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

Optimal. Leaf size=181 \[ -i \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{2} i \sin ^{-1}(a+b x)^2 \]

[Out]

-1/2*I*arcsin(b*x+a)^2+arcsin(b*x+a)*ln(1-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(I*a-(-a^2+1)^(1/2)))+arcsin(b*x+a)*
ln(1-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))-I*polylog(2,(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(I*a-(-
a^2+1)^(1/2)))-I*polylog(2,(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 = 0.28, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4805, 4741, 4521, 2190, 2279, 2391} \[ -i \text {PolyLog}\left (2,\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )-i \text {PolyLog}\left (2,\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{2} i \sin ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]/x,x]

[Out]

(-I/2)*ArcSin[a + b*x]^2 + ArcSin[a + b*x]*Log[1 - E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 - a^2])] + ArcSin[a + b
*x]*Log[1 - E^(I*ArcSin[a + b*x])/(I*a + Sqrt[1 - a^2])] - I*PolyLog[2, E^(I*ArcSin[a + b*x])/(I*a - Sqrt[1 -
a^2])] - I*PolyLog[2, 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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a+b x)}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \cos (x)}{-\frac {a}{b}+\frac {\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} i \sin ^{-1}(a+b x)^2+\frac {i \operatorname {Subst}\left (\int \frac {e^{i x} x}{-\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}{-\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}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-\operatorname {Subst}\left (\int \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 )-\operatorname {Subst}\left (\int \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}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )+i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )\\ &=-\frac {1}{2} i \sin ^{-1}(a+b x)^2+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-i \text {Li}_2\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 = 0.02, size = 197, normalized size = 1.09 \[ -i \text {Li}_2\left (-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}-i a}\right )-i \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x) \log \left (1+\frac {e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac {\sqrt {1-a^2}}{b}-\frac {i a}{b}\right )}\right )+\sin ^{-1}(a+b x) \log \left (1+\frac {e^{i \sin ^{-1}(a+b x)}}{b \left (\frac {\sqrt {1-a^2}}{b}-\frac {i a}{b}\right )}\right )-\frac {1}{2} i \sin ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]/x,x]

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (b x + a\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(b*x + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)/x, x)

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maple [B]  time = 0.26, size = 579, normalized size = 3.20 \[ -\frac {i \arcsin \left (b x +a \right )^{2}}{2}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right ) a^{2}}{a^{2}-1}+\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right ) a^{2}}{a^{2}-1}-\frac {i \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right ) a^{2}}{a^{2}-1}-\frac {i \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right ) a^{2}}{a^{2}-1}+\frac {i \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {i \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*arcsin(b*x+a)^2-arcsin(b*x+a)/(a^2-1)*ln((I*a-(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a-(-a^2+
1)^(1/2)))-arcsin(b*x+a)/(a^2-1)*ln((I*a+(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))+a
rcsin(b*x+a)/(a^2-1)*ln((I*a-(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a-(-a^2+1)^(1/2)))*a^2+arcsin(b*
x+a)/(a^2-1)*ln((I*a+(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))*a^2-I/(a^2-1)*dilog((
I*a-(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a-(-a^2+1)^(1/2)))*a^2-I/(a^2-1)*dilog((I*a+(-a^2+1)^(1/2
)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))*a^2+I/(a^2-1)*dilog((I*a+(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b
*x+a)^2)^(1/2))/(I*a+(-a^2+1)^(1/2)))+I/(a^2-1)*dilog((I*a-(-a^2+1)^(1/2)-I*(b*x+a)-(1-(b*x+a)^2)^(1/2))/(I*a-
(-a^2+1)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asin}\left (a+b\,x\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)/x,x)

[Out]

int(asin(a + b*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)/x, x)

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