### 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]

E^(4*a + 4*b*x)/(16*b) - x/4

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Rubi [A]  time = 0.0229656, antiderivative size = 23, normalized size of antiderivative = 1., 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 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]]]

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]]

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*}

Mathematica [A]  time = 0.0151823, 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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Maple [A]  time = 0.006, size = 33, normalized size = 1.4 \begin{align*} -{\frac{x}{4}}+{\frac{\sinh \left ( 4\,bx+4\,a \right ) }{16\,b}}+{\frac{\cosh \left ( 4\,bx+4\,a \right ) }{16\,b}} \end{align*}

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 = 1.00212, size = 32, normalized size = 1.39 \begin{align*} -\frac{1}{4} \, x - \frac{a}{4 \, b} + \frac{e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} \end{align*}

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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Fricas [B]  time = 1.74641, size = 250, normalized size = 10.87 \begin{align*} -\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 )}} \end{align*}

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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Sympy [A]  time = 9.81299, size = 117, normalized size = 5.09 \begin{align*} \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{\left (a \right )} \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.1319, size = 24, normalized size = 1.04 \begin{align*} -\frac{1}{4} \, x + \frac{e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} \end{align*}

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