3.115 \(\int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.193626, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {4713, 4709, 4183, 2279, 2391} \[ \frac{i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

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

Rule 4713

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[
Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, 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 \sqrt{d-c^2 d x^2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx}{\sqrt{d-c^2 d x^2}}\\ &=\frac{\sqrt{1-c^2 x^2} \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{\left (i b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{\left (i b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{i b \sqrt{1-c^2 x^2} \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{i b \sqrt{1-c^2 x^2} \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.290129, size = 146, normalized size = 1.01 \[ \frac{b \sqrt{1-c^2 x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )}{\sqrt{d \left (1-c^2 x^2\right )}}-\frac{a \log \left (\sqrt{d} \sqrt{-d \left (c^2 x^2-1\right )}+d\right )}{\sqrt{d}}+\frac{a \log (x)}{\sqrt{d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

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

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Maple [A]  time = 0.101, size = 180, normalized size = 1.2 \begin{align*} -{a\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{-{c}^{2}d{x}^{2}+d} \right ) } \right ){\frac{1}{\sqrt{d}}}}-{\frac{ib}{d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( i\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -i\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/x/(-c^2*d*x^2+d)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d x^{3} - d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^2*d*x^3 - d*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 \sqrt{- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/x/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))/(x*sqrt(-d*(c*x - 1)*(c*x + 1))), 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}{\sqrt{-c^{2} d x^{2} + d} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/(sqrt(-c^2*d*x^2 + d)*x), x)