3.2.26 \(\int \frac {\text {ArcSin}(a+b x)}{x} \, dx\) [126]

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

[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 = 0.19, 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 = {4889, 4825, 4617, 2221, 2317, 2438} \begin {gather*} -i \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\text {ArcSin}(a+b x) \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\text {ArcSin}(a+b x) \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{2} i \text {ArcSin}(a+b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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 - 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 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 2317

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 2438

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 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]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a+b x)}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {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 \text {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 \text {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 )-\text {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 )-\text {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 \text {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 \text {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 = 0.01, size = 197, normalized size = 1.09 \begin {gather*} -\frac {1}{2} i \text {ArcSin}(a+b x)^2+\text {ArcSin}(a+b x) \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) \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 )-i \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}(a+b x)}}{-i a+\sqrt {1-a^2}}\right )-i \text {PolyLog}\left (2,\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]/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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 578 vs. \(2 (204 ) = 408\).
time = 0.64, size = 579, normalized size = 3.20

method	result	size
derivativedivides	\(-\frac {i \arcsin \left (b x +a \right )^{2}}{2}-\frac {i a^{2} \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {i a^{2} \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {a^{2} \arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {a^{2} \arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {i \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {i \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}\)	\(579\)
default	\(-\frac {i \arcsin \left (b x +a \right )^{2}}{2}-\frac {i a^{2} \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {i a^{2} \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {a^{2} \arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {a^{2} \arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {i \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {i \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {\arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}\)	\(579\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)/x,x,method=_RETURNVERBOSE)

[Out]

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

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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)/x,x, algorithm="maxima")

[Out]

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

[Out]

integral(arcsin(b*x + a)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asin}{\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)/x,x)

[Out]

Integral(asin(a + b*x)/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)/x,x, algorithm="giac")

[Out]

integrate(arcsin(b*x + a)/x, x)

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

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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