3.154 \(\int e^{c (a+b x)} \sqrt {\text {sech}^2(a c+b c x)} \, dx\)

Optimal. Leaf size=44 \[ \frac {\log \left (e^{2 c (a+b x)}+1\right ) \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6720, 2282, 12, 260} \[ \frac {\log \left (e^{2 c (a+b x)}+1\right ) \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[a*c + b*c*x]*Log[1 + E^(2*c*(a + b*x))]*Sqrt[Sech[a*c + b*c*x]^2])/(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 260

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 2282

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 6720

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.95 \[ \frac {\log \left (e^{2 c (a+b x)}+1\right ) \cosh (c (a+b x)) \sqrt {\text {sech}^2(c (a+b x))}}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 42, normalized size = 0.95 \[ \frac {\log \left (\frac {2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(sech(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)

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giac [A]  time = 0.13, size = 20, normalized size = 0.45 \[ \frac {\log \left (e^{\left (2 \, b c x\right )} + e^{\left (-2 \, a c\right )}\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*b*c*x) + e^(-2*a*c))/(b*c)

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maple [A]  time = 0.72, size = 66, normalized size = 1.50 \[ \frac {\ln \left ({\mathrm e}^{2 b c x}+{\mathrm e}^{-2 a c}\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 21, normalized size = 0.48 \[ \frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*b*c*x + 2*a*c) + 1)/(b*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,\sqrt {\frac {1}{{\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int \sqrt {\operatorname {sech}^{2}{\left (a c + b c x \right )}} e^{b c x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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