\(\int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx\) [295]

Optimal result

Integrand size = 25, antiderivative size = 44 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\cosh (a c+b c x) \log \left (1+e^{2 c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}} \]

[Out]

cosh(b*c*x+a*c)*ln(1+exp(2*c*(b*x+a)))/b/c/(cosh(b*c*x+a*c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, 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} \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\log \left (e^{2 c (a+b x)}+1\right ) \cosh (a c+b c x)}{b c \sqrt {\cosh ^2(a c+b c x)}} \]

[In]

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

[Out]

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

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}& = \frac {\cosh (a c+b c x) \int e^{c (a+b x)} \text {sech}(a c+b c x) \, dx}{\sqrt {\cosh ^2(a c+b c x)}} \\ & = \frac {\cosh (a c+b c x) \text {Subst}\left (\int \frac {2 x}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}} \\ & = \frac {(2 \cosh (a c+b c x)) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}} \\ & = \frac {\cosh (a c+b c x) \log \left (1+e^{2 c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\cosh (c (a+b x)) \log \left (1+e^{2 c (a+b x)}\right )}{b c \sqrt {\cosh ^2(c (a+b x))}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66

method	result	size
default	\(\operatorname {csgn}\left (\cosh \left (c \left (b x +a \right )\right )\right ) \left (x +\frac {\ln \left (\cosh \left (c \left (b x +a \right )\right )\right )}{c b}\right )\)	\(29\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 b c x}+{\mathrm e}^{-2 a c}\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) {\mathrm e}^{-c \left (b x +a \right )}}{b c \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\log \left (\frac {2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.48 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{b c} \]

[In]

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

[Out]

log(e^(2*b*c*x + 2*a*c) + 1)/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.45 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\cosh ^2(a c+b c x)}} \, dx=\frac {\log \left (e^{\left (2 \, b c x\right )} + e^{\left (-2 \, a c\right )}\right )}{b c} \]

[In]

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

[Out]

log(e^(2*b*c*x) + e^(-2*a*c))/(b*c)

Mupad [F(-1)]

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

[In]

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

[Out]

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