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

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

[Out]

(E^(2*c*(a + b*x))*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(4*b*c) - (x*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c +
b*c*x]^2])/2

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Rubi [A]  time = 0.111542, 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} \[ \frac{e^{2 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{4 b c}-\frac{1}{2} x \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*c*(a + b*x))*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(4*b*c) - (x*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c +
b*c*x]^2])/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{\sinh ^2(a c+b c x)} \, dx &=\left (\text{csch}(a c+b c x) \sqrt{\sinh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \sinh (a c+b c x) \, dx\\ &=\frac{\left (\text{csch}(a c+b c x) \sqrt{\sinh ^2(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 (\text{csch}(a c+b c x) \sqrt{\sinh ^2(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 (\text{csch}(a c+b c x) \sqrt{\sinh ^2(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)} \text{csch}(a c+b c x) \sqrt{\sinh ^2(a c+b c x)}}{4 b c}-\frac{1}{2} x \text{csch}(a c+b c x) \sqrt{\sinh ^2(a c+b c x)}\\ \end{align*}

Mathematica [A]  time = 0.0378588, size = 48, normalized size = 0.65 \[ \frac{\left (e^{2 c (a+b x)}-2 b c x\right ) \sqrt{\sinh ^2(c (a+b x))} \text{csch}(c (a+b x))}{4 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((E^(2*c*(a + b*x)) - 2*b*c*x)*Csch[c*(a + b*x)]*Sqrt[Sinh[c*(a + b*x)]^2])/(4*b*c)

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Maple [A]  time = 0.112, size = 100, normalized size = 1.4 \begin{align*}{\frac{\cosh \left ( c \left ( bx+a \right ) \right ) }{2\,cb}\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}}}-{\frac{1}{2\,cb}\ln \left ( \cosh \left ( c \left ( bx+a \right ) \right ) +\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}} \right ) }+{\frac{ \left ( \cosh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}}{2\,cb\sinh \left ( c \left ( bx+a \right ) \right ) }\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.6531, size = 49, normalized size = 0.66 \begin{align*} -\frac{b c x + a c}{2 \, b c} + \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))*(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84701, size = 165, normalized size = 2.23 \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))*(sinh(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 [A]  time = 31.1821, size = 206, normalized size = 2.78 \begin{align*} \begin{cases} x \sqrt{\sinh ^{2}{\left (a c \right )}} e^{a c} & \text{for}\: b = 0 \\0 & \text{for}\: a = \frac{\log{\left (- e^{- b c x} \right )}}{c} \vee a = \frac{\log{\left (e^{- b c x} \right )}}{c} \vee c = 0 \\\frac{x \sqrt{\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2} - \frac{x \sqrt{\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \cosh{\left (a c + b c x \right )}}{2 \sinh{\left (a c + b c x \right )}} - \frac{\sqrt{\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2 b c} + \frac{\sqrt{\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \cosh{\left (a c + b c x \right )}}{b c \sinh{\left (a c + b c x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16266, size = 96, normalized size = 1.3 \begin{align*} -\frac{1}{2} \, x \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \frac{e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )}{4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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