\(\int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx\) [155]

Optimal result

Integrand size = 25, antiderivative size = 74 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {e^{2 c (a+b x)} \text {sech}(a c+b c x)}{4 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {x \text {sech}(a c+b c x)}{2 \sqrt {\text {sech}^2(a c+b c x)}} \]

[Out]

1/4*exp(2*c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+1/2*x*sech(b*c*x+a*c)/(sech(b*c*x+a*c)^2)^(
1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 14} \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {e^{2 c (a+b x)} \text {sech}(a c+b c x)}{4 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {x \text {sech}(a c+b c x)}{2 \sqrt {\text {sech}^2(a c+b c x)}} \]

[In]

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

[Out]

(E^(2*c*(a + b*x))*Sech[a*c + b*c*x])/(4*b*c*Sqrt[Sech[a*c + b*c*x]^2]) + (x*Sech[a*c + b*c*x])/(2*Sqrt[Sech[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 {\text {sech}(a c+b c x) \int e^{c (a+b x)} \cosh (a c+b c x) \, dx}{\sqrt {\text {sech}^2(a c+b c x)}} \\ & = \frac {\text {sech}(a c+b c x) \text {Subst}\left (\int \frac {1+x^2}{2 x} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\text {sech}^2(a c+b c x)}} \\ & = \frac {\text {sech}(a c+b c x) \text {Subst}\left (\int \frac {1+x^2}{x} \, dx,x,e^{c (a+b x)}\right )}{2 b c \sqrt {\text {sech}^2(a c+b c x)}} \\ & = \frac {\text {sech}(a c+b c x) \text {Subst}\left (\int \left (\frac {1}{x}+x\right ) \, dx,x,e^{c (a+b x)}\right )}{2 b c \sqrt {\text {sech}^2(a c+b c x)}} \\ & = \frac {e^{2 c (a+b x)} \text {sech}(a c+b c x)}{4 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {x \text {sech}(a c+b c x)}{2 \sqrt {\text {sech}^2(a c+b c x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {\left (e^{2 c (a+b x)}+2 b c x\right ) \text {sech}(c (a+b x))}{4 b c \sqrt {\text {sech}^2(c (a+b x))}} \]

[In]

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

[Out]

((E^(2*c*(a + b*x)) + 2*b*c*x)*Sech[c*(a + b*x)])/(4*b*c*Sqrt[Sech[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 = 0.43 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\operatorname {csgn}\left (\operatorname {sech}\left (c \left (b x +a \right )\right )\right ) \left (\frac {\cosh \left (b c x +a c \right )^{2}}{2}+\frac {\sinh \left (b c x +a c \right ) \cosh \left (b c x +a c \right )}{2}+\frac {b c x}{2}+\frac {a c}{2}\right )}{c b}\)	\(60\)
risch	\(\frac {x \,{\mathrm e}^{c \left (b x +a \right )}}{2 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {{\mathrm e}^{3 c \left (b x +a \right )}}{4 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}\)	\(106\)

[In]

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

[Out]

csgn(sech(c*(b*x+a)))/c/b*(1/2*cosh(b*c*x+a*c)^2+1/2*sinh(b*c*x+a*c)*cosh(b*c*x+a*c)+1/2*b*c*x+1/2*a*c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {{\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - {\left (2 \, b c x - 1\right )} \sinh \left (b c x + a c\right )}{4 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]

[In]

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

[Out]

1/4*((2*b*c*x + 1)*cosh(b*c*x + a*c) - (2*b*c*x - 1)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b*c*
x + a*c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.39 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {1}{2} \, x + \frac {a}{2 \, b} + \frac {e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \]

[In]

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

[Out]

1/2*x + 1/2*a/b + 1/4*e^(2*b*c*x + 2*a*c)/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.45 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\text {sech}^2(a c+b c x)}} \, dx=\frac {{\left (2 \, b c x e^{\left (-a c\right )} + e^{\left (2 \, b c x + a c\right )}\right )} e^{\left (a c\right )}}{4 \, b c} \]

[In]

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

[Out]

1/4*(2*b*c*x*e^(-a*c) + e^(2*b*c*x + a*c))*e^(a*c)/(b*c)

Mupad [F(-1)]

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

[In]

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

[Out]

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