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]

1/4*exp(2*b*x+2*a)/b+1/2*x

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, 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 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]]]

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*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {e^{2 a+2 b x}}{4 b}+\frac {x}{2} \]

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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fricas [B]  time = 0.63, size = 50, normalized size = 2.17 \[ \frac {{\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) - {\left (2 \, b x - 1\right )} \sinh \left (b x + a\right )}{4 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

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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giac [A]  time = 0.11, size = 22, normalized size = 0.96 \[ \frac {2 \, b x + 2 \, a + e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \]

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

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maple [A]  time = 0.04, size = 37, normalized size = 1.61 \[ \frac {\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{2}+\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}}{b} \]

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 = 0.31, size = 24, normalized size = 1.04 \[ \frac {1}{2} \, x + \frac {a}{2 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \]

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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mupad [B]  time = 0.92, size = 18, normalized size = 0.78 \[ \frac {x}{2}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*exp(a + b*x),x)

[Out]

x/2 + exp(2*a + 2*b*x)/(4*b)

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sympy [A]  time = 1.02, size = 63, normalized size = 2.74 \[ \begin {cases} - \frac {x e^{a} e^{b x} \sinh {\left (a + b x \right )}}{2} + \frac {x e^{a} e^{b x} \cosh {\left (a + b x \right )}}{2} + \frac {e^{a} e^{b x} \sinh {\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\x e^{a} \cosh {\relax (a )} & \text {otherwise} \end {cases} \]

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)/(2*b), Ne(b, 0)), (x*exp(a)*cosh(a), True))

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