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

Optimal. Leaf size=63 \[ -\frac{1}{2} i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.069963, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {4625, 3717, 2190, 2279, 2391} \[ -\frac{1}{2} i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/x,x]

[Out]

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

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x} \, dx &=\operatorname{Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} (i b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} i b \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0321451, size = 52, normalized size = 0.83 \[ -\frac{1}{2} i b \left (\sin ^{-1}(c x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )\right )+a \log (x)+b \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])/x,x]

[Out]

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

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Maple [A]  time = 0.03, size = 122, normalized size = 1.9 \begin{align*} a\ln \left ( cx \right ) -{\frac{i}{2}}b \left ( \arcsin \left ( cx \right ) \right ) ^{2}+b\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +b\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -ib{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -ib{\it polylog} \left ( 2,-icx-\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))/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x, x) + a*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsin(c*x) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x) + a)/x, x)