3.294 \(\int e^{c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \, dx\)

Optimal. Leaf size=74 \[ \frac{e^{2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{4 b c}+\frac{1}{2} x \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x) \]

[Out]

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

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Rubi [A]  time = 0.0993262, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6720, 2282, 12, 14} \[ \frac{e^{2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{4 b c}+\frac{1}{2} x \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]]

Rubi steps

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

Mathematica [A]  time = 0.0383602, size = 48, normalized size = 0.65 \[ \frac{\left (e^{2 c (a+b x)}+2 b c x\right ) \sqrt{\cosh ^2(c (a+b x))} \text{sech}(c (a+b x))}{4 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c \left ( bx+a \right ) }}\sqrt{ \left ( \cosh \left ( bcx+ac \right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15082, size = 39, normalized size = 0.53 \begin{align*} \frac{1}{2} \, x + \frac{a}{2 \, b} + \frac{e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.72602, size = 163, normalized size = 2.2 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.2278, size = 31, normalized size = 0.42 \begin{align*} \frac{1}{2} \, x + \frac{e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x + 1/4*e^(2*b*c*x + 2*a*c)/(b*c)