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

Optimal. Leaf size=39 \[ -\frac{1}{8} \log \left (x^2+1\right )-\frac{1}{4 (x+1)}+\frac{1}{4} \log (x+1)-\frac{\tan ^{-1}(x)}{2 (x+1)^2} \]

[Out]

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

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Rubi [A]  time = 0.0293697, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4862, 710, 801, 260} \[ -\frac{1}{8} \log \left (x^2+1\right )-\frac{1}{4 (x+1)}+\frac{1}{4} \log (x+1)-\frac{\tan ^{-1}(x)}{2 (x+1)^2} \]

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 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]

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 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]

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 = 0.0277444, size = 35, normalized size = 0.9 \[ \frac{1}{8} \left (-\log \left (x^2+1\right )-\frac{2}{x+1}+2 \log (x+1)-\frac{4 \tan ^{-1}(x)}{(x+1)^2}\right ) \]

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

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Maple [A]  time = 0.009, size = 32, normalized size = 0.8 \begin{align*} -{\frac{1}{4+4\,x}}-{\frac{\arctan \left ( x \right ) }{2\, \left ( 1+x \right ) ^{2}}}+{\frac{\ln \left ( 1+x \right ) }{4}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.41133, size = 42, normalized size = 1.08 \begin{align*} -\frac{1}{4 \,{\left (x + 1\right )}} - \frac{\arctan \left (x\right )}{2 \,{\left (x + 1\right )}^{2}} - \frac{1}{8} \, \log \left (x^{2} + 1\right ) + \frac{1}{4} \, \log \left (x + 1\right ) \end{align*}

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)

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Fricas [A]  time = 2.79862, size = 146, normalized size = 3.74 \begin{align*} -\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 \left (x\right ) + 2}{8 \,{\left (x^{2} + 2 \, x + 1\right )}} \end{align*}

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)

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Sympy [B]  time = 0.662382, size = 153, normalized size = 3.92 \begin{align*} \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{x^{2}}{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 \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}{\left (x \right )}}{8 x^{2} + 16 x + 8} - \frac{1}{8 x^{2} + 16 x + 8} \end{align*}

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) + x**2/(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*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) - 1/(8*x**2 + 16*x + 8)

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Giac [A]  time = 1.07559, size = 43, normalized size = 1.1 \begin{align*} -\frac{1}{4 \,{\left (x + 1\right )}} - \frac{\arctan \left (x\right )}{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 ) \end{align*}

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))