3}{8} x^{2}+\frac {1}{4}}{\left (-1+x \right ) \left (1+x \right )^{2}}-\frac {3 \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (1+x \right )}{16}\)	\(35\)
risch	\(\frac {-\frac {3}{8} x -\frac {3}{8} x^{2}+\frac {1}{4}}{\left (-1+x \right ) \left (1+x \right )^{2}}-\frac {3 \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (1+x \right )}{16}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 38, normalized size = 1.06 \[ -\frac {3 \, x^{2} + 3 \, x - 2}{8 \, {\left (x^{3} + x^{2} - x - 1\right )}} + \frac {3}{16} \, \log \left (x + 1\right ) - \frac {3}{16} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 31, normalized size = 0.86 \[ \frac {3\,\mathrm {atanh}\relax (x)}{8}+\frac {\frac {3\,x^2}{8}+\frac {3\,x}{8}-\frac {1}{4}}{-x^3-x^2+x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x - 1)^2*(x + 1)^3),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 41, normalized size = 1.14 \[ \frac {- 3 x^{2} - 3 x + 2}{8 x^{3} + 8 x^{2} - 8 x - 8} - \frac {3 \log {\left (x - 1 \right )}}{16} + \frac {3 \log {\left (x + 1 \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*x**2 - 3*x + 2)/(8*x**3 + 8*x**2 - 8*x - 8) - 3*log(x - 1)/16 + 3*log(x + 1)/16

________________________________________________________________________________________