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

Optimal. Leaf size=156 \[ \frac {2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac {2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4657, 4181, 2531, 2282, 6589} \[ \frac {2 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac {2 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac {2 b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac {2 b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c*d) + ((2*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*
E^(I*ArcSin[c*x])])/(c*d) - ((2*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c*d) - (2*b^2*PolyL
og[3, (-I)*E^(I*ArcSin[c*x])])/(c*d) + (2*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(c*d)

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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4657

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

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

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Mathematica [A]  time = 0.50, size = 207, normalized size = 1.33 \[ \frac {a^2 (-\log (1-c x))+a^2 \log (c x+1)+4 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-4 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+4 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-4 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-4 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )+4 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )-4 i b^2 \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{2 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4*I)*b^2*ArcSin[c*x]^2*ArcTan[E^(I*ArcSin[c*x])] + 4*a*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 4*a*b*A
rcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - a^2*Log[1 - c*x] + a^2*Log[1 + c*x] + (4*I)*b*(a + b*ArcSin[c*x])*Po
lyLog[2, (-I)*E^(I*ArcSin[c*x])] - (4*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])] - 4*b^2*PolyLog
[3, (-I)*E^(I*ArcSin[c*x])] + 4*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(2*c*d)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{c^{2} d x^{2} - d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 404, normalized size = 2.59 \[ \frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}-\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}+\frac {2 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d c}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}+\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}-\frac {2 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d c}+\frac {2 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}-\frac {2 i a b \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}+\frac {2 i a b \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c d}-\frac {2 i a^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c/d*b^2*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I/c/d*b^2*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^
2+1)^(1/2)))+2*b^2*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d/c-1/c/d*b^2*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2
+1)^(1/2)))+2*I/c/d*b^2*arcsin(c*x)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*b^2*polylog(3,-I*(I*c*x+(-c^2*x
^2+1)^(1/2)))/d/c+2/c*a*b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I/c/d*a*b*polylog(2,I*(I*c*x+(-c^
2*x^2+1)^(1/2)))-2/c*a*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I/c/d*a*b*polylog(2,-I*(I*c*x+(-c^
2*x^2+1)^(1/2)))-2*I/c/d*a^2*arctan(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 \[ \frac {1}{2} \, a^{2} {\left (\frac {\log \left (c x + 1\right )}{c d} - \frac {\log \left (c x - 1\right )}{c d}\right )} + \frac {b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) - b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, c d \int \frac {2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{2} d x^{2} - d}\,{d x}}{2 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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