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

Optimal. Leaf size=271 \[ -2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{3} i \sin ^{-1}(a+b x)^3 \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4805, 4741, 4521, 2190, 2531, 2282, 6589} \[ -2 i \sin ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )-2 i \sin ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )+2 \text {PolyLog}\left (3,\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+2 \text {PolyLog}\left (3,\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{3} i \sin ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^2/x,x]

[Out]

(-I/3)*ArcSin[a + b*x]^3 + ArcSin[a + b*x]^2*Log[1 - E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + ArcSin[a +
b*x]^2*Log[1 - E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] - (2*I)*ArcSin[a + b*x]*PolyLog[2, E^(I*ArcSin[a
+ b*x])/(I*a - Sqrt[1 - a^2])] - (2*I)*ArcSin[a + b*x]*PolyLog[2, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])]
+ 2*PolyLog[3, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + 2*PolyLog[3, E^(I*ArcSin[a + b*x])/(I*a + Sqrt[
1 - a^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 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 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 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 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 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 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)^2}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \cos (x)}{-\frac {a}{b}+\frac {\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{3} i \sin ^{-1}(a+b x)^3+\frac {i \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{-\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^2}{-\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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-2 \operatorname {Subst}\left (\int x \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 )-2 \operatorname {Subst}\left (\int x \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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+2 i \operatorname {Subst}\left (\int \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 )+2 i \operatorname {Subst}\left (\int \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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{i a-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{i a+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )\\ &=-\frac {1}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 309, normalized size = 1.14 \[ -2 i \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {e^{i \sin ^{-1}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )-2 i \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {e^{i \sin ^{-1}(a+b x)}}{\left (\frac {\sqrt {1-a^2}}{b}-\frac {i a}{b}\right ) b}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1+\frac {e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac {\sqrt {1-a^2}}{b}-\frac {i a}{b}\right )}\right )+\sin ^{-1}(a+b x)^2 \log \left (1+\frac {e^{i \sin ^{-1}(a+b x)}}{b \left (\frac {\sqrt {1-a^2}}{b}-\frac {i a}{b}\right )}\right )-\frac {1}{3} i \sin ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^2/x,x]

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (b x + a\right )^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(b*x + a)^2/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (b x + a\right )^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)^2/x, x)

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maple [F]  time = 1.70, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (b x +a \right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (b x + a\right )^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)^2/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {asin}\left (a+b\,x\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^2/x,x)

[Out]

int(asin(a + b*x)^2/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{2}{\left (a + b x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)**2/x, x)

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