### 3.903 $$\int e^{a+b x} \cosh (a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=35 $\frac{e^{-a-b x}}{4 b}+\frac{e^{3 a+3 b x}}{12 b}$

[Out]

E^(-a - b*x)/(4*b) + E^(3*a + 3*b*x)/(12*b)

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Rubi [A]  time = 0.0267362, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {2282, 12, 14} $\frac{e^{-a-b x}}{4 b}+\frac{e^{3 a+3 b x}}{12 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

E^(-a - b*x)/(4*b) + E^(3*a + 3*b*x)/(12*b)

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^{a+b x} \cosh (a+b x) \sinh (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^4}{4 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^4}{x^2} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^2}+x^2\right ) \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{e^{-a-b x}}{4 b}+\frac{e^{3 a+3 b x}}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.0098237, size = 28, normalized size = 0.8 $\frac{e^{-a-b x} \left (e^{4 (a+b x)}+3\right )}{12 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(E^(-a - b*x)*(3 + E^(4*(a + b*x))))/(12*b)

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Maple [A]  time = 0.005, size = 54, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ({\frac{\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}}-{\frac{\sinh \left ( bx+a \right ) }{3}}+{\frac{\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3}}+{\frac{\cosh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01073, size = 39, normalized size = 1.11 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )}}{12 \, b} + \frac{e^{\left (-b x - a\right )}}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*e^(3*b*x + 3*a)/b + 1/4*e^(-b*x - a)/b

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Fricas [A]  time = 1.45966, size = 144, normalized size = 4.11 \begin{align*} \frac{\cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2}}{3 \,{\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 8.2502, size = 76, normalized size = 2.17 \begin{align*} \begin{cases} \frac{e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )}}{3 b} - \frac{e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{3 b} + \frac{e^{a} e^{b x} \cosh ^{2}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x e^{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(b*x+a)*cosh(b*x+a)*sinh(b*x+a),x)

[Out]

Piecewise((exp(a)*exp(b*x)*sinh(a + b*x)**2/(3*b) - exp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)/(3*b) + exp(a)
*exp(b*x)*cosh(a + b*x)**2/(3*b), Ne(b, 0)), (x*exp(a)*sinh(a)*cosh(a), True))

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Giac [A]  time = 1.12734, size = 35, normalized size = 1. \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )} + 3 \, e^{\left (-b x - a\right )}}{12 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(e^(3*b*x + 3*a) + 3*e^(-b*x - a))/b