3.157 \(\int \frac{(a+b \sin ^{-1}(c x))^3}{x^2} \, dx\)

Optimal. Leaf size=137 \[ 6 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-6 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-6 b^3 c \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+6 b^3 c \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \]

[Out]

-((a + b*ArcSin[c*x])^3/x) - 6*b*c*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])] + (6*I)*b^2*c*(a + b*ArcSi
n[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])] - (6*I)*b^2*c*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])] - 6*b^3
*c*PolyLog[3, -E^(I*ArcSin[c*x])] + 6*b^3*c*PolyLog[3, E^(I*ArcSin[c*x])]

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Rubi [A]  time = 0.212895, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4627, 4709, 4183, 2531, 2282, 6589} \[ 6 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-6 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-6 b^3 c \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+6 b^3 c \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^3/x^2,x]

[Out]

-((a + b*ArcSin[c*x])^3/x) - 6*b*c*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])] + (6*I)*b^2*c*(a + b*ArcSi
n[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])] - (6*I)*b^2*c*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])] - 6*b^3
*c*PolyLog[3, -E^(I*ArcSin[c*x])] + 6*b^3*c*PolyLog[3, E^(I*ArcSin[c*x])]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, 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 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{\left (a+b \sin ^{-1}(c x)\right )^3}{x^2} \, dx &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}+(3 b c) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}+(3 b c) \operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )-\left (6 b^2 c\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\left (6 b^2 c\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-\left (6 i b^3 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\left (6 i b^3 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-\left (6 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )+\left (6 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^3}{x}-6 b c \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-6 i b^2 c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-6 b^3 c \text{Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )+6 b^3 c \text{Li}_3\left (e^{i \sin ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [B]  time = 0.314443, size = 283, normalized size = 2.07 \[ 3 a b^2 c \left (2 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \left (\frac{\sin ^{-1}(c x)}{c x}-2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )+2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )+b^3 c \left (6 i \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-6 i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-6 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+6 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )-\frac{\sin ^{-1}(c x)^3}{c x}+3 \sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-3 \sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )-3 a^2 b c \log \left (\sqrt{1-c^2 x^2}+1\right )+3 a^2 b c \log (x)-\frac{3 a^2 b \sin ^{-1}(c x)}{x}-\frac{a^3}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])^3/x^2,x]

[Out]

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

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Maple [B]  time = 0.042, size = 378, normalized size = 2.8 \begin{align*} -{\frac{{a}^{3}}{x}}-{\frac{{b}^{3} \left ( \arcsin \left ( cx \right ) \right ) ^{3}}{x}}-3\,c{b}^{3} \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +6\,ic{b}^{3}\arcsin \left ( cx \right ){\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -6\,{b}^{3}c{\it polylog} \left ( 3,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +3\,c{b}^{3} \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -6\,ic{b}^{3}\arcsin \left ( cx \right ){\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +6\,{b}^{3}c{\it polylog} \left ( 3,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -3\,{\frac{a{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{x}}+6\,ca{b}^{2}\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -6\,ca{b}^{2}\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -6\,ica{b}^{2}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +6\,ica{b}^{2}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -3\,{\frac{{a}^{2}b\arcsin \left ( cx \right ) }{x}}-3\,c{a}^{2}b{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^3/x^2,x)

[Out]

-a^3/x-b^3/x*arcsin(c*x)^3-3*c*b^3*arcsin(c*x)^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+6*I*c*b^3*arcsin(c*x)*polylog(
2,-I*c*x-(-c^2*x^2+1)^(1/2))-6*b^3*c*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))+3*c*b^3*arcsin(c*x)^2*ln(1-I*c*x-(-c
^2*x^2+1)^(1/2))-6*I*c*b^3*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+6*b^3*c*polylog(3,I*c*x+(-c^2*x^2+1
)^(1/2))-3*a*b^2/x*arcsin(c*x)^2+6*c*a*b^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-6*c*a*b^2*arcsin(c*x)*ln
(1+I*c*x+(-c^2*x^2+1)^(1/2))-6*I*c*a*b^2*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+6*I*c*a*b^2*polylog(2,-I*c*x-(-c^
2*x^2+1)^(1/2))-3*a^2*b/x*arcsin(c*x)-3*c*a^2*b*arctanh(1/(-c^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -3 \,{\left (c \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\arcsin \left (c x\right )}{x}\right )} a^{2} b - \frac{a^{3}}{x} - \frac{b^{3} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{3} + \frac{3}{2} \,{\left (a b^{2} c^{2}{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \arctan \left (\frac{c x}{\sqrt{c x + 1} \sqrt{-c x + 1}}\right )^{2} -{\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} a b^{2} \arctan \left (\frac{c x}{\sqrt{c x + 1} \sqrt{-c x + 1}}\right )^{2} + 8 \, b^{3} c \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2}}{4 \,{\left (c^{2} x^{3} - x\right )}}\,{d x}\right )} x}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^3/x^2,x, algorithm="maxima")

[Out]

-3*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*a^2*b - a^3/x - (b^3*arctan2(c*x, sqrt(c*x
+ 1)*sqrt(-c*x + 1))^3 + x*integrate(3*(sqrt(c*x + 1)*sqrt(-c*x + 1)*b^3*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-
c*x + 1))^2 - (a*b^2*c^2*x^2 - a*b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/(c^2*x^4 - x^2), x))/x

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arcsin \left (c x\right )^{3} + 3 \, a b^{2} \arcsin \left (c x\right )^{2} + 3 \, a^{2} b \arcsin \left (c x\right ) + a^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^3/x^2,x, algorithm="fricas")

[Out]

integral((b^3*arcsin(c*x)^3 + 3*a*b^2*arcsin(c*x)^2 + 3*a^2*b*arcsin(c*x) + a^3)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right )^{3}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**3/x**2,x)

[Out]

Integral((a + b*asin(c*x))**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^3/x^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^3/x^2, x)