3.6 \(\int \frac {\csc ^{-1}(\sqrt {x})}{x} \, dx\)

Optimal. Leaf size=56 \[ i \text {Li}_2\left (e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )+i \csc ^{-1}\left (\sqrt {x}\right )^2-2 \csc ^{-1}\left (\sqrt {x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right ) \]

[Out]

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)

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Rubi [A]  time = 0.08, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5219, 4625, 3717, 2190, 2279, 2391} \[ i \text {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )+i \csc ^{-1}\left (\sqrt {x}\right )^2-2 \csc ^{-1}\left (\sqrt {x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[Sqrt[x]]/x,x]

[Out]

I*ArcCsc[Sqrt[x]]^2 - 2*ArcCsc[Sqrt[x]]*Log[1 - E^((2*I)*ArcCsc[Sqrt[x]])] + I*PolyLog[2, E^((2*I)*ArcCsc[Sqrt
[x]])]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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]

Rule 2391

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]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[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]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5219

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcSin[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {\csc ^{-1}\left (\sqrt {x}\right )}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {\csc ^{-1}(x)}{x} \, dx,x,\sqrt {x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{x} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\left (2 \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\right )\\ &=i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2+4 i \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\\ &=i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )+2 \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\\ &=i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )-i \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )\\ &=i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )+i \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.96 \[ i \left (\text {Li}_2\left (e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )+\csc ^{-1}\left (\sqrt {x}\right ) \left (\csc ^{-1}\left (\sqrt {x}\right )+2 i \log \left (1-e^{2 i \csc ^{-1}\left (\sqrt {x}\right )}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCsc[Sqrt[x]]/x,x]

[Out]

I*(ArcCsc[Sqrt[x]]*(ArcCsc[Sqrt[x]] + (2*I)*Log[1 - E^((2*I)*ArcCsc[Sqrt[x]])]) + PolyLog[2, E^((2*I)*ArcCsc[S
qrt[x]])])

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccsc}\left (\sqrt {x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x^(1/2))/x,x, algorithm="fricas")

[Out]

integral(arccsc(sqrt(x))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccsc}\left (\sqrt {x}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x^(1/2))/x,x, algorithm="giac")

[Out]

integrate(arccsc(sqrt(x))/x, x)

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maple [A]  time = 0.11, size = 105, normalized size = 1.88 \[ i \mathrm {arccsc}\left (\sqrt {x}\right )^{2}-2 \,\mathrm {arccsc}\left (\sqrt {x}\right ) \ln \left (1+\frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right )-2 \,\mathrm {arccsc}\left (\sqrt {x}\right ) \ln \left (1-\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )+2 i \polylog \left (2, -\frac {i}{\sqrt {x}}-\sqrt {1-\frac {1}{x}}\right )+2 i \polylog \left (2, \frac {i}{\sqrt {x}}+\sqrt {1-\frac {1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsc(x^(1/2))/x,x)

[Out]

I*arccsc(x^(1/2))^2-2*arccsc(x^(1/2))*ln(1+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))+2*I*polylog(2,I/x^(1/2)+(1-1/x)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccsc}\left (\sqrt {x}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x^(1/2))/x,x, algorithm="maxima")

[Out]

integrate(arccsc(sqrt(x))/x, x)

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mupad [B]  time = 0.85, size = 42, normalized size = 0.75 \[ \mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )}^2\,1{}\mathrm {i}-2\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {1}{\sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(1/x^(1/2))/x,x)

[Out]

polylog(2, exp(asin(1/x^(1/2))*2i))*1i + asin(1/x^(1/2))^2*1i - 2*log(1 - exp(asin(1/x^(1/2))*2i))*asin(1/x^(1
/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acsc}{\left (\sqrt {x} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsc(x**(1/2))/x,x)

[Out]

Integral(acsc(sqrt(x))/x, x)

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