3.2.35 \(\int \frac {\text {ArcSin}(a+b x)^2}{x} \, dx\) [135]

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

[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 = 0.27, 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 = {4889, 4825, 4617, 2221, 2611, 2320, 6724} \begin {gather*} -2 i \text {ArcSin}(a+b x) \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-2 i \text {ArcSin}(a+b x) \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\text {ArcSin}(a+b x)^2 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\text {ArcSin}(a+b x)^2 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{3} i \text {ArcSin}(a+b x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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 - 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 + Sqr
t[1 - a^2])]

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 2320

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 2611

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 4617

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 4825

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 4889

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 6724

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 {\text {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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 = 0.03, size = 309, normalized size = 1.14 \begin {gather*} -\frac {1}{3} i \text {ArcSin}(a+b x)^3+\text {ArcSin}(a+b x)^2 \log \left (1+\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+\text {ArcSin}(a+b x)^2 \log \left (1+\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )-2 i \text {ArcSin}(a+b x) \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )-2 i \text {ArcSin}(a+b x) \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+2 \text {PolyLog}\left (3,\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+2 \text {PolyLog}\left (3,\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right ) \end {gather*}

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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Maple [F]
time = 1.90, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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