3.278 \(\int e^x \cosh (3 x) \, dx\)

Optimal. Leaf size=19 \[ \frac {e^{4 x}}{8}-\frac {1}{4} e^{-2 x} \]

[Out]

-1/4/exp(2*x)+1/8*exp(4*x)

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2282, 12, 14} \[ \frac {e^{4 x}}{8}-\frac {1}{4} e^{-2 x} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Cosh[3*x],x]

[Out]

-1/(4*E^(2*x)) + E^(4*x)/8

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^x \cosh (3 x) \, dx &=\operatorname {Subst}\left (\int \frac {1+x^6}{2 x^3} \, dx,x,e^x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x^6}{x^3} \, dx,x,e^x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x^3}+x^3\right ) \, dx,x,e^x\right )\\ &=-\frac {1}{4} e^{-2 x}+\frac {e^{4 x}}{8}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.84 \[ \frac {1}{8} e^{-2 x} \left (e^{6 x}-2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Cosh[3*x],x]

[Out]

(-2 + E^(6*x))/(8*E^(2*x))

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fricas [B]  time = 0.63, size = 38, normalized size = 2.00 \[ -\frac {\cosh \relax (x)^{3} - 9 \, \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} - 3 \, \sinh \relax (x)^{3}}{8 \, {\left (\cosh \relax (x) - \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*cosh(3*x),x, algorithm="fricas")

[Out]

-1/8*(cosh(x)^3 - 9*cosh(x)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 - 3*sinh(x)^3)/(cosh(x) - sinh(x))

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giac [A]  time = 0.11, size = 13, normalized size = 0.68 \[ \frac {1}{8} \, e^{\left (4 \, x\right )} - \frac {1}{4} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*cosh(3*x),x, algorithm="giac")

[Out]

1/8*e^(4*x) - 1/4*e^(-2*x)

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maple [A]  time = 0.10, size = 26, normalized size = 1.37 \[ \frac {\sinh \left (2 x \right )}{4}+\frac {\sinh \left (4 x \right )}{8}-\frac {\cosh \left (2 x \right )}{4}+\frac {\cosh \left (4 x \right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*cosh(3*x),x)

[Out]

1/4*sinh(2*x)+1/8*sinh(4*x)-1/4*cosh(2*x)+1/8*cosh(4*x)

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maxima [A]  time = 0.31, size = 13, normalized size = 0.68 \[ \frac {1}{8} \, e^{\left (4 \, x\right )} - \frac {1}{4} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*cosh(3*x),x, algorithm="maxima")

[Out]

1/8*e^(4*x) - 1/4*e^(-2*x)

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mupad [B]  time = 0.93, size = 12, normalized size = 0.63 \[ \frac {{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{6\,x}-2\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(3*x)*exp(x),x)

[Out]

(exp(-2*x)*(exp(6*x) - 2))/8

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sympy [A]  time = 0.25, size = 20, normalized size = 1.05 \[ \frac {3 e^{x} \sinh {\left (3 x \right )}}{8} - \frac {e^{x} \cosh {\left (3 x \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*cosh(3*x),x)

[Out]

3*exp(x)*sinh(3*x)/8 - exp(x)*cosh(3*x)/8

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