\(\int \sec ^{-1}(c e^{a+b x}) \, dx\) [42]

Optimal result

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

1/2*I*arcsec(c*exp(b*x+a))^2/b-arcsec(c*exp(b*x+a))*ln(1+(1/c/(exp(1)^(b*x 
+a))+I*(1-1/c^2/(exp(1)^(b*x+a))^2)^(1/2))^2)/b+1/2*I*polylog(2,-(1/c/(exp 
(1)^(b*x+a))+I*(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 = 0.81 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.29 \[ \int \sec ^{-1}\left (c e^{a+b x}\right ) \, dx=x \sec ^{-1}\left (c e^{a+b x}\right )-\frac {e^{-a-b x} \left (4 \sqrt {-1+c^2 e^{2 (a+b x)}} \arctan \left (\sqrt {-1+c^2 e^{2 (a+b x)}}\right ) \left (2 b x-\log \left (c^2 e^{2 (a+b x)}\right )\right )+\sqrt {1-c^2 e^{2 (a+b x)}} \left (\log ^2\left (c^2 e^{2 (a+b x)}\right )-4 \log \left (c^2 e^{2 (a+b x)}\right ) \log \left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )+2 \log ^2\left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )-4 \sqrt {1-c^2 e^{2 (a+b x)}} \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1-\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )}{8 b c \sqrt {1-\frac {e^{-2 (a+b x)}}{c^2}}} \] Input:

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

Output:

x*ArcSec[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 = 100, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {2720, 5741, 5137, 3042, 4202, 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.

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 5741

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

\(\Big \downarrow \) 5137

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \arccos \left (\frac {e^{-a-b x}}{c}\right ) \tan \left (\arccos \left (\frac {e^{-a-b x}}{c}\right )\right )d\arccos \left (\frac {e^{-a-b x}}{c}\right )}{b}\)

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5137

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

rule 5741

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

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {\frac {i \operatorname {arcsec}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arcsec}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\left (\frac {{\mathrm e}^{-b x -a}}{c}+i \sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )^{2}\right )+\frac {i \operatorname {polylog}\left (2, -\left (\frac {{\mathrm e}^{-b x -a}}{c}+i \sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )^{2}\right )}{2}}{b}\)	\(111\)
default	\(\frac {\frac {i \operatorname {arcsec}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arcsec}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\left (\frac {{\mathrm e}^{-b x -a}}{c}+i \sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )^{2}\right )+\frac {i \operatorname {polylog}\left (2, -\left (\frac {{\mathrm e}^{-b x -a}}{c}+i \sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )^{2}\right )}{2}}{b}\)	\(111\)

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate(arcsec(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]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^{-1}\left (c e^{a+b x}\right ) \, dx=\int \mathrm {acos}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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