Integrand size = 23, antiderivative size = 165 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\frac {2 \sqrt {1-x^2}}{35 x^5}+\frac {4 \sqrt {1-x^2}}{21 x^3}+\frac {88 \sqrt {1-x^2}}{105 x}-\frac {32 \arcsin (x)}{35}-\frac {\sqrt {1-x^2} \log \left (1-x^2\right )}{7 x^7}-\frac {6 \sqrt {1-x^2} \log \left (1-x^2\right )}{35 x^5}-\frac {8 \sqrt {1-x^2} \log \left (1-x^2\right )}{35 x^3}-\frac {16 \sqrt {1-x^2} \log \left (1-x^2\right )}{35 x} \] Output:
2/35*(-x^2+1)^(1/2)/x^5+4/21*(-x^2+1)^(1/2)/x^3+88/105*(-x^2+1)^(1/2)/x-32 /35*arcsin(x)-1/7*(-x^2+1)^(1/2)*ln(-x^2+1)/x^7-6/35*(-x^2+1)^(1/2)*ln(-x^ 2+1)/x^5-8/35*(-x^2+1)^(1/2)*ln(-x^2+1)/x^3-16/35*(-x^2+1)^(1/2)*ln(-x^2+1 )/x
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.41 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\frac {1}{105} \left (-96 \arcsin (x)+\frac {\sqrt {1-x^2} \left (6 x^2+20 x^4+88 x^6-3 \left (5+6 x^2+8 x^4+16 x^6\right ) \log \left (1-x^2\right )\right )}{x^7}\right ) \] Input:
Integrate[Log[1 - x^2]/(x^8*Sqrt[1 - x^2]),x]
Output:
(-96*ArcSin[x] + (Sqrt[1 - x^2]*(6*x^2 + 20*x^4 + 88*x^6 - 3*(5 + 6*x^2 + 8*x^4 + 16*x^6)*Log[1 - x^2]))/x^7)/105
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx\) |
\(\Big \downarrow \) 2929 |
\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}}dx\) |
Input:
Int[Log[1 - x^2]/(x^8*Sqrt[1 - x^2]),x]
Output:
$Aborted
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((h_.)* (x_))^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Unintegrable[(h*x) ^m*(f + g*x^s)^r*(a + b*Log[c*(d + e*x^n)^p])^q, x] /; FreeQ[{a, b, c, d, e , f, g, h, m, n, p, q, r, s}, x]
\[\int \frac {\ln \left (-x^{2}+1\right )}{x^{8} \sqrt {-x^{2}+1}}d x\]
Input:
int(ln(-x^2+1)/x^8/(-x^2+1)^(1/2),x)
Output:
int(ln(-x^2+1)/x^8/(-x^2+1)^(1/2),x)
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.48 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\frac {192 \, x^{7} \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) + {\left (88 \, x^{6} + 20 \, x^{4} + 6 \, x^{2} - 3 \, {\left (16 \, x^{6} + 8 \, x^{4} + 6 \, x^{2} + 5\right )} \log \left (-x^{2} + 1\right )\right )} \sqrt {-x^{2} + 1}}{105 \, x^{7}} \] Input:
integrate(log(-x^2+1)/x^8/(-x^2+1)^(1/2),x, algorithm="fricas")
Output:
1/105*(192*x^7*arctan((sqrt(-x^2 + 1) - 1)/x) + (88*x^6 + 20*x^4 + 6*x^2 - 3*(16*x^6 + 8*x^4 + 6*x^2 + 5)*log(-x^2 + 1))*sqrt(-x^2 + 1))/x^7
Timed out. \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\text {Timed out} \] Input:
integrate(ln(-x**2+1)/x**8/(-x**2+1)**(1/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.69 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=-\frac {1}{35} \, {\left (\frac {16 \, \sqrt {-x^{2} + 1}}{x} + \frac {8 \, \sqrt {-x^{2} + 1}}{x^{3}} + \frac {6 \, \sqrt {-x^{2} + 1}}{x^{5}} + \frac {5 \, \sqrt {-x^{2} + 1}}{x^{7}}\right )} \log \left (-x^{2} + 1\right ) + \frac {88 \, \sqrt {-x^{2} + 1}}{105 \, x} + \frac {4 \, \sqrt {-x^{2} + 1}}{21 \, x^{3}} + \frac {2 \, \sqrt {-x^{2} + 1}}{35 \, x^{5}} - \frac {32}{35} \, \arcsin \left (x\right ) \] Input:
integrate(log(-x^2+1)/x^8/(-x^2+1)^(1/2),x, algorithm="maxima")
Output:
-1/35*(16*sqrt(-x^2 + 1)/x + 8*sqrt(-x^2 + 1)/x^3 + 6*sqrt(-x^2 + 1)/x^5 + 5*sqrt(-x^2 + 1)/x^7)*log(-x^2 + 1) + 88/105*sqrt(-x^2 + 1)/x + 4/21*sqrt (-x^2 + 1)/x^3 + 2/35*sqrt(-x^2 + 1)/x^5 - 32/35*arcsin(x)
