\(\int \frac {\arcsin (\sqrt {x})}{x} \, dx\) [364]

Optimal result

Integrand size = 10, antiderivative size = 56 \[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=-i \arcsin \left (\sqrt {x}\right )^2+2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right )-i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\sqrt {x}\right )}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4914, 3798, 2221, 2317, 2438} \[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=-i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\sqrt {x}\right )}\right )-i \arcsin \left (\sqrt {x}\right )^2+2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right ) \]

[In]

Int[ArcSin[Sqrt[x]]/x,x]

[Out]

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

Rule 2221

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]
- 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 2317

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 2438

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 3798

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 4914

Int[ArcSin[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Cot[x], x], x, ArcSin[a*x^p]], x]
/; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x \cot (x) \, dx,x,\arcsin \left (\sqrt {x}\right )\right ) \\ & = -i \arcsin \left (\sqrt {x}\right )^2-4 i \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\arcsin \left (\sqrt {x}\right )\right ) \\ & = -i \arcsin \left (\sqrt {x}\right )^2+2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right )-2 \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\arcsin \left (\sqrt {x}\right )\right ) \\ & = -i \arcsin \left (\sqrt {x}\right )^2+2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right )+i \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \arcsin \left (\sqrt {x}\right )}\right ) \\ & = -i \arcsin \left (\sqrt {x}\right )^2+2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right )-i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\sqrt {x}\right )}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=2 \arcsin \left (\sqrt {x}\right ) \log \left (1-e^{2 i \arcsin \left (\sqrt {x}\right )}\right )-i \left (\arcsin \left (\sqrt {x}\right )^2+\operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\sqrt {x}\right )}\right )\right ) \]

[In]

Integrate[ArcSin[Sqrt[x]]/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(-i \arcsin \left (\sqrt {x}\right )^{2}+2 \arcsin \left (\sqrt {x}\right ) \ln \left (1+i \sqrt {x}+\sqrt {1-x}\right )-2 i \operatorname {polylog}\left (2, -i \sqrt {x}-\sqrt {1-x}\right )+2 \arcsin \left (\sqrt {x}\right ) \ln \left (1-i \sqrt {x}-\sqrt {1-x}\right )-2 i \operatorname {polylog}\left (2, i \sqrt {x}+\sqrt {1-x}\right )\)	\(97\)
default	\(-i \arcsin \left (\sqrt {x}\right )^{2}+2 \arcsin \left (\sqrt {x}\right ) \ln \left (1+i \sqrt {x}+\sqrt {1-x}\right )-2 i \operatorname {polylog}\left (2, -i \sqrt {x}-\sqrt {1-x}\right )+2 \arcsin \left (\sqrt {x}\right ) \ln \left (1-i \sqrt {x}-\sqrt {1-x}\right )-2 i \operatorname {polylog}\left (2, i \sqrt {x}+\sqrt {1-x}\right )\)	\(97\)

[In]

int(arcsin(x^(1/2))/x,x,method=_RETURNVERBOSE)

[Out]

-I*arcsin(x^(1/2))^2+2*arcsin(x^(1/2))*ln(1+I*x^(1/2)+(1-x)^(1/2))-2*I*polylog(2,-I*x^(1/2)-(1-x)^(1/2))+2*arc
sin(x^(1/2))*ln(1-I*x^(1/2)-(1-x)^(1/2))-2*I*polylog(2,I*x^(1/2)+(1-x)^(1/2))

Fricas [F]

\[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\arcsin \left (\sqrt {x}\right )}{x} \,d x } \]

[In]

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

[Out]

integral(arcsin(sqrt(x))/x, x)

Sympy [F]

\[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=\int \frac {\operatorname {asin}{\left (\sqrt {x} \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(asin(sqrt(x))/x, x)

Maxima [F]

\[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\arcsin \left (\sqrt {x}\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(arcsin(sqrt(x))/x, x)

Giac [F]

\[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\arcsin \left (\sqrt {x}\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(arcsin(sqrt(x))/x, x)

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {\arcsin \left (\sqrt {x}\right )}{x} \, dx=-\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\sqrt {x}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}-{\mathrm {asin}\left (\sqrt {x}\right )}^2\,1{}\mathrm {i}+2\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\sqrt {x}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

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