3.7.97 \(\int \frac {\tan ^{-1}(x)}{(1+x)^3} \, dx\) [697]

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

[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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Rubi [A]
time = 0.02, antiderivative size = 39, 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 = {4972, 724, 815, 266} \begin {gather*} -\frac {\text {ArcTan}(x)}{2 (x+1)^2}-\frac {1}{8} \log \left (x^2+1\right )-\frac {1}{4 (x+1)}+\frac {1}{4} \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]/(1 + x)^3,x]

[Out]

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

Rule 266

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 724

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 815

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 4972

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*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.90 \begin {gather*} \frac {1}{8} \left (-\frac {2}{1+x}-\frac {4 \tan ^{-1}(x)}{(1+x)^2}+2 \log (1+x)-\log \left (1+x^2\right )\right ) \end {gather*}

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 = 0.10, size = 32, normalized size = 0.82

method	result	size
default	\(-\frac {1}{4 \left (1+x \right )}-\frac {\arctan \left (x \right )}{2 \left (1+x \right )^{2}}+\frac {\ln \left (1+x \right )}{4}-\frac {\ln \left (x^{2}+1\right )}{8}\)	\(32\)
risch	\(\frac {i \ln \left (i x +1\right )}{4 \left (1+x \right )^{2}}-\frac {i \left (2 i \ln \left (1+x \right ) x^{2}-i \ln \left (x^{2}+1\right ) x^{2}+4 i \ln \left (1+x \right ) x -2 i \ln \left (x^{2}+1\right ) x +2 i \ln \left (1+x \right )-i \ln \left (x^{2}+1\right )-2 i x -2 i+2 \ln \left (-i x +1\right )\right )}{8 \left (1+x \right )^{2}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.80, size = 31, normalized size = 0.79 \begin {gather*} -\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 {gather*}

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 = 0.53, size = 50, normalized size = 1.28 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (31) = 62\).
time = 0.25, size = 153, normalized size = 3.92 \begin {gather*} \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}{\left (x \right )}}{8 x^{2} + 16 x + 8} - \frac {2}{8 x^{2} + 16 x + 8} \end {gather*}

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) - l
og(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)

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Giac [A]
time = 0.97, size = 32, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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))

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Mupad [B]
time = 0.36, size = 31, normalized size = 0.79 \begin {gather*} \frac {\ln \left (x+1\right )}{4}-\frac {\ln \left (x^2+1\right )}{8}-\frac {\frac {x}{4}+\frac {\mathrm {atan}\left (x\right )}{2}+\frac {1}{4}}{{\left (x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x)/(x + 1)^3,x)

[Out]

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

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