3.887 \(\int e^{a+b x} \cosh (c+d x) \, dx\)

Optimal. Leaf size=54 \[ \frac {b e^{a+b x} \cosh (c+d x)}{b^2-d^2}-\frac {d e^{a+b x} \sinh (c+d x)}{b^2-d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5475} \[ \frac {b e^{a+b x} \cosh (c+d x)}{b^2-d^2}-\frac {d e^{a+b x} \sinh (c+d x)}{b^2-d^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[c + d*x],x]

[Out]

(b*E^(a + b*x)*Cosh[c + d*x])/(b^2 - d^2) - (d*E^(a + b*x)*Sinh[c + d*x])/(b^2 - d^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^{a+b x} \cosh (c+d x) \, dx &=\frac {b e^{a+b x} \cosh (c+d x)}{b^2-d^2}-\frac {d e^{a+b x} \sinh (c+d x)}{b^2-d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 38, normalized size = 0.70 \[ \frac {e^{a+b x} (b \cosh (c+d x)-d \sinh (c+d x))}{(b-d) (b+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a + b*x)*(b*Cosh[c + d*x] - d*Sinh[c + d*x]))/((b - d)*(b + d))

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 66, normalized size = 1.22 \[ \frac {b \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + b \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) - {\left (d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{b^{2} - d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.11, size = 40, normalized size = 0.74 \[ \frac {e^{\left (b x + d x + a + c\right )}}{2 \, {\left (b + d\right )}} + \frac {e^{\left (b x - d x + a - c\right )}}{2 \, {\left (b - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.19, size = 78, normalized size = 1.44 \[ \frac {\sinh \left (a -c +\left (b -d \right ) x \right )}{2 b -2 d}+\frac {\sinh \left (a +c +\left (b +d \right ) x \right )}{2 b +2 d}+\frac {\cosh \left (a -c +\left (b -d \right ) x \right )}{2 b -2 d}+\frac {\cosh \left (a +c +\left (b +d \right ) x \right )}{2 b +2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(b*x+a)*cosh(d*x+c),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(-d/b>0)', see `assume?` for mo
re details)Is -d/b equal to -1?

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 53, normalized size = 0.98 \[ \frac {{\mathrm {e}}^{a-c+b\,x-d\,x}\,\left (b+d+b\,{\mathrm {e}}^{2\,c+2\,d\,x}-d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{2\,\left (b^2-d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)*exp(a + b*x),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 2.10, size = 167, normalized size = 3.09 \[ \begin {cases} x e^{a} \cosh {\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {x e^{a} e^{- d x} \sinh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{- d x} \cosh {\left (c + d x \right )}}{2} + \frac {e^{a} e^{- d x} \sinh {\left (c + d x \right )}}{2 d} & \text {for}\: b = - d \\- \frac {x e^{a} e^{d x} \sinh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{d x} \cosh {\left (c + d x \right )}}{2} + \frac {e^{a} e^{d x} \sinh {\left (c + d x \right )}}{2 d} & \text {for}\: b = d \\\frac {b e^{a} e^{b x} \cosh {\left (c + d x \right )}}{b^{2} - d^{2}} - \frac {d e^{a} e^{b x} \sinh {\left (c + d x \right )}}{b^{2} - d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(a)*cosh(c), Eq(b, 0) & Eq(d, 0)), (x*exp(a)*exp(-d*x)*sinh(c + d*x)/2 + x*exp(a)*exp(-d*x)*co
sh(c + d*x)/2 + exp(a)*exp(-d*x)*sinh(c + d*x)/(2*d), Eq(b, -d)), (-x*exp(a)*exp(d*x)*sinh(c + d*x)/2 + x*exp(
a)*exp(d*x)*cosh(c + d*x)/2 + exp(a)*exp(d*x)*sinh(c + d*x)/(2*d), Eq(b, d)), (b*exp(a)*exp(b*x)*cosh(c + d*x)
/(b**2 - d**2) - d*exp(a)*exp(b*x)*sinh(c + d*x)/(b**2 - d**2), True))

________________________________________________________________________________________