3.921 \(\int e^{2 (a+b x)} \cosh (a+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=23 \[ \frac {e^{4 a+4 b x}}{16 b}-\frac {x}{4} \]

[Out]

1/16*exp(4*b*x+4*a)/b-1/4*x

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2282, 12, 14} \[ \frac {e^{4 a+4 b x}}{16 b}-\frac {x}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Cosh[a + b*x]*Sinh[a + b*x],x]

[Out]

E^(4*a + 4*b*x)/(16*b) - x/4

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^{2 (a+b x)} \cosh (a+b x) \sinh (a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-1+x^4}{4 x} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {-1+x^4}{x} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{x}+x^3\right ) \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac {e^{4 a+4 b x}}{16 b}-\frac {x}{4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.09 \[ \frac {1}{4} \left (\frac {e^{4 a+4 b x}}{4 b}-x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(4*a + 4*b*x)/(4*b) - x)/4

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fricas [B]  time = 0.47, size = 91, normalized size = 3.96 \[ -\frac {{\left (4 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (4 \, b x + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (4 \, b x - 1\right )} \sinh \left (b x + a\right )^{2}}{16 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((4*b*x - 1)*cosh(b*x + a)^2 - 2*(4*b*x + 1)*cosh(b*x + a)*sinh(b*x + a) + (4*b*x - 1)*sinh(b*x + a)^2)/
(b*cosh(b*x + a)^2 - 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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giac [A]  time = 0.12, size = 18, normalized size = 0.78 \[ -\frac {1}{4} \, x + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x + 1/16*e^(4*b*x + 4*a)/b

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maple [A]  time = 0.10, size = 33, normalized size = 1.43 \[ -\frac {x}{4}+\frac {\sinh \left (4 b x +4 a \right )}{16 b}+\frac {\cosh \left (4 b x +4 a \right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*cosh(b*x+a)*sinh(b*x+a),x)

[Out]

-1/4*x+1/16/b*sinh(4*b*x+4*a)+1/16*cosh(4*b*x+4*a)/b

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maxima [A]  time = 0.32, size = 24, normalized size = 1.04 \[ -\frac {1}{4} \, x - \frac {a}{4 \, b} + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)*sinh(b*x+a),x, algorithm="maxima")

[Out]

-1/4*x - 1/4*a/b + 1/16*e^(4*b*x + 4*a)/b

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mupad [B]  time = 1.81, size = 18, normalized size = 0.78 \[ \frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{16\,b}-\frac {x}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*exp(2*a + 2*b*x)*sinh(a + b*x),x)

[Out]

exp(4*a + 4*b*x)/(16*b) - x/4

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sympy [A]  time = 4.56, size = 117, normalized size = 5.09 \[ \begin {cases} - \frac {x e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )}}{4} + \frac {x e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2} - \frac {x e^{2 a} e^{2 b x} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac {e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh {\relax (a )} \cosh {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**2/4 + x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)/2 - x
*exp(2*a)*exp(2*b*x)*cosh(a + b*x)**2/4 + exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)/(4*b), Ne(b, 0)), (x
*exp(2*a)*sinh(a)*cosh(a), True))

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