\(\int \frac {\arcsin (a x^5)}{x} \, dx\) [360]

Optimal result

Integrand size = 10, antiderivative size = 62 \[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=-\frac {1}{10} i \arcsin \left (a x^5\right )^2+\frac {1}{5} \arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right )-\frac {1}{10} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (a x^5\right )}\right ) \]

[Out]

-1/10*I*arcsin(a*x^5)^2+1/5*arcsin(a*x^5)*ln(1-(I*a*x^5+(-a^2*x^10+1)^(1/2))^2)-1/10*I*polylog(2,(I*a*x^5+(-a^
2*x^10+1)^(1/2))^2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, 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} \[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=-\frac {1}{10} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (a x^5\right )}\right )-\frac {1}{10} i \arcsin \left (a x^5\right )^2+\frac {1}{5} \arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right ) \]

[In]

Int[ArcSin[a*x^5]/x,x]

[Out]

(-1/10*I)*ArcSin[a*x^5]^2 + (ArcSin[a*x^5]*Log[1 - E^((2*I)*ArcSin[a*x^5])])/5 - (I/10)*PolyLog[2, E^((2*I)*Ar
cSin[a*x^5])]

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}& = \frac {1}{5} \text {Subst}\left (\int x \cot (x) \, dx,x,\arcsin \left (a x^5\right )\right ) \\ & = -\frac {1}{10} i \arcsin \left (a x^5\right )^2-\frac {2}{5} i \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\arcsin \left (a x^5\right )\right ) \\ & = -\frac {1}{10} i \arcsin \left (a x^5\right )^2+\frac {1}{5} \arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right )-\frac {1}{5} \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\arcsin \left (a x^5\right )\right ) \\ & = -\frac {1}{10} i \arcsin \left (a x^5\right )^2+\frac {1}{5} \arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right )+\frac {1}{10} i \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \arcsin \left (a x^5\right )}\right ) \\ & = -\frac {1}{10} i \arcsin \left (a x^5\right )^2+\frac {1}{5} \arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right )-\frac {1}{10} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (a x^5\right )}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=\frac {1}{5} \left (\arcsin \left (a x^5\right ) \log \left (1-e^{2 i \arcsin \left (a x^5\right )}\right )-\frac {1}{2} i \left (\arcsin \left (a x^5\right )^2+\operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (a x^5\right )}\right )\right )\right ) \]

[In]

Integrate[ArcSin[a*x^5]/x,x]

[Out]

(ArcSin[a*x^5]*Log[1 - E^((2*I)*ArcSin[a*x^5])] - (I/2)*(ArcSin[a*x^5]^2 + PolyLog[2, E^((2*I)*ArcSin[a*x^5])]
))/5

Maple [F]

\[\int \frac {\arcsin \left (a \,x^{5}\right )}{x}d x\]

[In]

int(arcsin(a*x^5)/x,x)

[Out]

int(arcsin(a*x^5)/x,x)

Fricas [F]

\[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=\int { \frac {\arcsin \left (a x^{5}\right )}{x} \,d x } \]

[In]

integrate(arcsin(a*x^5)/x,x, algorithm="fricas")

[Out]

integral(arcsin(a*x^5)/x, x)

Sympy [F]

\[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=\int \frac {\operatorname {asin}{\left (a x^{5} \right )}}{x}\, dx \]

[In]

integrate(asin(a*x**5)/x,x)

[Out]

Integral(asin(a*x**5)/x, x)

Maxima [F]

\[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=\int { \frac {\arcsin \left (a x^{5}\right )}{x} \,d x } \]

[In]

integrate(arcsin(a*x^5)/x,x, algorithm="maxima")

[Out]

integrate(arcsin(a*x^5)/x, x)

Giac [F]

\[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=\int { \frac {\arcsin \left (a x^{5}\right )}{x} \,d x } \]

[In]

integrate(arcsin(a*x^5)/x,x, algorithm="giac")

[Out]

integrate(arcsin(a*x^5)/x, x)

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \frac {\arcsin \left (a x^5\right )}{x} \, dx=-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (a\,x^5\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{10}+\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (a\,x^5\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (a\,x^5\right )}{5}-\frac {{\mathrm {asin}\left (a\,x^5\right )}^2\,1{}\mathrm {i}}{10} \]

[In]

int(asin(a*x^5)/x,x)

[Out]

(log(1 - exp(asin(a*x^5)*2i))*asin(a*x^5))/5 - (polylog(2, exp(asin(a*x^5)*2i))*1i)/10 - (asin(a*x^5)^2*1i)/10