3.886 \(\int e^{a+b x} \cosh ^2(c+d x) \, dx\)

Optimal. Leaf size=88 \[ \frac {b e^{a+b x} \cosh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \sinh (c+d x) \cosh (c+d x)}{b^2-4 d^2}-\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )} \]

[Out]

-2*d^2*exp(b*x+a)/b/(b^2-4*d^2)+b*exp(b*x+a)*cosh(d*x+c)^2/(b^2-4*d^2)-2*d*exp(b*x+a)*cosh(d*x+c)*sinh(d*x+c)/
(b^2-4*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5477, 2194} \[ \frac {b e^{a+b x} \cosh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \sinh (c+d x) \cosh (c+d x)}{b^2-4 d^2}-\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[c + d*x]^2,x]

[Out]

(-2*d^2*E^(a + b*x))/(b*(b^2 - 4*d^2)) + (b*E^(a + b*x)*Cosh[c + d*x]^2)/(b^2 - 4*d^2) - (2*d*E^(a + b*x)*Cosh
[c + d*x]*Sinh[c + d*x])/(b^2 - 4*d^2)

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 5477

Int[Cosh[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a +
b*x))*Cosh[d + e*x]^n)/(e^2*n^2 - b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 - b^2*c^2*Log[F]^2),
Int[F^(c*(a + b*x))*Cosh[d + e*x]^(n - 2), x], x] + Simp[(e*n*F^(c*(a + b*x))*Sinh[d + e*x]*Cosh[d + e*x]^(n -
1))/(e^2*n^2 - b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 - b^2*c^2*Log[F]^2, 0] &
& GtQ[n, 1]

Rubi steps

\begin {align*} \int e^{a+b x} \cosh ^2(c+d x) \, dx &=\frac {b e^{a+b x} \cosh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \cosh (c+d x) \sinh (c+d x)}{b^2-4 d^2}-\frac {\left (2 d^2\right ) \int e^{a+b x} \, dx}{b^2-4 d^2}\\ &=-\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )}+\frac {b e^{a+b x} \cosh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \cosh (c+d x) \sinh (c+d x)}{b^2-4 d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 56, normalized size = 0.64 \[ \frac {e^{a+b x} \left (b^2 \cosh (2 (c+d x))+b^2-2 b d \sinh (2 (c+d x))-4 d^2\right )}{2 \left (b^3-4 b d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cosh[c + d*x]^2,x]

[Out]

(E^(a + b*x)*(b^2 - 4*d^2 + b^2*Cosh[2*(c + d*x)] - 2*b*d*Sinh[2*(c + d*x)]))/(2*(b^3 - 4*b*d^2))

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 146, normalized size = 1.66 \[ \frac {b^{2} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} + {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{2} + {\left (b^{2} - 4 \, d^{2}\right )} \cosh \left (b x + a\right ) + {\left (b^{2} \cosh \left (d x + c\right )^{2} + b^{2} - 4 \, d^{2}\right )} \sinh \left (b x + a\right ) - 4 \, {\left (b d \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + b d \cosh \left (d x + c\right ) \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left (b^{3} - 4 \, b d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(d*x+c)^2,x, algorithm="fricas")

[Out]

1/2*(b^2*cosh(b*x + a)*cosh(d*x + c)^2 + (b^2*cosh(b*x + a) + b^2*sinh(b*x + a))*sinh(d*x + c)^2 + (b^2 - 4*d^
2)*cosh(b*x + a) + (b^2*cosh(d*x + c)^2 + b^2 - 4*d^2)*sinh(b*x + a) - 4*(b*d*cosh(b*x + a)*cosh(d*x + c) + b*
d*cosh(d*x + c)*sinh(b*x + a))*sinh(d*x + c))/(b^3 - 4*b*d^2)

________________________________________________________________________________________

giac [A]  time = 0.12, size = 56, normalized size = 0.64 \[ \frac {e^{\left (b x + 2 \, d x + a + 2 \, c\right )}}{4 \, {\left (b + 2 \, d\right )}} + \frac {e^{\left (b x - 2 \, d x + a - 2 \, c\right )}}{4 \, {\left (b - 2 \, d\right )}} + \frac {e^{\left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(d*x+c)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.29, size = 112, normalized size = 1.27 \[ \frac {\sinh \left (b x +a \right )}{2 b}+\frac {\sinh \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\sinh \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d}+\frac {\cosh \left (b x +a \right )}{2 b}+\frac {\cosh \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\cosh \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(d*x+c)^2,x)

[Out]

1/2*sinh(b*x+a)/b+1/4*sinh(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*sinh(a+2*c+(b+2*d)*x)/(b+2*d)+1/2*cosh(b*x+a)/b+1/4*co
sh(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*cosh(a+2*c+(b+2*d)*x)/(b+2*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)^2,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(-(2*d)/b>0)', see `assume?` fo
r more details)Is -(2*d)/b equal to -1?

