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

Optimal. Leaf size=287 \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{c d x+d} \sqrt{e-c e x}} \]

[Out]

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

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Rubi [A]  time = 0.579608, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4739, 4709, 4183, 2531, 2282, 6589} \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{c d x+d} \sqrt{e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4739

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(
q_), x_Symbol] :> Dist[((-((d^2*g)/e))^IntPart[q]*(d + e*x)^FracPart[q]*(f + g*x)^FracPart[q])/(1 - c^2*x^2)^F
racPart[q], Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]

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 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]

Rubi steps

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

Mathematica [A]  time = 1.50283, size = 336, normalized size = 1.17 \[ \frac{2 a b \sqrt{1-c^2 x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}+\frac{b^2 \sqrt{1-c^2 x^2} \left (2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-2 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+2 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt{c d x+d} \sqrt{e-c e x}}+\frac{a^2 \log (c x)}{\sqrt{d} \sqrt{e}}-\frac{a^2 \log \left (\sqrt{d} \sqrt{e} \sqrt{c d x+d} \sqrt{e-c e x}+d e\right )}{\sqrt{d} \sqrt{e}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Log[c*x])/(Sqrt[d]*Sqrt[e]) - (a^2*Log[d*e + Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]])/(Sqrt[d]*S
qrt[e]) + (2*a*b*Sqrt[1 - c^2*x^2]*(ArcSin[c*x]*(Log[1 - E^(I*ArcSin[c*x])] - Log[1 + E^(I*ArcSin[c*x])]) + I*
PolyLog[2, -E^(I*ArcSin[c*x])] - I*PolyLog[2, E^(I*ArcSin[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b^2*Sq
rt[1 - c^2*x^2]*(ArcSin[c*x]^2*Log[1 - E^(I*ArcSin[c*x])] - ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] + (2*I)*A
rcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)*ArcSin[c*x]*PolyLog[2, E^(I*ArcSin[c*x])] - 2*PolyLog[3, -E^
(I*ArcSin[c*x])] + 2*PolyLog[3, E^(I*ArcSin[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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Maple [F]  time = 0.289, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{x}{\frac{1}{\sqrt{cdx+d}}}{\frac{1}{\sqrt{-cex+e}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/x/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2),x)

[Out]

int((a+b*arcsin(c*x))^2/x/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2),x)

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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))^2/x/(c*d*x+d)^(1/2)/(-c*e*x+e)^(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{{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt{c d x + d} \sqrt{-c e x + e}}{c^{2} d e x^{3} - d e 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*d*x+d)^(1/2)/(-c*e*x+e)^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^2*d*e*x^3 - d*e*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))**2/x/(c*d*x+d)**(1/2)/(-c*e*x+e)**(1/2),x)

[Out]

Timed out

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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}}{\sqrt{c d x + d} \sqrt{-c e x + e} 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*d*x+d)^(1/2)/(-c*e*x+e)^(1/2),x, algorithm="giac")

[Out]

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