Time = 0.14 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.62 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\frac {1}{4480} \, {\left (\frac {x^{7} {\left (\frac {49 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + \frac {245 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{x^{4}} + \frac {1225 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{6}}{x^{6}} + 5\right )}}{{\left (\sqrt {-x^{2} + 1} - 1\right )}^{7}} - \frac {1225 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {245 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{x^{3}} - \frac {49 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}}{x^{5}} - \frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{7}}{x^{7}}\right )} \log \left (-x^{2} + 1\right ) - \frac {x^{5} {\left (\frac {49 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + \frac {750 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{x^{4}} + 3\right )}}{1680 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}} + \frac {25 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{56 \, x} + \frac {7 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{240 \, x^{3}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}}{560 \, x^{5}} - \frac {32}{35} \, \arcsin \left (x\right ) \] Input:
integrate(log(-x^2+1)/x^8/(-x^2+1)^(1/2),x, algorithm="giac")
Output:
1/4480*(x^7*(49*(sqrt(-x^2 + 1) - 1)^2/x^2 + 245*(sqrt(-x^2 + 1) - 1)^4/x^ 4 + 1225*(sqrt(-x^2 + 1) - 1)^6/x^6 + 5)/(sqrt(-x^2 + 1) - 1)^7 - 1225*(sq rt(-x^2 + 1) - 1)/x - 245*(sqrt(-x^2 + 1) - 1)^3/x^3 - 49*(sqrt(-x^2 + 1) - 1)^5/x^5 - 5*(sqrt(-x^2 + 1) - 1)^7/x^7)*log(-x^2 + 1) - 1/1680*x^5*(49* (sqrt(-x^2 + 1) - 1)^2/x^2 + 750*(sqrt(-x^2 + 1) - 1)^4/x^4 + 3)/(sqrt(-x^ 2 + 1) - 1)^5 + 25/56*(sqrt(-x^2 + 1) - 1)/x + 7/240*(sqrt(-x^2 + 1) - 1)^ 3/x^3 + 1/560*(sqrt(-x^2 + 1) - 1)^5/x^5 - 32/35*arcsin(x)
Timed out. \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\int \frac {\ln \left (1-x^2\right )}{x^8\,\sqrt {1-x^2}} \,d x \] Input:
int(log(1 - x^2)/(x^8*(1 - x^2)^(1/2)),x)
Output:
int(log(1 - x^2)/(x^8*(1 - x^2)^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.58 \[ \int \frac {\log \left (1-x^2\right )}{x^8 \sqrt {1-x^2}} \, dx=\frac {-96 \mathit {asin} \left (x \right ) x^{7}-48 \sqrt {-x^{2}+1}\, \mathrm {log}\left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}-2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}{\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}+2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}\right ) x^{6}-24 \sqrt {-x^{2}+1}\, \mathrm {log}\left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}-2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}{\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}+2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}\right ) x^{4}-18 \sqrt {-x^{2}+1}\, \mathrm {log}\left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}-2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}{\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}+2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}\right ) x^{2}-15 \sqrt {-x^{2}+1}\, \mathrm {log}\left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}-2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}{\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{4}+2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}\right )+88 \sqrt {-x^{2}+1}\, x^{6}+20 \sqrt {-x^{2}+1}\, x^{4}+6 \sqrt {-x^{2}+1}\, x^{2}}{105 x^{7}} \] Input:
int(log(-x^2+1)/x^8/(-x^2+1)^(1/2),x)
Output:
( - 96*asin(x)*x**7 - 48*sqrt( - x**2 + 1)*log((tan(asin(x)/2)**4 - 2*tan( asin(x)/2)**2 + 1)/(tan(asin(x)/2)**4 + 2*tan(asin(x)/2)**2 + 1))*x**6 - 2 4*sqrt( - x**2 + 1)*log((tan(asin(x)/2)**4 - 2*tan(asin(x)/2)**2 + 1)/(tan (asin(x)/2)**4 + 2*tan(asin(x)/2)**2 + 1))*x**4 - 18*sqrt( - x**2 + 1)*log ((tan(asin(x)/2)**4 - 2*tan(asin(x)/2)**2 + 1)/(tan(asin(x)/2)**4 + 2*tan( asin(x)/2)**2 + 1))*x**2 - 15*sqrt( - x**2 + 1)*log((tan(asin(x)/2)**4 - 2 *tan(asin(x)/2)**2 + 1)/(tan(asin(x)/2)**4 + 2*tan(asin(x)/2)**2 + 1)) + 8 8*sqrt( - x**2 + 1)*x**6 + 20*sqrt( - x**2 + 1)*x**4 + 6*sqrt( - x**2 + 1) *x**2)/(105*x**7)