3.299 \(\int e^x \cosh (a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac {e^x \cosh (a+b x)}{1-b^2}-\frac {b e^x \sinh (a+b x)}{1-b^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5475} \[ \frac {e^x \cosh (a+b x)}{1-b^2}-\frac {b e^x \sinh (a+b x)}{1-b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Cosh[a + b*x],x]

[Out]

(E^x*Cosh[a + b*x])/(1 - b^2) - (b*E^x*Sinh[a + b*x])/(1 - b^2)

Rule 5475

Int[Cosh[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b*x))
*Cosh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sinh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2)
, x] /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 - b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin {align*} \int e^x \cosh (a+b x) \, dx &=\frac {e^x \cosh (a+b x)}{1-b^2}-\frac {b e^x \sinh (a+b x)}{1-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.68 \[ \frac {e^x (b \sinh (a+b x)-\cosh (a+b x))}{b^2-1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Cosh[a + b*x],x]

[Out]

(E^x*(-Cosh[a + b*x] + b*Sinh[a + b*x]))/(-1 + b^2)

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fricas [A]  time = 0.61, size = 45, normalized size = 1.10 \[ -\frac {\cosh \left (b x + a\right ) \cosh \relax (x) - {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right ) \sinh \relax (x)}{b^{2} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 32, normalized size = 0.78 \[ \frac {e^{\left (b x + a + x\right )}}{2 \, {\left (b + 1\right )}} - \frac {e^{\left (-b x - a + x\right )}}{2 \, {\left (b - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 62, normalized size = 1.51 \[ \frac {\sinh \left (\left (b -1\right ) x +a \right )}{2 b -2}+\frac {\sinh \left (\left (1+b \right ) x +a \right )}{2+2 b}-\frac {\cosh \left (\left (b -1\right ) x +a \right )}{2 \left (b -1\right )}+\frac {\cosh \left (\left (1+b \right ) x +a \right )}{2+2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-b>0)', see `assume?` for more
details)Is -b equal to -1?

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mupad [B]  time = 0.09, size = 45, normalized size = 1.10 \[ -\frac {{\mathrm {e}}^{x-a-b\,x}\,\left (b+{\mathrm {e}}^{2\,a+2\,b\,x}-b\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}{2\,\left (b^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*exp(x),x)

[Out]

-(exp(x - a - b*x)*(b + exp(2*a + 2*b*x) - b*exp(2*a + 2*b*x) + 1))/(2*(b^2 - 1))

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sympy [A]  time = 0.65, size = 99, normalized size = 2.41 \[ \begin {cases} \frac {x e^{x} \sinh {\left (a - x \right )}}{2} + \frac {x e^{x} \cosh {\left (a - x \right )}}{2} - \frac {e^{x} \sinh {\left (a - x \right )}}{2} & \text {for}\: b = -1 \\- \frac {x e^{x} \sinh {\left (a + x \right )}}{2} + \frac {x e^{x} \cosh {\left (a + x \right )}}{2} + \frac {e^{x} \sinh {\left (a + x \right )}}{2} & \text {for}\: b = 1 \\\frac {b e^{x} \sinh {\left (a + b x \right )}}{b^{2} - 1} - \frac {e^{x} \cosh {\left (a + b x \right )}}{b^{2} - 1} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(x)*sinh(a - x)/2 + x*exp(x)*cosh(a - x)/2 - exp(x)*sinh(a - x)/2, Eq(b, -1)), (-x*exp(x)*sinh
(a + x)/2 + x*exp(x)*cosh(a + x)/2 + exp(x)*sinh(a + x)/2, Eq(b, 1)), (b*exp(x)*sinh(a + b*x)/(b**2 - 1) - exp
(x)*cosh(a + b*x)/(b**2 - 1), True))

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