Integrand size = 10, antiderivative size = 84 \[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=-\frac {i \arcsin \left (c e^{a+b x}\right )^2}{2 b}+\frac {\arcsin \left (c e^{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )}{2 b} \]
-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*ex p(b*x+a)^2)^(1/2))^2)/b
\[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=\int \arcsin \left (c e^{a+b x}\right ) \, dx \]
Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2720, 5136, 3042, 25, 4200, 25, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \arcsin \left (c e^{a+b x}\right ) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int e^{-a-b x} \arcsin \left (c e^{a+b x}\right )de^{a+b x}}{b}\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle \frac {\int \frac {e^{-a-b x} \sqrt {1-c^2 e^{2 a+2 b x}} \arcsin \left (c e^{a+b x}\right )}{c}d\arcsin \left (c e^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\arcsin \left (c e^{a+b x}\right ) \tan \left (\arcsin \left (c e^{a+b x}\right )+\frac {\pi }{2}\right )d\arcsin \left (c e^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \arcsin \left (c e^{a+b x}\right ) \tan \left (\arcsin \left (c e^{a+b x}\right )+\frac {\pi }{2}\right )d\arcsin \left (c e^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle \frac {2 i \int -\frac {e^{a+b x+2 i \arcsin \left (c e^{a+b x}\right )}}{1-e^{2 i \arcsin \left (c e^{a+b x}\right )}}d\arcsin \left (c e^{a+b x}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-2 i \int \frac {e^{a+b x+2 i \arcsin \left (c e^{a+b x}\right )}}{1-e^{2 i \arcsin \left (c e^{a+b x}\right )}}d\arcsin \left (c e^{a+b x}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {-2 i \left (\frac {1}{2} i \arcsin \left (c e^{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )d\arcsin \left (c e^{a+b x}\right )\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {-2 i \left (\frac {1}{2} i \arcsin \left (c e^{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )-\frac {1}{4} \int e^{-a-b x} \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )de^{2 i \arcsin \left (c e^{a+b x}\right )}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {-2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )+\frac {1}{2} i \arcsin \left (c e^{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (c e^{a+b x}\right )}\right )\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\) |
((-1/2*I)*E^(2*a + 2*b*x) - (2*I)*((I/2)*ArcSin[c*E^(a + b*x)]*Log[1 - E^( (2*I)*ArcSin[c*E^(a + b*x)])] + PolyLog[2, E^((2*I)*ArcSin[c*E^(a + b*x)]) ]/4))/b
3.5.38.3.1 Defintions of rubi rules used
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol ] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Time = 0.92 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.02
method | result | size |
derivativedivides | \(\frac {-\frac {i \arcsin \left (c \,{\mathrm e}^{b x +a}\right )^{2}}{2}+\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1+i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )-i \operatorname {polylog}\left (2, -i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )+\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1-i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )-i \operatorname {polylog}\left (2, i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b}\) | \(170\) |
default | \(\frac {-\frac {i \arcsin \left (c \,{\mathrm e}^{b x +a}\right )^{2}}{2}+\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1+i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )-i \operatorname {polylog}\left (2, -i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )+\arcsin \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1-i c \,{\mathrm e}^{b x +a}-\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )-i \operatorname {polylog}\left (2, i c \,{\mathrm e}^{b x +a}+\sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )}{b}\) | \(170\) |
1/b*(-1/2*I*arcsin(c*exp(b*x+a))^2+arcsin(c*exp(b*x+a))*ln(1+I*c*exp(b*x+a )+(1-c^2*exp(b*x+a)^2)^(1/2))-I*polylog(2,-I*c*exp(b*x+a)-(1-c^2*exp(b*x+a )^2)^(1/2))+arcsin(c*exp(b*x+a))*ln(1-I*c*exp(b*x+a)-(1-c^2*exp(b*x+a)^2)^ (1/2))-I*polylog(2,I*c*exp(b*x+a)+(1-c^2*exp(b*x+a)^2)^(1/2)))
Exception generated. \[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=\int \operatorname {asin}{\left (c e^{a + b x} \right )}\, dx \]
\[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=\int { \arcsin \left (c e^{\left (b x + a\right )}\right ) \,d x } \]
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*l og(-c*e^(b*x + a) + 1) + I*dilog(c*e^(b*x + a)) + I*dilog(-c*e^(b*x + a))) /b
\[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=\int { \arcsin \left (c e^{\left (b x + a\right )}\right ) \,d x } \]
Time = 0.79 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \arcsin \left (c e^{a+b x}\right ) \, dx=-\frac {{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )}^2\,1{}\mathrm {i}}{2\,b}-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2\,b}+\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (c\,{\mathrm {e}}^{a+b\,x}\right )}{b} \]