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

Optimal. Leaf size=257 \[ \frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}} \]

[Out]

-2*(a+b*arcsin(c*x))^2*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+2*I*b*(a+b*ar
csin(c*x))*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)-2*I*b*(a+b*arcsin(c*x)
)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)-2*b^2*polylog(3,-I*c*x-(-c^2*x^2
+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+2*b^2*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2
)/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4713, 4709, 4183, 2531, 2282, 6589} \[ \frac {2 i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 4713

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

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^2 d x^2}} \, 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^2 d x^2}}\\ &=\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^2 d x^2}}\\ &=-\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^2 d x^2}}-\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^2 d x^2}}+\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^2 d x^2}}\\ &=-\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^2 d x^2}}+\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^2 d x^2}}-\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^2 d x^2}}-\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^2 d x^2}}+\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^2 d x^2}}\\ &=-\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^2 d x^2}}+\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^2 d x^2}}-\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^2 d x^2}}-\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^2 d x^2}}+\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^2 d x^2}}\\ &=-\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^2 d x^2}}+\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^2 d x^2}}-\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^2 d x^2}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 301, normalized size = 1.17 \[ -\frac {a^2 \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )}{\sqrt {d}}+\frac {a^2 \log (c x)}{\sqrt {d}}+\frac {2 a b \sqrt {1-c^2 x^2} \left (i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )}{\sqrt {d-c^2 d x^2}}+\frac {b^2 \sqrt {1-c^2 x^2} \left (2 i \sin ^{-1}(c x) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-2 i \sin ^{-1}(c x) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-2 \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )+2 \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt {d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Log[c*x])/Sqrt[d] - (a^2*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]])/Sqrt[d] + (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^2*d*x^2] + (b^2*Sqrt[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)*ArcSin[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])]))/Sqr
t[d - c^2*d*x^2]

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{2} d x^{3} - d x}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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)^(1/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.28, size = 388, normalized size = 1.51 \[ -\frac {a^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}-\frac {b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 i \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-\arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}+\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-\polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+\polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \left (c^{2} x^{2}-1\right )} \]

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)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{2} \log \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {d}}{{\left | x \right |}} + \frac {2 \, d}{{\left | x \right |}}\right )}{\sqrt {d}} - \sqrt {d} \int \frac {{\left (b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{2} d x^{3} - d x}\,{d x} \]

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)^(1/2),x, algorithm="maxima")

[Out]

-a^2*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x))/sqrt(d) - sqrt(d)*integrate((b^2*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1)/
(c^2*d*x^3 - d*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x\,\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

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)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))**2/(x*sqrt(-d*(c*x - 1)*(c*x + 1))), x)

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