________________________________________________________________________________________

mupad [B]  time = 0.23, size = 68, normalized size = 0.77 \[ \frac {2\,d^2\,{\mathrm {e}}^{a+b\,x}-b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^2\,{\mathrm {e}}^{a+b\,x}+2\,b\,d\,\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {e}}^{a+b\,x}\,\mathrm {sinh}\left (c+d\,x\right )}{4\,b\,d^2-b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^2*exp(a + b*x),x)

[Out]

(2*d^2*exp(a + b*x) - b^2*cosh(c + d*x)^2*exp(a + b*x) + 2*b*d*cosh(c + d*x)*exp(a + b*x)*sinh(c + d*x))/(4*b*
d^2 - b^3)

________________________________________________________________________________________

sympy [A]  time = 9.02, size = 432, normalized size = 4.91 \[ \begin {cases} x e^{a} \cosh ^{2}{\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\\left (- \frac {x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {\sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d}\right ) e^{a} & \text {for}\: b = 0 \\\frac {x e^{a} e^{- 2 d x} \sinh ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{- 2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{- 2 d x} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {e^{a} e^{- 2 d x} \sinh ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 e^{a} e^{- 2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} & \text {for}\: b = - 2 d \\\frac {x e^{a} e^{2 d x} \sinh ^{2}{\left (c + d x \right )}}{4} - \frac {x e^{a} e^{2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{2 d x} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {e^{a} e^{2 d x} \sinh ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 e^{a} e^{2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} & \text {for}\: b = 2 d \\\frac {b^{2} e^{a} e^{b x} \cosh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 b d e^{a} e^{b x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} e^{a} e^{b x} \sinh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 d^{2} e^{a} e^{b x} \cosh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(d*x+c)**2,x)

[Out]

Piecewise((x*exp(a)*cosh(c)**2, Eq(b, 0) & Eq(d, 0)), ((-x*sinh(c + d*x)**2/2 + x*cosh(c + d*x)**2/2 + sinh(c
+ d*x)*cosh(c + d*x)/(2*d))*exp(a), Eq(b, 0)), (x*exp(a)*exp(-2*d*x)*sinh(c + d*x)**2/4 + x*exp(a)*exp(-2*d*x)
*sinh(c + d*x)*cosh(c + d*x)/2 + x*exp(a)*exp(-2*d*x)*cosh(c + d*x)**2/4 + exp(a)*exp(-2*d*x)*sinh(c + d*x)**2
/(2*d) + 3*exp(a)*exp(-2*d*x)*sinh(c + d*x)*cosh(c + d*x)/(4*d), Eq(b, -2*d)), (x*exp(a)*exp(2*d*x)*sinh(c + d
*x)**2/4 - x*exp(a)*exp(2*d*x)*sinh(c + d*x)*cosh(c + d*x)/2 + x*exp(a)*exp(2*d*x)*cosh(c + d*x)**2/4 - exp(a)
*exp(2*d*x)*sinh(c + d*x)**2/(2*d) + 3*exp(a)*exp(2*d*x)*sinh(c + d*x)*cosh(c + d*x)/(4*d), Eq(b, 2*d)), (b**2
*exp(a)*exp(b*x)*cosh(c + d*x)**2/(b**3 - 4*b*d**2) - 2*b*d*exp(a)*exp(b*x)*sinh(c + d*x)*cosh(c + d*x)/(b**3
- 4*b*d**2) + 2*d**2*exp(a)*exp(b*x)*sinh(c + d*x)**2/(b**3 - 4*b*d**2) - 2*d**2*exp(a)*exp(b*x)*cosh(c + d*x)
**2/(b**3 - 4*b*d**2), True))

________________________________________________________________________________________