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

Optimal. Leaf size=211 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.365164, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4705, 4679, 4419, 4183, 2531, 2282, 6589, 4651, 260} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*x*(a + b*ArcSin[c*x]))/(d^2*Sqrt[1 - c^2*x^2])) + (a + b*ArcSin[c*x])^2/(2*d^2*(1 - c^2*x^2)) - (2*(a +
b*ArcSin[c*x])^2*ArcTanh[E^((2*I)*ArcSin[c*x])])/d^2 - (b^2*Log[1 - c^2*x^2])/(2*d^2) + (I*b*(a + b*ArcSin[c*
x])*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/d^2 - (I*b*(a + b*ArcSin[c*x])*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d^2
- (b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(2*d^2) + (b^2*PolyLog[3, E^((2*I)*ArcSin[c*x])])/(2*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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )^2} \, dx &=\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{(b c) \int \frac{a+b \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d^2}+\frac{\int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )} \, dx}{d}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac{\left (b^2 c^2\right ) \int \frac{x}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{2 \operatorname{Subst}\left (\int (a+b x)^2 \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac{b c x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 \left (1-c^2 x^2\right )}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac{b^2 \text{Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{b^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 1.27137, size = 365, normalized size = 1.73 \[ \frac{2 a b \left (i \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{c x}{\sqrt{1-c^2 x^2}}+\frac{\sin ^{-1}(c x)}{1-c^2 x^2}+2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+2 b^2 \left (i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c x)}\right )+i \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} \log \left (1-c^2 x^2\right )+\frac{\sin ^{-1}(c x)^2}{2-2 c^2 x^2}-\frac{c x \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}+\frac{2}{3} i \sin ^{-1}(c x)^3+\sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+\frac{a^2}{1-c^2 x^2}-a^2 \log \left (1-c^2 x^2\right )+2 a^2 \log (c x)}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2/(1 - c^2*x^2) + 2*a^2*Log[c*x] - a^2*Log[1 - c^2*x^2] + 2*a*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*b^2*((-I/24)*Pi^3 - (c*x*ArcSin
[c*x])/Sqrt[1 - c^2*x^2] + ArcSin[c*x]^2/(2 - 2*c^2*x^2) + ((2*I)/3)*ArcSin[c*x]^3 + ArcSin[c*x]^2*Log[1 - E^(
(-2*I)*ArcSin[c*x])] - ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[c*x])] - Log[1 - c^2*x^2]/2 + I*ArcSin[c*x]*PolyL
og[2, E^((-2*I)*ArcSin[c*x])] + I*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + PolyLog[3, E^((-2*I)*ArcSin
[c*x])]/2 - PolyLog[3, -E^((2*I)*ArcSin[c*x])]/2))/(2*d^2)

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Maple [B]  time = 0.244, size = 829, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-1/2*a^2*(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) + integrate((b^2*arctan2
(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*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^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{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))^2/x/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} x^{5} - 2 c^{2} x^{3} + x}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{c^{4} x^{5} - 2 c^{2} x^{3} + x}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c x \right )}}{c^{4} x^{5} - 2 c^{2} x^{3} + x}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**4*x**5 - 2*c**2*x**3 + x), x) + Integral(b**2*asin(c*x)**2/(c**4*x**5 - 2*c**2*x**3 + x), x
) + Integral(2*a*b*asin(c*x)/(c**4*x**5 - 2*c**2*x**3 + x), x))/d**2

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

[Out]

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