\(\int e^{c+3 \coth ^{-1}(a+b x)} \, dx\) [5]

Optimal result

Integrand size = 12, antiderivative size = 1 \[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=0 \] Output:

0
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 78.00 \[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=\frac {e^c \left (\frac {\sqrt {1-\frac {1}{(a+b x)^2}} \left (a^2+b x (-5+b x)+a (-5+2 b x)\right )}{-1+a+b x}+3 \log \left ((a+b x) \left (1+\sqrt {1-\frac {1}{(a+b x)^2}}\right )\right )\right )}{b} \] Input:

Integrate[E^(c + 3*ArcCoth[a + b*x]),x]
 

Output:

(E^c*((Sqrt[1 - (a + b*x)^(-2)]*(a^2 + b*x*(-5 + b*x) + a*(-5 + 2*b*x)))/( 
-1 + a + b*x) + 3*Log[(a + b*x)*(1 + Sqrt[1 - (a + b*x)^(-2)])]))/b
 

Rubi [F]

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.

\(\displaystyle \int e^{3 \coth ^{-1}(a+b x)+c} \, dx\)

\(\Big \downarrow \) 7281

\(\displaystyle \frac {\int e^{c+3 \coth ^{-1}(a+b x)}d(a+b x)}{b}\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {\int e^{c+3 \coth ^{-1}(a+b x)}d(a+b x)}{b}\)

Input:

Int[E^(c + 3*ArcCoth[a + b*x]),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7281

Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] 
 Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /;  !FalseQ[lst]]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.19 (sec) , antiderivative size = 168, normalized size of antiderivative = 168.00

method	result	size
risch	\(\frac {\sqrt {b x +a -1}\, \sqrt {b x +a +1}\, {\mathrm e}^{c}}{b}+\frac {\left (\frac {3 \ln \left (\frac {\frac {b \left (a +1\right )}{2}+\frac {\left (-1+a \right ) b}{2}+b^{2} x}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+\left (b \left (a +1\right )+\left (-1+a \right ) b \right ) x +\left (a +1\right ) \left (-1+a \right )}\right )}{\sqrt {b^{2}}}-\frac {4 \sqrt {\left (x +\frac {-1+a}{b}\right )^{2} b^{2}+2 \left (x +\frac {-1+a}{b}\right ) b}}{b^{2} \left (x +\frac {-1+a}{b}\right )}\right ) {\mathrm e}^{c} \sqrt {\left (b x +a -1\right ) \left (b x +a +1\right )}}{\sqrt {b x +a -1}\, \sqrt {b x +a +1}}\)	\(168\)

Input:

int(exp(c+3*arccoth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*exp(c)+(3*ln((1/2*b*(a+1)+1/2*(-1+a)*b 
+b^2*x)/(b^2)^(1/2)+(b^2*x^2+(b*(a+1)+(-1+a)*b)*x+(a+1)*(-1+a))^(1/2))/(b^ 
2)^(1/2)-4/b^2/(x+(-1+a)/b)*((x+(-1+a)/b)^2*b^2+2*(x+(-1+a)/b)*b)^(1/2))*e 
xp(c)*((b*x+a-1)*(b*x+a+1))^(1/2)/(b*x+a-1)^(1/2)/(b*x+a+1)^(1/2)
 

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 79.00 \[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=\frac {{\left (b x + a - 5\right )} \sqrt {\frac {b x + a + 1}{b x + a - 1}} e^{c} + 3 \, e^{c} \log \left (\sqrt {\frac {b x + a + 1}{b x + a - 1}} + 1\right ) - 3 \, e^{c} \log \left (\sqrt {\frac {b x + a + 1}{b x + a - 1}} - 1\right )}{b} \] Input:

integrate(exp(c+3*arccoth(b*x+a)),x, algorithm="fricas")
 

Output:

((b*x + a - 5)*sqrt((b*x + a + 1)/(b*x + a - 1))*e^c + 3*e^c*log(sqrt((b*x 
 + a + 1)/(b*x + a - 1)) + 1) - 3*e^c*log(sqrt((b*x + a + 1)/(b*x + a - 1) 
) - 1))/b
 

Sympy [F]

\[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=e^{c} \int e^{3 \operatorname {acoth}{\left (a + b x \right )}}\, dx \] Input:

integrate(exp(c+3*acoth(b*x+a)),x)
                                                                                    
                                                                                    
 

Output:

exp(c)*Integral(exp(3*acoth(a + b*x)), x)
 

Maxima [F]

\[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=\int { e^{\left (c + 3 \, \operatorname {arcoth}\left (b x + a\right )\right )} \,d x } \] Input:

integrate(exp(c+3*arccoth(b*x+a)),x, algorithm="maxima")
 

Output:

integrate(e^(c + 3*arccoth(b*x + a)), x)
 

Giac [F]

\[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=\int { e^{\left (c + 3 \, \operatorname {arcoth}\left (b x + a\right )\right )} \,d x } \] Input:

integrate(exp(c+3*arccoth(b*x+a)),x, algorithm="giac")
 

Output:

integrate(e^(c + 3*arccoth(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=\int {\mathrm {e}}^{c+3\,\mathrm {acoth}\left (a+b\,x\right )} \,d x \] Input:

int(exp(c + 3*acoth(a + b*x)),x)
 

Output:

int(exp(c + 3*acoth(a + b*x)), x)
 

Reduce [F]

\[ \int e^{c+3 \coth ^{-1}(a+b x)} \, dx=e^{c} \left (\int e^{3 \mathit {acoth} \left (b x +a \right )}d x \right ) \] Input:

int(exp(c+3*acoth(b*x+a)),x)
 

Output:

e**c*int(e**(3*acoth(a + b*x)),x)