Integrand size = 10, antiderivative size = 62 \[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} i \csc ^{-1}\left (a x^5\right )^2-\frac {1}{5} \csc ^{-1}\left (a x^5\right ) \log \left (1-e^{2 i \csc ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (a x^5\right )}\right ) \]
1/10*I*arccsc(a*x^5)^2-1/5*arccsc(a*x^5)*ln(1-(I/a/x^5+(1-1/a^2/x^10)^(1/2 ))^2)+1/10*I*polylog(2,(I/a/x^5+(1-1/a^2/x^10)^(1/2))^2)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} i \left (\csc ^{-1}\left (a x^5\right ) \left (\csc ^{-1}\left (a x^5\right )+2 i \log \left (1-e^{2 i \csc ^{-1}\left (a x^5\right )}\right )\right )+\operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (a x^5\right )}\right )\right ) \]
(I/10)*(ArcCsc[a*x^5]*(ArcCsc[a*x^5] + (2*I)*Log[1 - E^((2*I)*ArcCsc[a*x^5 ])]) + PolyLog[2, E^((2*I)*ArcCsc[a*x^5])])
Time = 0.47 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {7282, 5742, 5136, 3042, 25, 4200, 25, 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.
\(\displaystyle \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx\) |
\(\Big \downarrow \) 7282 |
\(\displaystyle \frac {1}{5} \int \frac {\csc ^{-1}\left (a x^5\right )}{x^5}dx^5\) |
\(\Big \downarrow \) 5742 |
\(\displaystyle -\frac {1}{5} \int \frac {\arcsin \left (\frac {1}{a x^5}\right )}{x^5}d\frac {1}{x^5}\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle -\frac {1}{5} \int a \sqrt {1-\frac {1}{a^2 x^{10}}} x^5 \arcsin \left (\frac {1}{a x^5}\right )d\arcsin \left (\frac {1}{a x^5}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{5} \int -\arcsin \left (\frac {1}{a x^5}\right ) \tan \left (\arcsin \left (\frac {1}{a x^5}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{a x^5}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int \arcsin \left (\frac {1}{a x^5}\right ) \tan \left (\arcsin \left (\frac {1}{a x^5}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{a x^5}\right )\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle \frac {1}{5} \left (\frac {i x^{10}}{2}-2 i \int -\frac {e^{2 i \arcsin \left (\frac {1}{a x^5}\right )} \arcsin \left (\frac {1}{a x^5}\right )}{1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}}d\arcsin \left (\frac {1}{a x^5}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (2 i \int \frac {e^{2 i \arcsin \left (\frac {1}{a x^5}\right )} \arcsin \left (\frac {1}{a x^5}\right )}{1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}}d\arcsin \left (\frac {1}{a x^5}\right )+\frac {i x^{10}}{2}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{5} \left (2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{a x^5}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )d\arcsin \left (\frac {1}{a x^5}\right )\right )+\frac {i x^{10}}{2}\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {1}{5} \left (2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{a x^5}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )-\frac {1}{4} \int e^{2 i \arcsin \left (\frac {1}{a x^5}\right )} \log \left (1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )de^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )+\frac {i x^{10}}{2}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {1}{5} \left (2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )+\frac {1}{2} i \arcsin \left (\frac {1}{a x^5}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a x^5}\right )}\right )\right )+\frac {i x^{10}}{2}\right )\) |
((I/2)*x^10 + (2*I)*((I/2)*ArcSin[1/(a*x^5)]*Log[1 - E^((2*I)*ArcSin[1/(a* x^5)])] + PolyLog[2, E^((2*I)*ArcSin[1/(a*x^5)])]/4))/5
3.1.1.3.1 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_.) + Pi*(k_.) + (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*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] , x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b *ArcSin[x/c])/x, x], x, 1/x] /; FreeQ[{a, b, c}, x]
Int[(u_)/(x_), x_Symbol] :> With[{lst = PowerVariableExpn[u, 0, x]}, Simp[1 /lst[[2]] Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x], x], x, ( lst[[3]]*x)^lst[[2]]], x] /; !FalseQ[lst] && NeQ[lst[[2]], 0]] /; NonsumQ[ u] && !RationalFunctionQ[u, x]
\[\int \frac {\operatorname {arccsc}\left (a \,x^{5}\right )}{x}d x\]
\[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arccsc}\left (a x^{5}\right )}{x} \,d x } \]
\[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\int \frac {\operatorname {acsc}{\left (a x^{5} \right )}}{x}\, dx \]
\[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arccsc}\left (a x^{5}\right )}{x} \,d x } \]
5*a^2*integrate(sqrt(a*x^5 + 1)*sqrt(a*x^5 - 1)*log(x)/(a^4*x^11 - a^2*x), x) - 5*I*a^2*integrate(log(x)/(a^4*x^11 - a^2*x), x) + (arctan2(1, sqrt(a *x^5 + 1)*sqrt(a*x^5 - 1)) + I*log(a))*log(x) - 1/2*I*log(a^2*x^10)*log(x) + 1/2*I*log(a*x^5 + 1)*log(x) + 1/2*I*log(-a*x^5 + 1)*log(x) + 5/2*I*log( x)^2 + 1/10*I*dilog(a*x^5) + 1/10*I*dilog(-a*x^5)
\[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arccsc}\left (a x^{5}\right )}{x} \,d x } \]
Time = 0.89 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^{-1}\left (a x^5\right )}{x} \, dx=-\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{a\,x^5}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {1}{a\,x^5}\right )}{5}+\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{a\,x^5}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{10}+\frac {{\mathrm {asin}\left (\frac {1}{a\,x^5}\right )}^2\,1{}\mathrm {i}}{10} \]