### 3.926 $$\int e^{2 (a+b x)} \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=52 $-\frac{e^{-2 a-2 b x}}{32 b}-\frac{e^{2 a+2 b x}}{16 b}+\frac{e^{6 a+6 b x}}{96 b}$

[Out]

-E^(-2*a - 2*b*x)/(32*b) - E^(2*a + 2*b*x)/(16*b) + E^(6*a + 6*b*x)/(96*b)

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Rubi [A]  time = 0.0592756, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {2282, 12, 270} $-\frac{e^{-2 a-2 b x}}{32 b}-\frac{e^{2 a+2 b x}}{16 b}+\frac{e^{6 a+6 b x}}{96 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-E^(-2*a - 2*b*x)/(32*b) - E^(2*a + 2*b*x)/(16*b) + E^(6*a + 6*b*x)/(96*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int e^{2 (a+b x)} \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^4\right )^2}{16 x^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^4\right )^2}{x^3} \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x^3}-2 x+x^5\right ) \, dx,x,e^{a+b x}\right )}{16 b}\\ &=-\frac{e^{-2 a-2 b x}}{32 b}-\frac{e^{2 a+2 b x}}{16 b}+\frac{e^{6 a+6 b x}}{96 b}\\ \end{align*}

Mathematica [A]  time = 0.0378905, size = 38, normalized size = 0.73 $\frac{e^{-2 (a+b x)} \left (-6 e^{4 (a+b x)}+e^{8 (a+b x)}-3\right )}{96 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3 - 6*E^(4*(a + b*x)) + E^(8*(a + b*x)))/(96*b*E^(2*(a + b*x)))

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Maple [A]  time = 0.008, size = 58, normalized size = 1.1 \begin{align*} -{\frac{\sinh \left ( 2\,bx+2\,a \right ) }{32\,b}}+{\frac{\sinh \left ( 6\,bx+6\,a \right ) }{96\,b}}-{\frac{3\,\cosh \left ( 2\,bx+2\,a \right ) }{32\,b}}+{\frac{\cosh \left ( 6\,bx+6\,a \right ) }{96\,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)^2*sinh(b*x+a)^2,x)

[Out]

-1/32*sinh(2*b*x+2*a)/b+1/96/b*sinh(6*b*x+6*a)-3/32*cosh(2*b*x+2*a)/b+1/96*cosh(6*b*x+6*a)/b

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Maxima [A]  time = 1.01396, size = 57, normalized size = 1.1 \begin{align*} -\frac{{\left (6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{96 \, b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, 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)^2*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/96*(6*e^(-4*b*x - 4*a) - 1)*e^(6*b*x + 6*a)/b - 1/32*e^(-2*b*x - 2*a)/b

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Fricas [B]  time = 1.75634, size = 304, normalized size = 5.85 \begin{align*} -\frac{\cosh \left (b x + a\right )^{4} - 8 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 3}{48 \,{\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)^2*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/48*(cosh(b*x + a)^4 - 8*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 - 8*cosh(b*x + a)
*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 3)/(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 = 111.514, size = 70, normalized size = 1.35 \begin{align*} \begin{cases} \frac{e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac{e^{2 a} e^{2 b x} \cosh ^{4}{\left (a + b x \right )}}{6 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (a \right )} \cosh ^{2}{\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)**2*sinh(b*x+a)**2,x)

[Out]

Piecewise((exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**3/(3*b) - exp(2*a)*exp(2*b*x)*cosh(a + b*x)**4/(6*
b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**2*cosh(a)**2, True))

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Giac [A]  time = 1.1451, size = 58, normalized size = 1.12 \begin{align*} \frac{{\left (e^{\left (6 \, b x + 12 \, a\right )} - 6 \, e^{\left (2 \, b x + 8 \, a\right )}\right )} e^{\left (-6 \, a\right )} - 3 \, e^{\left (-2 \, b x - 2 \, a\right )}}{96 \, 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)^2*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

1/96*((e^(6*b*x + 12*a) - 6*e^(2*b*x + 8*a))*e^(-6*a) - 3*e^(-2*b*x - 2*a))/b