3.141 \(\int \frac{\sin ^{-1}(a+b x)^3}{x} \, dx\)

Optimal. Leaf size=365 \[ -3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{4} i \sin ^{-1}(a+b x)^4 \]

[Out]

(-I/4)*ArcSin[a + b*x]^4 + ArcSin[a + b*x]^3*Log[1 - E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + ArcSin[a +
b*x]^3*Log[1 - E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] - (3*I)*ArcSin[a + b*x]^2*PolyLog[2, E^(I*ArcSin[
a + b*x])/(I*a - Sqrt[1 - a^2])] - (3*I)*ArcSin[a + b*x]^2*PolyLog[2, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^
2])] + 6*ArcSin[a + b*x]*PolyLog[3, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + 6*ArcSin[a + b*x]*PolyLog[3
, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] + (6*I)*PolyLog[4, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])]
+ (6*I)*PolyLog[4, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])]

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Rubi [A]  time = 0.450544, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4805, 4741, 4521, 2190, 2531, 6609, 2282, 6589} \[ -3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{4} i \sin ^{-1}(a+b x)^4 \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^3/x,x]

[Out]

(-I/4)*ArcSin[a + b*x]^4 + ArcSin[a + b*x]^3*Log[1 - E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + ArcSin[a +
b*x]^3*Log[1 - E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] - (3*I)*ArcSin[a + b*x]^2*PolyLog[2, E^(I*ArcSin[
a + b*x])/(I*a - Sqrt[1 - a^2])] - (3*I)*ArcSin[a + b*x]^2*PolyLog[2, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^
2])] + 6*ArcSin[a + b*x]*PolyLog[3, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + 6*ArcSin[a + b*x]*PolyLog[3
, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] + (6*I)*PolyLog[4, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])]
+ (6*I)*PolyLog[4, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4741

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

Rule 4521

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a - Rt[-a^2 + b^
2, 2] + b*E^(I*(c + d*x))), x], x] + Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*a + Rt[-a^2 + b^2, 2] + b*E^
(I*(c + d*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NegQ[a^2 - b^2]

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

Mathematica [A]  time = 0.0379962, size = 424, normalized size = 1.16 \[ -3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )-3 i \sin ^{-1}(a+b x)^2 \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+6 \sin ^{-1}(a+b x) \text{PolyLog}\left (3,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+6 i \text{PolyLog}\left (4,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^3 \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+\sin ^{-1}(a+b x)^3 \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )-\frac{1}{4} i \sin ^{-1}(a+b x)^4 \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^3/x,x]

[Out]

(-I/4)*ArcSin[a + b*x]^4 + ArcSin[a + b*x]^3*Log[1 + E^(I*ArcSin[a + b*x])/((((-I)*a)/b - Sqrt[1 - a^2]/b)*b)]
+ ArcSin[a + b*x]^3*Log[1 + E^(I*ArcSin[a + b*x])/((((-I)*a)/b + Sqrt[1 - a^2]/b)*b)] - (3*I)*ArcSin[a + b*x]
^2*PolyLog[2, -(E^(I*ArcSin[a + b*x])/((((-I)*a)/b - Sqrt[1 - a^2]/b)*b))] - (3*I)*ArcSin[a + b*x]^2*PolyLog[2
, -(E^(I*ArcSin[a + b*x])/((((-I)*a)/b + Sqrt[1 - a^2]/b)*b))] + 6*ArcSin[a + b*x]*PolyLog[3, -(E^(I*ArcSin[a
+ b*x])/((((-I)*a)/b - Sqrt[1 - a^2]/b)*b))] + 6*ArcSin[a + b*x]*PolyLog[3, -(E^(I*ArcSin[a + b*x])/((((-I)*a)
/b + Sqrt[1 - a^2]/b)*b))] + (6*I)*PolyLog[4, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + (6*I)*PolyLog[4,
E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])]

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Maple [F]  time = 0.879, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)^3/x,x)

[Out]

int(arcsin(b*x+a)^3/x,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arcsin \left (b x + a\right )^{3}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3/x,x, algorithm="fricas")

[Out]

integral(arcsin(b*x + a)^3/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}^{3}{\left (a + b x \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**3/x,x)

[Out]

Integral(asin(a + b*x)**3/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (b x + a\right )^{3}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3/x,x, algorithm="giac")

[Out]

integrate(arcsin(b*x + a)^3/x, x)