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

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Rubi [A]  time = 0.0869514, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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])

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 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]

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*}

Mathematica [A]  time = 0.0371545, 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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Maple [A]  time = 0.207, size = 66, normalized size = 1.5 \begin{align*}{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \ln \left ({{\rm e}^{2\,bcx}}+{{\rm e}^{-2\,ac}} \right ){{\rm e}^{-c \left ( bx+a \right ) }}}{cb}\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}} \end{align*}

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]

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

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Maxima [A]  time = 1.67855, size = 28, normalized size = 0.64 \begin{align*} \frac{\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{b c} \end{align*}

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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Fricas [A]  time = 3.22516, size = 97, normalized size = 2.2 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a c} \int \sqrt{\operatorname{sech}^{2}{\left (a c + b c x \right )}} e^{b c x}\, dx \end{align*}

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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Giac [A]  time = 1.13834, size = 27, normalized size = 0.61 \begin{align*} \frac{\log \left (e^{\left (2 \, b c x\right )} + e^{\left (-2 \, a c\right )}\right )}{b c} \end{align*}

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)