Integrand size = 10, antiderivative size = 59 \[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=-\frac {1}{2} i \arccos \left (\frac {x}{a}\right )^2+\arccos \left (\frac {x}{a}\right ) \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \arccos \left (\frac {x}{a}\right )}\right ) \]
-1/2*I*arccos(x/a)^2+arccos(x/a)*ln(1+(x/a+I*(1-x^2/a^2)^(1/2))^2)-1/2*I*p olylog(2,-(x/a+I*(1-x^2/a^2)^(1/2))^2)
Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=-\frac {1}{2} i \sec ^{-1}\left (\frac {a}{x}\right )^2+\sec ^{-1}\left (\frac {a}{x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right ) \]
(-1/2*I)*ArcSec[a/x]^2 + ArcSec[a/x]*Log[1 + E^((2*I)*ArcSec[a/x])] - (I/2 )*PolyLog[2, -E^((2*I)*ArcSec[a/x])]
Time = 0.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5787, 5137, 3042, 4202, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx\) |
\(\Big \downarrow \) 5787 |
\(\displaystyle \int \frac {\arccos \left (\frac {x}{a}\right )}{x}dx\) |
\(\Big \downarrow \) 5137 |
\(\displaystyle -\int \frac {a \sqrt {1-\frac {x^2}{a^2}} \arccos \left (\frac {x}{a}\right )}{x}d\arccos \left (\frac {x}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \arccos \left (\frac {x}{a}\right ) \tan \left (\arccos \left (\frac {x}{a}\right )\right )d\arccos \left (\frac {x}{a}\right )\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle 2 i \int \frac {e^{2 i \arccos \left (\frac {x}{a}\right )} \arccos \left (\frac {x}{a}\right )}{1+e^{2 i \arccos \left (\frac {x}{a}\right )}}d\arccos \left (\frac {x}{a}\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right )^2\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 2 i \left (\frac {1}{2} i \int \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )d\arccos \left (\frac {x}{a}\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right ) \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right )^2\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle 2 i \left (\frac {1}{4} \int e^{-2 i \arccos \left (\frac {x}{a}\right )} \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )de^{2 i \arccos \left (\frac {x}{a}\right )}-\frac {1}{2} i \arccos \left (\frac {x}{a}\right ) \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right )^2\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle 2 i \left (-\frac {1}{4} \operatorname {PolyLog}\left (2,-e^{2 i \arccos \left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right ) \log \left (1+e^{2 i \arccos \left (\frac {x}{a}\right )}\right )\right )-\frac {1}{2} i \arccos \left (\frac {x}{a}\right )^2\) |
(-1/2*I)*ArcCos[x/a]^2 + (2*I)*((-1/2*I)*ArcCos[x/a]*Log[1 + E^((2*I)*ArcC os[x/a])] - PolyLog[2, -E^((2*I)*ArcCos[x/a])]/4)
3.1.13.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[ (a + b*x)^n*Tan[x], x], x, ArcCos[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0 ]
Int[ArcSec[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[ u*ArcCos[a/c + b*(x^n/c)]^m, x] /; FreeQ[{a, b, c, n, m}, x]
Time = 1.77 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29
method | result | size |
derivativedivides | \(-\frac {i \operatorname {arcsec}\left (\frac {a}{x}\right )^{2}}{2}+\operatorname {arcsec}\left (\frac {a}{x}\right ) \ln \left (1+{\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )}^{2}\right )-\frac {i \operatorname {polylog}\left (2, -{\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )}^{2}\right )}{2}\) | \(76\) |
default | \(-\frac {i \operatorname {arcsec}\left (\frac {a}{x}\right )^{2}}{2}+\operatorname {arcsec}\left (\frac {a}{x}\right ) \ln \left (1+{\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )}^{2}\right )-\frac {i \operatorname {polylog}\left (2, -{\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )}^{2}\right )}{2}\) | \(76\) |
-1/2*I*arcsec(a/x)^2+arcsec(a/x)*ln(1+(x/a+I*(1-x^2/a^2)^(1/2))^2)-1/2*I*p olylog(2,-(x/a+I*(1-x^2/a^2)^(1/2))^2)
\[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x} \,d x } \]
\[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=\int \frac {\operatorname {asec}{\left (\frac {a}{x} \right )}}{x}\, dx \]
\[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x} \,d x } \]
\[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx=\int \frac {\mathrm {acos}\left (\frac {x}{a}\right )}{x} \,d x \]