3.38 \(\int \frac{\sin ^{-1}(a x)^4}{x} \, dx\)

Optimal. Leaf size=113 \[ -2 i \sin ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(a x)}\right )+3 i \sin ^{-1}(a x) \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{2 i \sin ^{-1}(a x)}\right )-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

[Out]

(-I/5)*ArcSin[a*x]^5 + ArcSin[a*x]^4*Log[1 - E^((2*I)*ArcSin[a*x])] - (2*I)*ArcSin[a*x]^3*PolyLog[2, E^((2*I)*
ArcSin[a*x])] + 3*ArcSin[a*x]^2*PolyLog[3, E^((2*I)*ArcSin[a*x])] + (3*I)*ArcSin[a*x]*PolyLog[4, E^((2*I)*ArcS
in[a*x])] - (3*PolyLog[5, E^((2*I)*ArcSin[a*x])])/2

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Rubi [A]  time = 0.122287, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ -2 i \sin ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(a x)}\right )+3 i \sin ^{-1}(a x) \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{2 i \sin ^{-1}(a x)}\right )-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^4/x,x]

[Out]

(-I/5)*ArcSin[a*x]^5 + ArcSin[a*x]^4*Log[1 - E^((2*I)*ArcSin[a*x])] - (2*I)*ArcSin[a*x]^3*PolyLog[2, E^((2*I)*
ArcSin[a*x])] + 3*ArcSin[a*x]^2*PolyLog[3, E^((2*I)*ArcSin[a*x])] + (3*I)*ArcSin[a*x]*PolyLog[4, E^((2*I)*ArcS
in[a*x])] - (3*PolyLog[5, E^((2*I)*ArcSin[a*x])])/2

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 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 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\sin ^{-1}(a x)^4}{x} \, dx &=\operatorname{Subst}\left (\int x^4 \cot (x) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x^4}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-4 \operatorname{Subst}\left (\int x^3 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )+6 i \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(a x)}\right )-6 \operatorname{Subst}\left (\int x \text{Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(a x)}\right )+3 i \sin ^{-1}(a x) \text{Li}_4\left (e^{2 i \sin ^{-1}(a x)}\right )-3 i \operatorname{Subst}\left (\int \text{Li}_4\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(a x)}\right )+3 i \sin ^{-1}(a x) \text{Li}_4\left (e^{2 i \sin ^{-1}(a x)}\right )-\frac{3}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_4(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(a x)}\right )\\ &=-\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(a x)}\right )+3 i \sin ^{-1}(a x) \text{Li}_4\left (e^{2 i \sin ^{-1}(a x)}\right )-\frac{3}{2} \text{Li}_5\left (e^{2 i \sin ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0442051, size = 113, normalized size = 1. \[ 2 i \sin ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(a x)}\right )-3 i \sin ^{-1}(a x) \text{PolyLog}\left (4,e^{-2 i \sin ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{-2 i \sin ^{-1}(a x)}\right )+\frac{1}{5} i \sin ^{-1}(a x)^5+\sin ^{-1}(a x)^4 \log \left (1-e^{-2 i \sin ^{-1}(a x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^4/x,x]

[Out]

(I/5)*ArcSin[a*x]^5 + ArcSin[a*x]^4*Log[1 - E^((-2*I)*ArcSin[a*x])] + (2*I)*ArcSin[a*x]^3*PolyLog[2, E^((-2*I)
*ArcSin[a*x])] + 3*ArcSin[a*x]^2*PolyLog[3, E^((-2*I)*ArcSin[a*x])] - (3*I)*ArcSin[a*x]*PolyLog[4, E^((-2*I)*A
rcSin[a*x])] - (3*PolyLog[5, E^((-2*I)*ArcSin[a*x])])/2

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Maple [A]  time = 0.046, size = 287, normalized size = 2.5 \begin{align*} -{\frac{i}{5}} \left ( \arcsin \left ( ax \right ) \right ) ^{5}+ \left ( \arcsin \left ( ax \right ) \right ) ^{4}\ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) -4\,i \left ( \arcsin \left ( ax \right ) \right ) ^{3}{\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +12\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 3,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +24\,i\arcsin \left ( ax \right ){\it polylog} \left ( 4,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -24\,{\it polylog} \left ( 5,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) + \left ( \arcsin \left ( ax \right ) \right ) ^{4}\ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -4\,i \left ( \arcsin \left ( ax \right ) \right ) ^{3}{\it polylog} \left ( 2,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +12\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 3,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +24\,i\arcsin \left ( ax \right ){\it polylog} \left ( 4,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) -24\,{\it polylog} \left ( 5,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)^4/x,x)

[Out]

-1/5*I*arcsin(a*x)^5+arcsin(a*x)^4*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))-4*I*arcsin(a*x)^3*polylog(2,-I*a*x-(-a^2*x^2
+1)^(1/2))+12*arcsin(a*x)^2*polylog(3,-I*a*x-(-a^2*x^2+1)^(1/2))+24*I*arcsin(a*x)*polylog(4,-I*a*x-(-a^2*x^2+1
)^(1/2))-24*polylog(5,-I*a*x-(-a^2*x^2+1)^(1/2))+arcsin(a*x)^4*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-4*I*arcsin(a*x)^
3*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))+12*arcsin(a*x)^2*polylog(3,I*a*x+(-a^2*x^2+1)^(1/2))+24*I*arcsin(a*x)*po
lylog(4,I*a*x+(-a^2*x^2+1)^(1/2))-24*polylog(5,I*a*x+(-a^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (a x\right )^{4}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^4/x,x, algorithm="maxima")

[Out]

integrate(arcsin(a*x)^4/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arcsin \left (a x\right )^{4}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^4/x,x, algorithm="fricas")

[Out]

integral(arcsin(a*x)^4/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}^{4}{\left (a x \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**4/x,x)

[Out]

Integral(asin(a*x)**4/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (a x\right )^{4}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^4/x,x, algorithm="giac")

[Out]

integrate(arcsin(a*x)^4/x, x)