\(\int \frac {1}{2} \log ((-a^2+x^2)^2) \, dx\) [8]

Optimal result

Integrand size = 16, antiderivative size = 30 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=-2 x+2 a \text {arctanh}\left (\frac {x}{a}\right )+\frac {1}{2} x \log \left (\left (-a^2+x^2\right )^2\right ) \]

[Out]

-2*x+2*a*arctanh(x/a)+1/2*x*ln((-a^2+x^2)^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 2498, 327, 213} \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=\frac {1}{2} x \log \left (\left (x^2-a^2\right )^2\right )+2 a \text {arctanh}\left (\frac {x}{a}\right )-2 x \]

[In]

Int[Log[(-a^2 + x^2)^2]/2,x]

[Out]

-2*x + 2*a*ArcTanh[x/a] + (x*Log[(-a^2 + x^2)^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 213

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \log \left (\left (-a^2+x^2\right )^2\right ) \, dx \\ & = \frac {1}{2} x \log \left (\left (-a^2+x^2\right )^2\right )-2 \int \frac {x^2}{-a^2+x^2} \, dx \\ & = -2 x+\frac {1}{2} x \log \left (\left (-a^2+x^2\right )^2\right )-\left (2 a^2\right ) \int \frac {1}{-a^2+x^2} \, dx \\ & = -2 x+2 a \text {arctanh}\left (\frac {x}{a}\right )+\frac {1}{2} x \log \left (\left (-a^2+x^2\right )^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=\frac {1}{2} \left (-4 x+4 a \text {arctanh}\left (\frac {x}{a}\right )+x \log \left (\left (a^2-x^2\right )^2\right )\right ) \]

[In]

Integrate[Log[(-a^2 + x^2)^2]/2,x]

[Out]

(-4*x + 4*a*ArcTanh[x/a] + x*Log[(a^2 - x^2)^2])/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {x \ln \left (\left (a^{2}-x^{2}\right )^{2}\right )}{2}-2 x -a \ln \left (a -x \right )+a \ln \left (a +x \right )\)	\(35\)
risch	\(\frac {x \ln \left (\left (-a^{2}+x^{2}\right )^{2}\right )}{2}-2 x +a \ln \left (a +x \right )-a \ln \left (-a +x \right )\)	\(35\)
parts	\(\frac {x \ln \left (\left (-a^{2}+x^{2}\right )^{2}\right )}{2}-2 x -a \ln \left (a -x \right )+a \ln \left (a +x \right )\)	\(35\)
parallelrisch	\(-2 a \ln \left (-a +x \right )+\frac {x \ln \left (\left (a^{2}-x^{2}\right )^{2}\right )}{2}+\frac {\ln \left (\left (a^{2}-x^{2}\right )^{2}\right ) a}{2}-2 x\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=\frac {1}{2} \, x \log \left (a^{4} - 2 \, a^{2} x^{2} + x^{4}\right ) + a \log \left (a + x\right ) - a \log \left (-a + x\right ) - 2 \, x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=- 2 a \left (\frac {\log {\left (- a + x \right )}}{2} - \frac {\log {\left (a + x \right )}}{2}\right ) + \frac {x \log {\left (\left (- a^{2} + x^{2}\right )^{2} \right )}}{2} - 2 x \]

[In]

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

[Out]

-2*a*(log(-a + x)/2 - log(a + x)/2) + x*log((-a**2 + x**2)**2)/2 - 2*x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=\frac {1}{2} \, x \log \left ({\left (a^{2} - x^{2}\right )}^{2}\right ) + a \log \left (a + x\right ) - a \log \left (-a + x\right ) - 2 \, x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=\frac {1}{2} \, x \log \left ({\left (a^{2} - x^{2}\right )}^{2}\right ) + a \log \left ({\left | a + x \right |}\right ) - a \log \left ({\left | -a + x \right |}\right ) - 2 \, x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {1}{2} \log \left (\left (-a^2+x^2\right )^2\right ) \, dx=2\,a\,\mathrm {atanh}\left (\frac {x}{a}\right )-2\,x+\frac {x\,\ln \left ({\left (a^2-x^2\right )}^2\right )}{2} \]

[In]

int(log((a^2 - x^2)^2)/2,x)

[Out]

2*a*atanh(x/a) - 2*x + (x*log((a^2 - x^2)^2))/2