3.267 \(\int e^{a+b x} \cosh (a+b x) \, dx\)

Optimal. Leaf size=23 \[ \frac{e^{2 a+2 b x}}{4 b}+\frac{x}{2} \]

[Out]

E^(2*a + 2*b*x)/(4*b) + x/2

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Rubi [A]  time = 0.0151755, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2282, 12, 14} \[ \frac{e^{2 a+2 b x}}{4 b}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[a + b*x],x]

[Out]

E^(2*a + 2*b*x)/(4*b) + x/2

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

Mathematica [A]  time = 0.0111907, size = 23, normalized size = 1. \[ \frac{e^{2 a+2 b x}}{4 b}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*a + 2*b*x)/(4*b) + x/2

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Maple [A]  time = 0.007, size = 37, normalized size = 1.6 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.11335, size = 32, normalized size = 1.39 \begin{align*} \frac{1}{2} \, x + \frac{a}{2 \, b} + \frac{e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.79648, size = 131, normalized size = 5.7 \begin{align*} \frac{{\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) -{\left (2 \, b x - 1\right )} \sinh \left (b x + a\right )}{4 \,{\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a),x, algorithm="fricas")

[Out]

1/4*((2*b*x + 1)*cosh(b*x + a) - (2*b*x - 1)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

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Sympy [A]  time = 1.25577, size = 80, normalized size = 3.48 \begin{align*} \begin{cases} - \frac{x e^{a} e^{b x} \sinh{\left (a + b x \right )}}{2} + \frac{x e^{a} e^{b x} \cosh{\left (a + b x \right )}}{2} + \frac{e^{a} e^{b x} \sinh{\left (a + b x \right )}}{b} - \frac{e^{a} e^{b x} \cosh{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\x e^{a} \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x*exp(a)*exp(b*x)*sinh(a + b*x)/2 + x*exp(a)*exp(b*x)*cosh(a + b*x)/2 + exp(a)*exp(b*x)*sinh(a + b
*x)/b - exp(a)*exp(b*x)*cosh(a + b*x)/(2*b), Ne(b, 0)), (x*exp(a)*cosh(a), True))

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Giac [A]  time = 1.19928, size = 30, normalized size = 1.3 \begin{align*} \frac{2 \, b x + 2 \, a + e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a),x, algorithm="giac")

[Out]

1/4*(2*b*x + 2*a + e^(2*b*x + 2*a))/b