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

Optimal. Leaf size=173 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \left (1-c^2 x^2\right )}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-(b*c*x)/(12*d^3*(1 - c^2*x^2)^(3/2)) - (2*b*c*x)/(3*d^3*Sqrt[1 - c^2*x^2]) + (a + b*ArcSin[c*x])/(4*d^3*(1 -
c^2*x^2)^2) + (a + b*ArcSin[c*x])/(2*d^3*(1 - c^2*x^2)) - (2*(a + b*ArcSin[c*x])*ArcTanh[E^((2*I)*ArcSin[c*x])
])/d^3 + ((I/2)*b*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/d^3 - ((I/2)*b*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.251842, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4705, 4679, 4419, 4183, 2279, 2391, 191, 192} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \left (1-c^2 x^2\right )}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x)/(12*d^3*(1 - c^2*x^2)^(3/2)) - (2*b*c*x)/(3*d^3*Sqrt[1 - c^2*x^2]) + (a + b*ArcSin[c*x])/(4*d^3*(1 -
c^2*x^2)^2) + (a + b*ArcSin[c*x])/(2*d^3*(1 - c^2*x^2)) - (2*(a + b*ArcSin[c*x])*ArcTanh[E^((2*I)*ArcSin[c*x])
])/d^3 + ((I/2)*b*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/d^3 - ((I/2)*b*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d^3

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]

Rule 192

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

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^3} \, dx &=\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{(b c) \int \frac{1}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac{\int \frac{a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \left (1-c^2 x^2\right )}-\frac{(b c) \int \frac{1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{6 d^3}-\frac{(b c) \int \frac{1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^3}+\frac{\int \frac{a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d^2}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \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^3}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \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^3}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \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^3}-\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \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^3}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ &=-\frac{b c x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{a+b \sin ^{-1}(c x)}{2 d^3 \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^3}+\frac{i b \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac{i b \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.936411, size = 201, normalized size = 1.16 \[ -\frac{b \left (-6 i \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )+6 i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac{8 c x}{\sqrt{1-c^2 x^2}}+\frac{c x}{\left (1-c^2 x^2\right )^{3/2}}+\frac{6 \sin ^{-1}(c x)}{c^2 x^2-1}-\frac{3 \sin ^{-1}(c x)}{\left (c^2 x^2-1\right )^2}-12 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+12 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+\frac{6 a}{c^2 x^2-1}-\frac{3 a}{\left (c^2 x^2-1\right )^2}+6 a \log \left (1-c^2 x^2\right )-12 a \log (x)}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-3*a)/(-1 + c^2*x^2)^2 + (6*a)/(-1 + c^2*x^2) - 12*a*Log[x] + 6*a*Log[1 - c^2*x^2] + b*((c*x)/(1 - c^2*x^2)
^(3/2) + (8*c*x)/Sqrt[1 - c^2*x^2] - (3*ArcSin[c*x])/(-1 + c^2*x^2)^2 + (6*ArcSin[c*x])/(-1 + c^2*x^2) - 12*Ar
cSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + 12*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - (6*I)*PolyLog[2, -E
^((2*I)*ArcSin[c*x])] + (6*I)*PolyLog[2, E^((2*I)*ArcSin[c*x])]))/(12*d^3)

________________________________________________________________________________________

Maple [B]  time = 0.199, size = 503, normalized size = 2.9 \begin{align*}{\frac{a}{16\,{d}^{3} \left ( cx-1 \right ) ^{2}}}-{\frac{5\,a}{16\,{d}^{3} \left ( cx-1 \right ) }}-{\frac{a\ln \left ( cx-1 \right ) }{2\,{d}^{3}}}+{\frac{a}{16\,{d}^{3} \left ( cx+1 \right ) ^{2}}}+{\frac{5\,a}{16\,{d}^{3} \left ( cx+1 \right ) }}-{\frac{a\ln \left ( cx+1 \right ) }{2\,{d}^{3}}}+{\frac{a\ln \left ( cx \right ) }{{d}^{3}}}+{\frac{{\frac{4\,i}{3}}b{c}^{2}{x}^{2}}{{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{2\,b{c}^{3}{x}^{3}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ){c}^{2}{x}^{2}}{2\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{{\frac{2\,i}{3}}b{c}^{4}{x}^{4}}{{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{3\,xbc}{4\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{3\,b\arcsin \left ( cx \right ) }{4\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{{\frac{i}{2}}b}{{d}^{3}}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{b\arcsin \left ( cx \right ) }{{d}^{3}}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{{\frac{2\,i}{3}}b}{{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{b\arcsin \left ( cx \right ) }{{d}^{3}}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib}{{d}^{3}}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{b\arcsin \left ( cx \right ) }{{d}^{3}}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{ib}{{d}^{3}}{\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/(-c^2*d*x^2+d)^3,x)

[Out]

1/16*a/d^3/(c*x-1)^2-5/16*a/d^3/(c*x-1)-1/2*a/d^3*ln(c*x-1)+1/16*a/d^3/(c*x+1)^2+5/16*a/d^3/(c*x+1)-1/2*a/d^3*
ln(c*x+1)+a/d^3*ln(c*x)+4/3*I*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c^2*x^2+2/3*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c^3*x^3*(-c^
2*x^2+1)^(1/2)-1/2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^2*x^2-2/3*I*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c^4*x^4-3
/4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c*x*(-c^2*x^2+1)^(1/2)+3/4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)+1/2*I*b*poly
log(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3+b/d^3*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2/3*I*b/d^3/(c^4*x^4
-2*c^2*x^2+1)+b/d^3*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-I*b/d^3*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-b/d
^3*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-I*b/d^3*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, a{\left (\frac{2 \, c^{2} x^{2} - 3}{c^{4} d^{3} x^{4} - 2 \, c^{2} d^{3} x^{2} + d^{3}} + \frac{2 \, \log \left (c x + 1\right )}{d^{3}} + \frac{2 \, \log \left (c x - 1\right )}{d^{3}} - \frac{4 \, \log \left (x\right )}{d^{3}}\right )} - b \int \frac{\arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{c^{6} d^{3} x^{7} - 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} - d^{3} 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)^3,x, algorithm="maxima")

[Out]

-1/4*a*((2*c^2*x^2 - 3)/(c^4*d^3*x^4 - 2*c^2*d^3*x^2 + d^3) + 2*log(c*x + 1)/d^3 + 2*log(c*x - 1)/d^3 - 4*log(
x)/d^3) - b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^6*d^3*x^7 - 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3
- d^3*x), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \arcsin \left (c x\right ) + a}{c^{6} d^{3} x^{7} - 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} - d^{3} 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)^3,x, algorithm="fricas")

[Out]

integral(-(b*arcsin(c*x) + a)/(c^6*d^3*x^7 - 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 - d^3*x), x)

________________________________________________________________________________________

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{3} 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)^3,x, algorithm="giac")

[Out]

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