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

Optimal. Leaf size=116 \[ \frac{i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{a+b \sin ^{-1}(c x)}{d x}-\frac{2 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.151488, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4701, 4657, 4181, 2279, 2391, 266, 63, 208} \[ \frac{i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{a+b \sin ^{-1}(c x)}{d x}-\frac{2 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4657

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [B]  time = 0.34398, size = 259, normalized size = 2.23 \[ -\frac{-2 i b c x \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+2 i b c x \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+a c x \log (1-c x)-a c x \log (c x+1)+2 a+2 b c x \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )+2 b \sin ^{-1}(c x)+i \pi b c x \sin ^{-1}(c x)-\pi b c x \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-2 b c x \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\pi b c x \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+2 b c x \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi b c x \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\pi b c x \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(2*a + 2*b*ArcSin[c*x] + I*b*c*Pi*x*ArcSin[c*x] + 2*b*c*x*ArcTanh[Sqrt[1 - c^2*x^2]] - b*c*Pi*x*Log[1 - I*E^(
I*ArcSin[c*x])] - 2*b*c*x*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - b*c*Pi*x*Log[1 + I*E^(I*ArcSin[c*x])] + 2
*b*c*x*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + a*c*x*Log[1 - c*x] - a*c*x*Log[1 + c*x] + b*c*Pi*x*Log[-Cos[
(Pi + 2*ArcSin[c*x])/4]] + b*c*Pi*x*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*b*c*x*PolyLog[2, (-I)*E^(I*ArcSin
[c*x])] + (2*I)*b*c*x*PolyLog[2, I*E^(I*ArcSin[c*x])])/(2*d*x)

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Maple [A]  time = 0.133, size = 236, normalized size = 2. \begin{align*} -{\frac{ca\ln \left ( cx-1 \right ) }{2\,d}}+{\frac{ca\ln \left ( cx+1 \right ) }{2\,d}}-{\frac{a}{dx}}-{\frac{b\arcsin \left ( cx \right ) }{dx}}-{\frac{bc\arcsin \left ( cx \right ) }{d}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{bc\arcsin \left ( cx \right ) }{d}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{bc}{d}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{bc}{d}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-1 \right ) }-{\frac{icb}{d}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{icb}{d}{\it dilog} \left ( 1+i \left ( 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^2/(-c^2*d*x^2+d),x)

[Out]

-1/2*c*a/d*ln(c*x-1)+1/2*c*a/d*ln(c*x+1)-a/d/x-b/d*arcsin(c*x)/x-c*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^
(1/2)))+c*b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-c*b/d*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+c*b/d*ln(I*c
*x+(-c^2*x^2+1)^(1/2)-1)-I*c*b/d*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+I*c*b/d*dilog(1+I*(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*} \frac{1}{2} \, a{\left (\frac{c \log \left (c x + 1\right )}{d} - \frac{c \log \left (c x - 1\right )}{d} - \frac{2}{d x}\right )} + \frac{{\left (c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) + d x \int \frac{{\left (c^{2} x \log \left (c x + 1\right ) - c^{2} x \log \left (-c x + 1\right ) - 2 \, c\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{2} d x^{3} - d x}\,{d x} - 2 \, \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} b}{2 \, d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(c*log(c*x + 1)/d - c*log(c*x - 1)/d - 2/(d*x)) + 1/2*(c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*lo
g(c*x + 1) - c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) + 2*d*x*integrate(1/2*(c^2*x*log(c*x
+ 1) - c^2*x*log(-c*x + 1) - 2*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^2*d*x^3 - d*x), x) - 2*arctan2(c*x, sqrt(c*
x + 1)*sqrt(-c*x + 1)))*b/(d*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}{c^{2} d x^{4} - d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^2*d*x^4 - d*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a}{c^{2} x^{4} - x^{2}}\, dx + \int \frac{b \operatorname{asin}{\left (c x \right )}}{c^{2} x^{4} - x^{2}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**2*x**4 - x**2), x) + Integral(b*asin(c*x)/(c**2*x**4 - x**2), x))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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