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

Optimal. Leaf size=89 \[ -\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e} \]

[Out]

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

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Rubi [A]  time = 0.104629, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4805, 12, 4625, 3717, 2190, 2279, 2391} \[ -\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*(a + b*ArcSin[c + d*x])^2)/(b*d*e) + ((a + b*ArcSin[c + d*x])*Log[1 - E^((2*I)*ArcSin[c + d*x])])/(d*e
) - ((I/2)*b*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.0589282, size = 71, normalized size = 0.8 \[ \frac{-\frac{1}{2} i b \left (\sin ^{-1}(c+d x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )\right )+a \log (c+d x)+b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 182, normalized size = 2. \begin{align*}{\frac{a\ln \left ( dx+c \right ) }{de}}-{\frac{{\frac{i}{2}}b \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{de}}+{\frac{b\arcsin \left ( dx+c \right ) }{de}\ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+{\frac{b\arcsin \left ( dx+c \right ) }{de}\ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{ib}{de}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{ib}{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))/(d*e*x+c*e),x)

[Out]

1/d*a/e*ln(d*x+c)-1/2*I/d*b/e*arcsin(d*x+c)^2+1/d*b/e*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+1/d*b/
e*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-I/d*b/e*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-I/d*b/e*p
olylog(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))/(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 \arcsin \left (d x + c\right ) + a}{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))/(d*e*x+c*e),x, algorithm="fricas")

[Out]

integral((b*arcsin(d*x + c) + a)/(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}{c + d x}\, dx + \int \frac{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))/(d*e*x+c*e),x)

[Out]

(Integral(a/(c + d*x), x) + Integral(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{b \arcsin \left (d x + c\right ) + a}{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))/(d*e*x+c*e),x, algorithm="giac")

[Out]

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