3.3.76 \(\int \frac {\log (\frac {1-x^2}{1+x^2})}{(1+x)^2} \, dx\) [276]

Optimal. Leaf size=57 \[ -\frac {1}{1+x}-\tan ^{-1}(x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2605, 12, 2099, 266, 649, 209} \begin {gather*} -\text {ArcTan}(x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{x^2+1}\right )}{x+1}-\frac {1}{2} \log \left (x^2+1\right )-\frac {1}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[(1 - x^2)/(1 + x^2)]/(1 + x)^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

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

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 649

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

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx &=-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+\int \frac {4 x}{-1-x+x^4+x^5} \, dx\\ &=-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+4 \int \frac {x}{-1-x+x^4+x^5} \, dx\\ &=-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+4 \int \left (\frac {1}{4 (1+x)^2}+\frac {x}{4 \left (-1+x^2\right )}+\frac {-1-x}{4 \left (1+x^2\right )}\right ) \, dx\\ &=-\frac {1}{1+x}-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+\int \frac {x}{-1+x^2} \, dx+\int \frac {-1-x}{1+x^2} \, dx\\ &=-\frac {1}{1+x}+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\int \frac {1}{1+x^2} \, dx-\int \frac {x}{1+x^2} \, dx\\ &=-\frac {1}{1+x}-\tan ^{-1}(x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 57, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (-\frac {2}{1+x}-2 \tan ^{-1}(x)+\log (1-x)+\log (1+x)-\frac {2 \log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\log \left (1+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[(1 - x^2)/(1 + x^2)]/(1 + x)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.08, size = 112, normalized size = 1.96

method	result	size
risch	\(-\frac {\ln \left (\frac {-x^{2}+1}{x^{2}+1}\right )}{1+x}+\frac {i \ln \left (x -i\right ) x -i \ln \left (x +i\right ) x +i \ln \left (x -i\right )-i \ln \left (x +i\right )-\ln \left (x -i\right ) x -\ln \left (x +i\right ) x +\ln \left (x^{2}-1\right ) x -\ln \left (x -i\right )-\ln \left (x +i\right )+\ln \left (x^{2}-1\right )-2}{2+2 x}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln((-x^2+1)/(x^2+1))/(1+x)+1/2*(I*ln(x-I)*x-I*ln(x+I)*x+I*ln(x-I)-I*ln(x+I)-ln(x-I)*x-ln(x+I)*x+ln(x^2-1)*x-l
n(x-I)-ln(x+I)+ln(x^2-1)-2)/(1+x)

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 54, normalized size = 0.95 \begin {gather*} -\frac {\log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right )}{x + 1} - \frac {1}{x + 1} - \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 54, normalized size = 0.95 \begin {gather*} -\frac {2 \, {\left (x + 1\right )} \arctan \left (x\right ) + {\left (x + 1\right )} \log \left (x^{2} + 1\right ) - {\left (x + 1\right )} \log \left (x^{2} - 1\right ) + 2 \, \log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right ) + 2}{2 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.21, size = 41, normalized size = 0.72 \begin {gather*} \frac {\log {\left (x^{2} - 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \operatorname {atan}{\left (x \right )} - \frac {4}{4 x + 4} - \frac {\log {\left (\frac {1 - x^{2}}{x^{2} + 1} \right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 6.30, size = 56, normalized size = 0.98 \begin {gather*} -\frac {\log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right )}{x + 1} - \frac {1}{x + 1} - \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.42, size = 55, normalized size = 0.96 \begin {gather*} \frac {\ln \left (x^2-1\right )}{2}-\frac {\ln \left (x^2+1\right )}{2}-\mathrm {atan}\left (x\right )-\frac {1}{x+1}+\frac {\ln \left (x^2+1\right )}{x+1}-\frac {\ln \left (1-x^2\right )}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(-(x^2 - 1)/(x^2 + 1))/(x + 1)^2,x)

[Out]

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

________________________________________________________________________________________