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

Optimal. Leaf size=203 \[ \frac{i b \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}-\frac{i b \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}+\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}-\frac{b c x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.209705, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4697, 4709, 4183, 2279, 2391, 8} \[ \frac{i b \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}-\frac{i b \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}+\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}-\frac{b c x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4697

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.51766, size = 187, normalized size = 0.92 \[ \frac{b \sqrt{d-c^2 d x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \sin ^{-1}(c x)-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt{1-c^2 x^2}}+a \sqrt{d-c^2 d x^2}-a \sqrt{d} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )+a \sqrt{d} \log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.154, size = 413, normalized size = 2. \begin{align*} -\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{-{c}^{2}d{x}^{2}+d} \right ) } \right ) a+\sqrt{-{c}^{2}d{x}^{2}+d}a+{\frac{b\arcsin \left ( cx \right ){x}^{2}{c}^{2}}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{xbc}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) }{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b\arcsin \left ( cx \right ) }{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{b\arcsin \left ( cx \right ) }{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{ib}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\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((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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