3.117 \(\int \frac{a+b \sin ^{-1}(c x)}{x^3 \sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=229 \[ \frac{i b c^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}-\frac{i b c^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}-\frac{c^2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}} \]

[Out]

-(b*c*Sqrt[1 - c^2*x^2])/(2*x*Sqrt[d - c^2*d*x^2]) - (Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(2*d*x^2) - (c^
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/2)*b*c^2*Sqrt[1
- c^2*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] - ((I/2)*b*c^2*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 = 0.299153, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {4701, 4713, 4709, 4183, 2279, 2391, 30} \[ \frac{i b c^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}-\frac{i b c^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}-\frac{c^2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^3*Sqrt[d - c^2*d*x^2]),x]

[Out]

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

Rule 4701

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

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]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x^3 \sqrt{d-c^2 d x^2}} \, dx &=-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}+\frac{1}{2} c^2 \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{d-c^2 d x^2}} \, dx+\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \int \frac{1}{x^2} \, dx}{2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}+\frac{\left (c^2 \sqrt{1-c^2 x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx}{2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}+\frac{\left (c^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}-\frac{c^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 c^2 \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 )}{2 \sqrt{d-c^2 d x^2}}+\frac{\left (b c^2 \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 )}{2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}-\frac{c^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 c^2 \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 )}{2 \sqrt{d-c^2 d x^2}}-\frac{\left (i b c^2 \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 )}{2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{2 x \sqrt{d-c^2 d x^2}}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 d x^2}-\frac{c^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 c^2 \sqrt{1-c^2 x^2} \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}-\frac{i b c^2 \sqrt{1-c^2 x^2} \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 2.29856, size = 244, normalized size = 1.07 \[ \frac{\frac{b c^2 d^2 \left (1-c^2 x^2\right )^{3/2} \left (4 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-4 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac{1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{\left (d-c^2 d x^2\right )^{3/2}}-\frac{4 a \sqrt{d-c^2 d x^2}}{x^2}-4 a c^2 \sqrt{d} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )+4 a c^2 \sqrt{d} \log (x)}{8 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^3*Sqrt[d - c^2*d*x^2]),x]

[Out]

((-4*a*Sqrt[d - c^2*d*x^2])/x^2 + 4*a*c^2*Sqrt[d]*Log[x] - 4*a*c^2*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]
] + (b*c^2*d^2*(1 - c^2*x^2)^(3/2)*(-2*Cot[ArcSin[c*x]/2] - ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 + 4*ArcSin[c*x]*L
og[1 - E^(I*ArcSin[c*x])] - 4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] + (4*I)*PolyLog[2, -E^(I*ArcSin[c*x])] -
(4*I)*PolyLog[2, E^(I*ArcSin[c*x])] + ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 - 2*Tan[ArcSin[c*x]/2]))/(d - c^2*d*x^2
)^(3/2))/(8*d)

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Maple [B]  time = 0.221, size = 461, normalized size = 2. \begin{align*} -{\frac{a}{2\,d{x}^{2}}\sqrt{-{c}^{2}d{x}^{2}+d}}-{\frac{a{c}^{2}}{2}\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{b\arcsin \left ( cx \right ){c}^{2}}{2\,d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bc}{2\, \left ({c}^{2}{x}^{2}-1 \right ) dx}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ) }{2\,d{x}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b\arcsin \left ( cx \right ){c}^{2}}{2\,d \left ({c}^{2}{x}^{2}-1 \right ) }\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}}{2\,d \left ({c}^{2}{x}^{2}-1 \right ) }\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{{\frac{i}{2}}b{c}^{2}}{d \left ({c}^{2}{x}^{2}-1 \right ) }\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{{\frac{i}{2}}b{c}^{2}}{d \left ({c}^{2}{x}^{2}-1 \right ) }\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((a+b*arcsin(c*x))/x^3/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-1/2*a/d/x^2*(-c^2*d*x^2+d)^(1/2)-1/2*a*c^2/d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-1/2*b*(-d*(c^2*
x^2-1))^(1/2)/d/(c^2*x^2-1)*arcsin(c*x)*c^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)/x/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+
1/2*b*(-d*(c^2*x^2-1))^(1/2)/x^2/d/(c^2*x^2-1)*arcsin(c*x)+1/2*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(
c^2*x^2-1)*c^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-1/2*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c
^2*x^2-1)*c^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-1/2*I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(
c^2*x^2-1)*c^2*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+1/2*I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x
^2-1)*c^2*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((a+b*arcsin(c*x))/x^3/(-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^{5} - d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^3/(-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^5 - d*x^3), 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^{3} \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**3/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))/(x**3*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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^3/(-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^3), x)