\(\int \csc ^{-1}(c e^{a+b x}) \, dx\) [40]

Optimal result

Integrand size = 10, antiderivative size = 85 \[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\frac {i \csc ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\csc ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{b}+\frac {i \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(85)=170\).

Time = 0.39 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.29 \[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=x \csc ^{-1}\left (c e^{a+b x}\right )+\frac {e^{-a-b x} \left (4 \sqrt {-1+c^2 e^{2 (a+b x)}} \arctan \left (\sqrt {-1+c^2 e^{2 (a+b x)}}\right ) \left (2 b x-\log \left (c^2 e^{2 (a+b x)}\right )\right )+\sqrt {1-c^2 e^{2 (a+b x)}} \left (\log ^2\left (c^2 e^{2 (a+b x)}\right )-4 \log \left (c^2 e^{2 (a+b x)}\right ) \log \left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )+2 \log ^2\left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )-4 \sqrt {1-c^2 e^{2 (a+b x)}} \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1-\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )}{8 b c \sqrt {1-\frac {e^{-2 (a+b x)}}{c^2}}} \] Input:

Integrate[ArcCsc[c*E^(a + b*x)],x]
 

Output:

x*ArcCsc[c*E^(a + b*x)] + (E^(-a - b*x)*(4*Sqrt[-1 + c^2*E^(2*(a + b*x))]* 
ArcTan[Sqrt[-1 + c^2*E^(2*(a + b*x))]]*(2*b*x - Log[c^2*E^(2*(a + b*x))]) 
+ Sqrt[1 - c^2*E^(2*(a + b*x))]*(Log[c^2*E^(2*(a + b*x))]^2 - 4*Log[c^2*E^ 
(2*(a + b*x))]*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2] + 2*Log[(1 + Sqr 
t[1 - c^2*E^(2*(a + b*x))])/2]^2) - 4*Sqrt[1 - c^2*E^(2*(a + b*x))]*PolyLo 
g[2, (1 - Sqrt[1 - c^2*E^(2*(a + b*x))])/2]))/(8*b*c*Sqrt[1 - 1/(c^2*E^(2* 
(a + b*x)))])
 

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2720, 5742, 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.

\(\displaystyle \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int e^{-a-b x} \csc ^{-1}\left (c e^{a+b x}\right )de^{a+b x}}{b}\)

\(\Big \downarrow \) 5742

\(\displaystyle -\frac {\int e^{-a-b x} \arcsin \left (\frac {e^{-a-b x}}{c}\right )de^{-a-b x}}{b}\)

\(\Big \downarrow \) 5136

\(\displaystyle -\frac {\int c e^{a+b x} \sqrt {1-\frac {e^{-2 a-2 b x}}{c^2}} \arcsin \left (\frac {e^{-a-b x}}{c}\right )d\arcsin \left (\frac {e^{-a-b x}}{c}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int -\arcsin \left (\frac {e^{-a-b x}}{c}\right ) \tan \left (\arcsin \left (\frac {e^{-a-b x}}{c}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {e^{-a-b x}}{c}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \arcsin \left (\frac {e^{-a-b x}}{c}\right ) \tan \left (\arcsin \left (\frac {e^{-a-b x}}{c}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {e^{-a-b x}}{c}\right )}{b}\)

\(\Big \downarrow \) 4200

\(\displaystyle -\frac {2 i \int -\frac {e^{a+b x+2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}}{1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}}d\arcsin \left (\frac {e^{-a-b x}}{c}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-2 i \int \frac {e^{a+b x+2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}}{1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}}d\arcsin \left (\frac {e^{-a-b x}}{c}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \arcsin \left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )d\arcsin \left (\frac {e^{-a-b x}}{c}\right )\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \arcsin \left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )-\frac {1}{4} \int e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )} \log \left (1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )de^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {-2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )+\frac {1}{2} i \arcsin \left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {e^{-a-b x}}{c}\right )}\right )\right )-\frac {1}{2} i e^{2 a+2 b x}}{b}\)

Input:

Int[ArcCsc[c*E^(a + b*x)],x]
 

Output:

-(((-1/2*I)*E^(2*a + 2*b*x) - (2*I)*((I/2)*ArcSin[E^(-a - b*x)/c]*Log[1 - 
E^((2*I)*ArcSin[E^(-a - b*x)/c])] + PolyLog[2, E^((2*I)*ArcSin[E^(-a - b*x 
)/c])]/4))/b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

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]
 

rule 2715

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]
 

rule 2720

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]]]
 

rule 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4200

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]
 

rule 5136

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]
 

rule 5742

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b 
*ArcSin[x/c])/x, x], x, 1/x] /; FreeQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.21

method	result	size
derivativedivides	\(\frac {\frac {i \operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \operatorname {polylog}\left (2, -\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )-\operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1-\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \operatorname {polylog}\left (2, \frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )}{b}\)	\(188\)
default	\(\frac {\frac {i \operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \operatorname {polylog}\left (2, -\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )-\operatorname {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1-\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \operatorname {polylog}\left (2, \frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )}{b}\)	\(188\)

Input:

int(arccsc(exp(b*x+a)*c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\int \operatorname {acsc}{\left (c e^{a + b x} \right )}\, dx \] Input:

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

Output:

Integral(acsc(c*exp(a + b*x)), x)
 

Maxima [F]

\[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\int { \operatorname {arccsc}\left (c e^{\left (b x + a\right )}\right ) \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\int { \operatorname {arccsc}\left (c e^{\left (b x + a\right )}\right ) \,d x } \] Input:

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

Output:

integrate(arccsc(c*e^(b*x + a)), x)
 

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07 \[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2\,b}+\frac {{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )}^2\,1{}\mathrm {i}}{2\,b}-\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )}{b} \] Input:

int(asin(exp(- a - b*x)/c),x)
 

Output:

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

Reduce [F]

\[ \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx=\int \mathit {acsc} \left (e^{b x +a} c \right )d x \] Input:

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

Output:

int(acsc(e**(a + b*x)*c),x)