Integrand size = 10, antiderivative size = 56 \[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=i \csc ^{-1}\left (\sqrt {x}\right )^2-2 \csc ^{-1}\left (\sqrt {x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )+i \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right ) \]
I*arccsc(x^(1/2))^2-2*arccsc(x^(1/2))*ln(1-(I/x^(1/2)+(1-1/x)^(1/2))^2)+I* polylog(2,(I/x^(1/2)+(1-1/x)^(1/2))^2)
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=i \left (\csc ^{-1}\left (\sqrt {x}\right ) \left (\csc ^{-1}\left (\sqrt {x}\right )+2 i \log \left (1-e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )\right )+\operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )\right ) \]
I*(ArcCsc[Sqrt[x]]*(ArcCsc[Sqrt[x]] + (2*I)*Log[1 - E^((2*I)*ArcCsc[Sqrt[x ]])]) + PolyLog[2, E^((2*I)*ArcCsc[Sqrt[x]])])
Time = 0.47 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {7267, 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.
\(\displaystyle \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}}d\sqrt {x}\) |
\(\Big \downarrow \) 5742 |
\(\displaystyle -2 \int \frac {\arcsin \left (\frac {1}{\sqrt {x}}\right )}{\sqrt {x}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle -2 \int \sqrt {1-\frac {1}{x}} \sqrt {x} \arcsin \left (\frac {1}{\sqrt {x}}\right )d\arcsin \left (\frac {1}{\sqrt {x}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 \int -\arcsin \left (\frac {1}{\sqrt {x}}\right ) \tan \left (\arcsin \left (\frac {1}{\sqrt {x}}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{\sqrt {x}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 \int \arcsin \left (\frac {1}{\sqrt {x}}\right ) \tan \left (\arcsin \left (\frac {1}{\sqrt {x}}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{\sqrt {x}}\right )\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle -2 \left (2 i \int -\frac {e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )} \arcsin \left (\frac {1}{\sqrt {x}}\right )}{1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}}d\arcsin \left (\frac {1}{\sqrt {x}}\right )-\frac {i x}{2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \left (-2 i \int \frac {e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )} \arcsin \left (\frac {1}{\sqrt {x}}\right )}{1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}}d\arcsin \left (\frac {1}{\sqrt {x}}\right )-\frac {i x}{2}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -2 \left (-2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )d\arcsin \left (\frac {1}{\sqrt {x}}\right )\right )-\frac {i x}{2}\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -2 \left (-2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )-\frac {1}{4} \int e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )} \log \left (1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )de^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )-\frac {i x}{2}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 \left (-2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )+\frac {1}{2} i \arcsin \left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{\sqrt {x}}\right )}\right )\right )-\frac {i x}{2}\right )\) |
-2*((-1/2*I)*x - (2*I)*((I/2)*ArcSin[1/Sqrt[x]]*Log[1 - E^((2*I)*ArcSin[1/ Sqrt[x]])] + PolyLog[2, E^((2*I)*ArcSin[1/Sqrt[x]])]/4))
3.1.6.3.1 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_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.81 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.88
method | result | size |
derivativedivides | \(i \operatorname {arccsc}\left (\sqrt {x}\right )^{2}-2 \,\operatorname {arccsc}\left (\sqrt {x}\right ) \ln \left (1-\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )+2 i \operatorname {polylog}\left (2, \frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right )-2 \,\operatorname {arccsc}\left (\sqrt {x}\right ) \ln \left (1+\frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right )+2 i \operatorname {polylog}\left (2, -\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )\) | \(105\) |
default | \(i \operatorname {arccsc}\left (\sqrt {x}\right )^{2}-2 \,\operatorname {arccsc}\left (\sqrt {x}\right ) \ln \left (1-\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )+2 i \operatorname {polylog}\left (2, \frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right )-2 \,\operatorname {arccsc}\left (\sqrt {x}\right ) \ln \left (1+\frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right )+2 i \operatorname {polylog}\left (2, -\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )\) | \(105\) |
I*arccsc(x^(1/2))^2-2*arccsc(x^(1/2))*ln(1-I/x^(1/2)-(1-1/x)^(1/2))+2*I*po lylog(2,I/x^(1/2)+(1-1/x)^(1/2))-2*arccsc(x^(1/2))*ln(1+I/x^(1/2)+(1-1/x)^ (1/2))+2*I*polylog(2,-I/x^(1/2)-(1-1/x)^(1/2))
\[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arccsc}\left (\sqrt {x}\right )}{x} \,d x } \]
\[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=\int \frac {\operatorname {acsc}{\left (\sqrt {x} \right )}}{x}\, dx \]
\[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arccsc}\left (\sqrt {x}\right )}{x} \,d x } \]
Exception generated. \[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=\text {Exception raised: NotImplementedError} \]
Exception raised: NotImplementedError >> unable to parse Giac output: Inva lid series expansion: non tractable function asin at +infinity
Time = 1.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx=\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )}^2\,1{}\mathrm {i}-2\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right ) \]