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

Optimal. Leaf size=66 \[ \frac {b e^{c+d x} \cosh (2 a+2 b x)}{4 b^2-d^2}-\frac {d e^{c+d x} \sinh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5509, 12, 5474} \[ \frac {b e^{c+d x} \cosh (2 a+2 b x)}{4 b^2-d^2}-\frac {d e^{c+d x} \sinh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*E^(c + d*x)*Cosh[2*a + 2*b*x])/(4*b^2 - d^2) - (d*E^(c + d*x)*Sinh[2*a + 2*b*x])/(2*(4*b^2 - d^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 5474

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

Rule 5509

Int[Cosh[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol]
:> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sinh[d + e*x]^m*Cosh[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e,
f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int e^{c+d x} \cosh (a+b x) \sinh (a+b x) \, dx &=\int \frac {1}{2} e^{c+d x} \sinh (2 a+2 b x) \, dx\\ &=\frac {1}{2} \int e^{c+d x} \sinh (2 a+2 b x) \, dx\\ &=\frac {b e^{c+d x} \cosh (2 a+2 b x)}{4 b^2-d^2}-\frac {d e^{c+d x} \sinh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 47, normalized size = 0.71 \[ \frac {e^{c+d x} (2 b \cosh (2 (a+b x))-d \sinh (2 (a+b x)))}{2 \left (4 b^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 142, normalized size = 2.15 \[ \frac {b \cosh \left (b x + a\right )^{2} \cosh \left (d x + c\right ) - d \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) + b \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{2} + {\left (b \cosh \left (b x + a\right )^{2} - d \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )} \sinh \left (d x + c\right )}{{\left (4 \, b^{2} - d^{2}\right )} \cosh \left (b x + a\right )^{2} - {\left (4 \, b^{2} - d^{2}\right )} \sinh \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 47, normalized size = 0.71 \[ \frac {e^{\left (2 \, b x + d x + 2 \, a + c\right )}}{4 \, {\left (2 \, b + d\right )}} + \frac {e^{\left (-2 \, b x + d x - 2 \, a + c\right )}}{4 \, {\left (2 \, b - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 102, normalized size = 1.55 \[ -\frac {\sinh \left (2 a -c +\left (2 b -d \right ) x \right )}{4 \left (2 b -d \right )}+\frac {\sinh \left (2 a +c +\left (2 b +d \right ) x \right )}{8 b +4 d}+\frac {\cosh \left (2 a -c +\left (2 b -d \right ) x \right )}{8 b -4 d}+\frac {\cosh \left (2 a +c +\left (2 b +d \right ) x \right )}{8 b +4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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mupad [B]  time = 1.91, size = 58, normalized size = 0.88 \[ \frac {{\mathrm {e}}^{c+d\,x}\,{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left (2\,b+d+2\,b\,{\mathrm {e}}^{4\,a+4\,b\,x}-d\,{\mathrm {e}}^{4\,a+4\,b\,x}\right )}{4\,\left (4\,b^2-d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.45, size = 304, normalized size = 4.61 \[ \begin {cases} x e^{c} \sinh {\relax (a )} \cosh {\relax (a )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {x e^{c} e^{d x} \sinh ^{2}{\left (a - \frac {d x}{2} \right )}}{4} + \frac {x e^{c} e^{d x} \sinh {\left (a - \frac {d x}{2} \right )} \cosh {\left (a - \frac {d x}{2} \right )}}{2} + \frac {x e^{c} e^{d x} \cosh ^{2}{\left (a - \frac {d x}{2} \right )}}{4} + \frac {e^{c} e^{d x} \sinh {\left (a - \frac {d x}{2} \right )} \cosh {\left (a - \frac {d x}{2} \right )}}{2 d} & \text {for}\: b = - \frac {d}{2} \\- \frac {x e^{c} e^{d x} \sinh ^{2}{\left (a + \frac {d x}{2} \right )}}{4} + \frac {x e^{c} e^{d x} \sinh {\left (a + \frac {d x}{2} \right )} \cosh {\left (a + \frac {d x}{2} \right )}}{2} - \frac {x e^{c} e^{d x} \cosh ^{2}{\left (a + \frac {d x}{2} \right )}}{4} + \frac {e^{c} e^{d x} \sinh {\left (a + \frac {d x}{2} \right )} \cosh {\left (a + \frac {d x}{2} \right )}}{2 d} & \text {for}\: b = \frac {d}{2} \\\frac {b e^{c} e^{d x} \sinh ^{2}{\left (a + b x \right )}}{4 b^{2} - d^{2}} + \frac {b e^{c} e^{d x} \cosh ^{2}{\left (a + b x \right )}}{4 b^{2} - d^{2}} - \frac {d e^{c} e^{d x} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b^{2} - d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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