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

Optimal. Leaf size=117 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d} \]

[Out]

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

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Rubi [A]  time = 0.172138, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4675, 3719, 2190, 2531, 2282, 6589} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4675

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

Rule 3719

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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{x \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 \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\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 )}{c^2 d}\\ &=\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\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 )}{c^2 d}\\ &=\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ &=\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac{b^2 \text{Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0801326, size = 143, normalized size = 1.22 \[ \frac{6 i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-3 b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )-3 a^2 \log \left (1-c^2 x^2\right )+6 i a b \sin ^{-1}(c x)^2-12 a b \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+2 i b^2 \sin ^{-1}(c x)^3-6 b^2 \sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{6 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 258, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}\ln \left ( cx-1 \right ) }{2\,{c}^{2}d}}-{\frac{{a}^{2}\ln \left ( cx+1 \right ) }{2\,{c}^{2}d}}+{\frac{{\frac{i}{3}}{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{3}}{{c}^{2}d}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{{c}^{2}d}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{i{b}^{2}\arcsin \left ( cx \right ) }{{c}^{2}d}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{b}^{2}}{2\,{c}^{2}d}{\it polylog} \left ( 3,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{iab \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{{c}^{2}d}}-2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{{c}^{2}d}}+{\frac{iab}{{c}^{2}d}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} x \arcsin \left (c x\right )^{2} + 2 \, a b x \arcsin \left (c x\right ) + a^{2} x}{c^{2} d x^{2} - d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a^{2} x}{c^{2} x^{2} - 1}\, dx + \int \frac{b^{2} x \operatorname{asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac{2 a b x \operatorname{asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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} x}{c^{2} d x^{2} - d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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