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

Optimal. Leaf size=84 \[ -\frac {i \text {Li}_2\left (e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac {i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2282, 4625, 3717, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac {i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[c*E^(a + b*x)],x]

[Out]

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

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 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 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]

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 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]

Rubi steps

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

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Mathematica [F]  time = 1.14, size = 0, normalized size = 0.00 \[ \int \sin ^{-1}\left (c e^{a+b x}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[ArcSin[c*E^(a + b*x)],x]

[Out]

Integrate[ArcSin[c*E^(a + b*x)], x]

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(c*exp(b*x+a)),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(c*exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(arcsin(c*e^(b*x + a)), x)

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maple [A]  time = 0.03, size = 181, normalized size = 2.15 \[ -\frac {i \arcsin \left (c \,{\mathrm e}^{b x +a}\right )^{2}}{2 b}+\frac {\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1+i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b}+\frac {\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1-i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b}-\frac {i \polylog \left (2, -i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b}-\frac {i \polylog \left (2, i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(c*exp(b*x+a)),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {-i \, b^{2} c^{2} {\left (\frac {b x \log \left (c e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-c e^{\left (b x + a\right )}\right )}{b^{2} c^{2}} + \frac {b x \log \left (-c e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (c e^{\left (b x + a\right )}\right )}{b^{2} c^{2}}\right )} + 2 \, b^{2} c e^{a} \int \frac {\sqrt {-c e^{\left (b x + a\right )} + 1} x e^{\left (b x\right )}}{\sqrt {c e^{\left (b x + a\right )} + 1} c e^{\left (b x + a\right )} - \sqrt {c e^{\left (b x + a\right )} + 1}}\,{d x} + 2 \, b x \arctan \left (c e^{\left (b x + a\right )}, \sqrt {c e^{\left (b x + a\right )} + 1} \sqrt {-c e^{\left (b x + a\right )} + 1}\right ) + i \, b x \log \left (c e^{\left (b x + a\right )} + 1\right ) + i \, b x \log \left (-c e^{\left (b x + a\right )} + 1\right ) + i \, {\rm Li}_2\left (c e^{\left (b x + a\right )}\right ) + i \, {\rm Li}_2\left (-c e^{\left (b x + a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(c*exp(b*x+a)),x, algorithm="maxima")

[Out]

1/2*(-2*I*b^2*c^2*integrate(x*e^(2*b*x + 2*a)/(c^4*e^(4*b*x + 4*a) - c^2*e^(2*b*x + 2*a) + (c^2*e^(2*b*x + 2*a
) - 1)*e^(log(c*e^(b*x + a) + 1) + log(-c*e^(b*x + a) + 1))), x) + 2*b^2*c*integrate(x*e^(b*x + a + 1/2*log(c*
e^(b*x + a) + 1) + 1/2*log(-c*e^(b*x + a) + 1))/(c^4*e^(4*b*x + 4*a) - c^2*e^(2*b*x + 2*a) + (c^2*e^(2*b*x + 2
*a) - 1)*e^(log(c*e^(b*x + a) + 1) + log(-c*e^(b*x + a) + 1))), x) + 2*b*x*arctan2(c*e^(b*x + a), sqrt(c*e^(b*
x + a) + 1)*sqrt(-c*e^(b*x + a) + 1)) + I*b*x*log(c*e^(b*x + a) + 1) + I*b*x*log(-c*e^(b*x + a) + 1) + I*dilog
(c*e^(b*x + a)) + I*dilog(-c*e^(b*x + a)))/b

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mupad [B]  time = 0.74, size = 70, normalized size = 0.83 \[ -\frac {{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )}^2\,1{}\mathrm {i}}{2\,b}-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2\,b}+\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(c*exp(a + b*x)),x)

[Out]

(log(1 - exp(asin(c*exp(a + b*x))*2i))*asin(c*exp(a + b*x)))/b - (polylog(2, exp(asin(c*exp(a + b*x))*2i))*1i)
/(2*b) - (asin(c*exp(a + b*x))^2*1i)/(2*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(c*exp(b*x+a)),x)

[Out]

Integral(asin(c*exp(a + b*x)), x)

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