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]

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

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Rubi [A]  time = 0.10, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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 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 {\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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.52, size = 66, normalized size = 0.89 \[ \frac {{\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - {\left (2 \, b c x - 1\right )} \sinh \left (b c x + a c\right )}{4 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]

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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giac [A]  time = 0.11, size = 23, normalized size = 0.31 \[ \frac {1}{2} \, x + \frac {e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \]

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)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{c \left (b x +a \right )} \sqrt {\frac {\cosh \left (2 b c x +2 a c \right )}{2}+\frac {1}{2}}\, dx \]

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 = 0.34, size = 29, normalized size = 0.39 \[ \frac {1}{2} \, x + \frac {a}{2 \, b} + \frac {e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \]

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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mupad [B]  time = 0.12, size = 76, normalized size = 1.03 \[ \frac {\left (x\,{\mathrm {e}}^{a\,c+b\,c\,x}+\frac {{\mathrm {e}}^{3\,a\,c+3\,b\,c\,x}}{2\,b\,c}\right )\,\sqrt {{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}+\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}{{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*exp(a*c + b*c*x) + exp(3*a*c + 3*b*c*x)/(2*b*c))*((exp(a*c + b*c*x)/2 + exp(- a*c - b*c*x)/2)^2)^(1/2))/(e
xp(2*a*c + 2*b*c*x) + 1)

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sympy [A]  time = 16.61, size = 204, normalized size = 2.76 \[ \begin {cases} 0 & \text {for}\: a = \frac {\log {\left (- i e^{- b c x} \right )}}{c} \vee a = \frac {\log {\left (i e^{- b c x} \right )}}{c} \\x & \text {for}\: c = 0 \\x \sqrt {\cosh ^{2}{\left (a c \right )}} e^{a c} & \text {for}\: b = 0 \\- \frac {x \sqrt {\cosh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \sinh {\left (a c + b c x \right )}}{2 \cosh {\left (a c + b c x \right )}} + \frac {x \sqrt {\cosh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2} + \frac {\sqrt {\cosh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \sinh {\left (a c + b c x \right )}}{b c \cosh {\left (a c + b c x \right )}} - \frac {\sqrt {\cosh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2 b c} & \text {otherwise} \end {cases} \]

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]

Piecewise((0, Eq(a, log(I*exp(-b*c*x))/c) | Eq(a, log(-I*exp(-b*c*x))/c)), (x, Eq(c, 0)), (x*sqrt(cosh(a*c)**2
)*exp(a*c), Eq(b, 0)), (-x*sqrt(cosh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)*sinh(a*c + b*c*x)/(2*cosh(a*c + b*c*
x)) + x*sqrt(cosh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)/2 + sqrt(cosh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)*sinh
(a*c + b*c*x)/(b*c*cosh(a*c + b*c*x)) - sqrt(cosh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)/(2*b*c), True))

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