3.120 \(\int \frac {\tan ^{-1}(x)}{(-1+x)^3} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4862, 710, 801, 260} \[ \frac {1}{8} \log \left (x^2+1\right )+\frac {1}{4 (1-x)}-\frac {1}{4} \log (1-x)-\frac {\tan ^{-1}(x)}{2 (1-x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]/(-1 + x)^3,x]

[Out]

1/(4*(1 - x)) - ArcTan[x]/(2*(1 - x)^2) - Log[1 - x]/4 + Log[1 + x^2]/8

Rule 260

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

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 35, normalized size = 0.78 \[ \frac {1}{8} \left (\log \left (x^2+1\right )-\frac {2}{x-1}-2 \log (1-x)-\frac {4 \tan ^{-1}(x)}{(x-1)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x]/(-1 + x)^3,x]

[Out]

(-2/(-1 + x) - (4*ArcTan[x])/(-1 + x)^2 - 2*Log[1 - x] + Log[1 + x^2])/8

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 50, normalized size = 1.11 \[ \frac {{\left (x^{2} - 2 \, x + 1\right )} \log \left (x^{2} + 1\right ) - 2 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 1\right ) - 2 \, x - 4 \, \arctan \relax (x) + 2}{8 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((x^2 - 2*x + 1)*log(x^2 + 1) - 2*(x^2 - 2*x + 1)*log(x - 1) - 2*x - 4*arctan(x) + 2)/(x^2 - 2*x + 1)

________________________________________________________________________________________

giac [A]  time = 0.11, size = 32, normalized size = 0.71 \[ -\frac {1}{4 \, {\left (x - 1\right )}} - \frac {\arctan \relax (x)}{2 \, {\left (x - 1\right )}^{2}} + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(x - 1) - 1/2*arctan(x)/(x - 1)^2 + 1/8*log(x^2 + 1) - 1/4*log(abs(x - 1))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 32, normalized size = 0.71 \[ -\frac {\arctan \relax (x )}{2 \left (x -1\right )^{2}}+\frac {\ln \left (x^{2}+1\right )}{8}-\frac {1}{4 \left (x -1\right )}-\frac {\ln \left (x -1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)/(x-1)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 31, normalized size = 0.69 \[ -\frac {1}{4 \, {\left (x - 1\right )}} - \frac {\arctan \relax (x)}{2 \, {\left (x - 1\right )}^{2}} + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(x - 1) - 1/2*arctan(x)/(x - 1)^2 + 1/8*log(x^2 + 1) - 1/4*log(x - 1)

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 31, normalized size = 0.69 \[ \frac {\ln \left (x^2+1\right )}{8}-\frac {\ln \left (x-1\right )}{4}-\frac {\frac {x}{4}+\frac {\mathrm {atan}\relax (x)}{2}-\frac {1}{4}}{{\left (x-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x)/(x - 1)^3,x)

[Out]

log(x^2 + 1)/8 - log(x - 1)/4 - (x/4 + atan(x)/2 - 1/4)/(x - 1)^2

________________________________________________________________________________________

sympy [B]  time = 0.52, size = 153, normalized size = 3.40 \[ - \frac {2 x^{2} \log {\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} + \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} + \frac {4 x \log {\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} - \frac {2 x \log {\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} - \frac {2 x}{8 x^{2} - 16 x + 8} - \frac {2 \log {\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} + \frac {\log {\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} - \frac {4 \operatorname {atan}{\relax (x )}}{8 x^{2} - 16 x + 8} + \frac {2}{8 x^{2} - 16 x + 8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x**2*log(x - 1)/(8*x**2 - 16*x + 8) + x**2*log(x**2 + 1)/(8*x**2 - 16*x + 8) + 4*x*log(x - 1)/(8*x**2 - 16*
x + 8) - 2*x*log(x**2 + 1)/(8*x**2 - 16*x + 8) - 2*x/(8*x**2 - 16*x + 8) - 2*log(x - 1)/(8*x**2 - 16*x + 8) +
log(x**2 + 1)/(8*x**2 - 16*x + 8) - 4*atan(x)/(8*x**2 - 16*x + 8) + 2/(8*x**2 - 16*x + 8)

________________________________________________________________________________________