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

Optimal. Leaf size=122 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b c x}{2 d^2 \sqrt{1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.172984, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4705, 4679, 4419, 4183, 2279, 2391, 191} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b c x}{2 d^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4705

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(2*f*(p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 4679

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

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.356829, size = 153, normalized size = 1.25 \[ \frac{b \left (i \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{c x}{\sqrt{1-c^2 x^2}}+\frac{\sin ^{-1}(c x)}{1-c^2 x^2}+2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+\frac{a}{1-c^2 x^2}-a \log \left (1-c^2 x^2\right )+2 a \log (x)}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.169, size = 335, normalized size = 2.8 \begin{align*} -{\frac{a}{4\,{d}^{2} \left ( cx-1 \right ) }}-{\frac{a\ln \left ( cx-1 \right ) }{2\,{d}^{2}}}+{\frac{a}{4\,{d}^{2} \left ( cx+1 \right ) }}-{\frac{a\ln \left ( cx+1 \right ) }{2\,{d}^{2}}}+{\frac{a\ln \left ( cx \right ) }{{d}^{2}}}-{\frac{{\frac{i}{2}}b{c}^{2}{x}^{2}}{{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{xbc}{2\,{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) }{2\,{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{{\frac{i}{2}}b}{{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b\arcsin \left ( cx \right ) }{{d}^{2}}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib}{{d}^{2}}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b\arcsin \left ( cx \right ) }{{d}^{2}}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib}{{d}^{2}}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{b\arcsin \left ( cx \right ) }{{d}^{2}}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{\frac{i}{2}}b}{{d}^{2}}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \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)^2,x)

[Out]

-1/4*a/d^2/(c*x-1)-1/2*a/d^2*ln(c*x-1)+1/4*a/d^2/(c*x+1)-1/2*a/d^2*ln(c*x+1)+a/d^2*ln(c*x)-1/2*I*b/d^2/(c^2*x^
2-1)*c^2*x^2+1/2*b/d^2/(c^2*x^2-1)*c*x*(-c^2*x^2+1)^(1/2)-1/2*b/d^2/(c^2*x^2-1)*arcsin(c*x)+1/2*I*b/d^2/(c^2*x
^2-1)+b/d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b/d^2*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+b/d^2*arcs
in(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-I*b/d^2*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-b/d^2*arcsin(c*x)*ln(1+(I*c
*x+(-c^2*x^2+1)^(1/2))^2)+1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{1}{c^{2} d^{2} x^{2} - d^{2}} + \frac{\log \left (c x + 1\right )}{d^{2}} + \frac{\log \left (c x - 1\right )}{d^{2}} - \frac{2 \, \log \left (x\right )}{d^{2}}\right )} + b \int \frac{\arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} 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)^2,x, algorithm="maxima")

[Out]

-1/2*a*(1/(c^2*d^2*x^2 - d^2) + log(c*x + 1)/d^2 + log(c*x - 1)/d^2 - 2*log(x)/d^2) + b*integrate(arctan2(c*x,
sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), 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^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} 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)^2,x, algorithm="fricas")

[Out]

integral((b*arcsin(c*x) + a)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)

[Out]

Timed out

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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 )}^{2} 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)^2,x, algorithm="giac")

[Out]

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