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

Optimal. Leaf size=126 \[ -\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]

[Out]

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

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Rubi [A]  time = 0.183361, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4805, 12, 4625, 3717, 2190, 2531, 2282, 6589} \[ -\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4625

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

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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{\left (a+b \sin ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}-\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+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{i b \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{i b \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{i b \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{b^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ \end{align*}

Mathematica [A]  time = 0.163338, size = 170, normalized size = 1.35 \[ \frac{-i a b \left (\sin ^{-1}(c+d x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )\right )+b^2 \left (i \sin ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac{1}{3} i \sin ^{-1}(c+d x)^3+\sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac{i \pi ^3}{24}\right )+a^2 \log (c+d x)+2 a b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*b*ArcSin[c + d*x]*Log[1 - E^((2*I)*ArcSin[c + d*x])] + a^2*Log[c + d*x] - I*a*b*(ArcSin[c + d*x]^2 + Poly
Log[2, E^((2*I)*ArcSin[c + d*x])]) + b^2*((-I/24)*Pi^3 + (I/3)*ArcSin[c + d*x]^3 + ArcSin[c + d*x]^2*Log[1 - E
^((-2*I)*ArcSin[c + d*x])] + I*ArcSin[c + d*x]*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])] + PolyLog[3, E^((-2*I)*A
rcSin[c + d*x])]/2))/(d*e)

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Maple [B]  time = 0.039, size = 455, normalized size = 3.6 \begin{align*}{\frac{{a}^{2}\ln \left ( dx+c \right ) }{de}}-{\frac{{\frac{i}{3}}{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{3}}{de}}+{\frac{{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{de}\ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,i{b}^{2}\arcsin \left ( dx+c \right ) }{de}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+2\,{\frac{{b}^{2}{\it polylog} \left ( 3,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{de}}+{\frac{{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{de}\ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,i{b}^{2}\arcsin \left ( dx+c \right ) }{de}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+2\,{\frac{{b}^{2}{\it polylog} \left ( 3,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{de}}-{\frac{iab \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{de}}+2\,{\frac{ab\arcsin \left ( dx+c \right ) \ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{de}}+2\,{\frac{ab\arcsin \left ( dx+c \right ) \ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{de}}-{\frac{2\,iab}{de}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,iab}{de}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(d*x+c))^2/(d*e*x+c*e),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{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d e x + c e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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