\(\int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx\) [128]

Optimal result

Integrand size = 25, antiderivative size = 46 \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\sqrt {\text {csch}^2(a c+b c x)} \log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c} \]

[Out]

ln(1-exp(2*c*(b*x+a)))*sinh(b*c*x+a*c)*(csch(b*c*x+a*c)^2)^(1/2)/b/c

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 266} \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{b c} \]

[In]

Int[E^(c*(a + b*x))*Sqrt[Csch[a*c + b*c*x]^2],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2320

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 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \int e^{c (a+b x)} \text {csch}(a c+b c x) \, dx \\ & = \frac {\left (\sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \text {Subst}\left (\int \frac {2 x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c} \\ & = \frac {\left (2 \sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \text {Subst}\left (\int \frac {x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c} \\ & = \frac {\sqrt {\text {csch}^2(a c+b c x)} \log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\sqrt {\text {csch}^2(c (a+b x))} \log \left (1-e^{2 c (a+b x)}\right ) \sinh (c (a+b x))}{b c} \]

[In]

Integrate[E^(c*(a + b*x))*Sqrt[Csch[a*c + b*c*x]^2],x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.63

method	result	size
default	\(\operatorname {csgn}\left (\operatorname {csch}\left (c \left (b x +a \right )\right )\right ) \left (x +\frac {\ln \left (\sinh \left (c \left (b x +a \right )\right )\right )}{c b}\right )\)	\(29\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 b c x}-{\mathrm e}^{-2 a c}\right ) \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{b c}\)	\(68\)

[In]

int(exp(c*(b*x+a))*(csch(b*c*x+a*c)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

csgn(csch(c*(b*x+a)))*(x+1/c/b*ln(sinh(c*(b*x+a))))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\log \left (\frac {2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)^2)^(1/2),x, algorithm="fricas")

[Out]

log(2*sinh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c)))/(b*c)

Sympy [F]

\[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=e^{a c} \int \sqrt {\operatorname {csch}^{2}{\left (a c + b c x \right )}} e^{b c x}\, dx \]

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)**2)**(1/2),x)

[Out]

exp(a*c)*Integral(sqrt(csch(a*c + b*c*x)**2)*exp(b*c*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\frac {\log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\right )}{b c \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )} \]

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)^2)^(1/2),x, algorithm="giac")

[Out]

log(abs(e^(2*b*c*x + 2*a*c) - 1))/(b*c*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)))

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \sqrt {\text {csch}^2(a c+b c x)} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,\sqrt {\frac {1}{{\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \]

[In]

int(exp(c*(a + b*x))*(1/sinh(a*c + b*c*x)^2)^(1/2),x)

[Out]

int(exp(c*(a + b*x))*(1/sinh(a*c + b*c*x)^2)^(1/2), x)