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

Optimal. Leaf size=347 \[ -\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}+\frac{2 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}+\frac{2 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e} \]

[Out]

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

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Rubi [A]  time = 0.508978, antiderivative size = 347, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4741, 4519, 2190, 2531, 2282, 6589} \[ -\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}+\frac{2 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}+\frac{2 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4741

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

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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

Mathematica [A]  time = 0.332467, size = 332, normalized size = 0.96 \[ \frac{-6 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )-6 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )+6 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )+6 b^2 \text{PolyLog}\left (3,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )+3 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}-c d}\right )+3 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{b}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.421, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \log \left (e x + d\right )}{e} + \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 )}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(e*x + d)/e + 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)))/(e*x + d), 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}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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