### 3.908 $$\int e^{a+b x} \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=49 $-\frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{a+b x}}{8 b}+\frac{e^{5 a+5 b x}}{80 b}$

[Out]

-E^(-3*a - 3*b*x)/(48*b) - E^(a + b*x)/(8*b) + E^(5*a + 5*b*x)/(80*b)

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Rubi [A]  time = 0.0530814, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {2282, 12, 270} $-\frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{a+b x}}{8 b}+\frac{e^{5 a+5 b x}}{80 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-E^(-3*a - 3*b*x)/(48*b) - E^(a + b*x)/(8*b) + E^(5*a + 5*b*x)/(80*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^{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^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^4\right )^2}{x^4} \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2+\frac{1}{x^4}+x^4\right ) \, dx,x,e^{a+b x}\right )}{16 b}\\ &=-\frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{a+b x}}{8 b}+\frac{e^{5 a+5 b x}}{80 b}\\ \end{align*}

Mathematica [A]  time = 0.025766, size = 40, normalized size = 0.82 $\frac{e^{-3 (a+b x)} \left (-30 e^{4 (a+b x)}+3 e^{8 (a+b x)}-5\right )}{240 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-5 - 30*E^(4*(a + b*x)) + 3*E^(8*(a + b*x)))/(240*b*E^(3*(a + b*x)))

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Maple [B]  time = 0.01, size = 84, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{5}}-{\frac{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{2\,\cosh \left ( bx+a \right ) }{15}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{5}}-{\frac{\sinh \left ( bx+a \right ) }{5} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/5*cosh(b*x+a)^3*sinh(b*x+a)^2-2/15*cosh(b*x+a)*sinh(b*x+a)^2-2/15*cosh(b*x+a)+1/5*cosh(b*x+a)^4*sinh(b*
x+a)-1/5*(2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a))

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Maxima [A]  time = 0.992282, size = 51, normalized size = 1.04 \begin{align*} \frac{e^{\left (5 \, b x + 5 \, a\right )} - 10 \, e^{\left (b x + a\right )}}{80 \, b} - \frac{e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(e^(5*b*x + 5*a) - 10*e^(b*x + a))/b - 1/48*e^(-3*b*x - 3*a)/b

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Fricas [B]  time = 1.55048, size = 258, normalized size = 5.27 \begin{align*} -\frac{\cosh \left (b x + a\right )^{4} - 16 \, \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} - 16 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 15}{120 \,{\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)^2*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/120*(cosh(b*x + a)^4 - 16*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 - 16*cosh(b*x +
a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 15)/(b*cosh(b*x + a) - b*sinh(b*x + a))

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Sympy [A]  time = 137.372, size = 144, normalized size = 2.94 \begin{align*} \begin{cases} - \frac{2 e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{15 b} + \frac{2 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b} + \frac{e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac{2 e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b} - \frac{2 e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x e^{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(b*x+a)*cosh(b*x+a)**2*sinh(b*x+a)**2,x)

[Out]

Piecewise((-2*exp(a)*exp(b*x)*sinh(a + b*x)**4/(15*b) + 2*exp(a)*exp(b*x)*sinh(a + b*x)**3*cosh(a + b*x)/(15*b
) + exp(a)*exp(b*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(5*b) + 2*exp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)**3
/(15*b) - 2*exp(a)*exp(b*x)*cosh(a + b*x)**4/(15*b), Ne(b, 0)), (x*exp(a)*sinh(a)**2*cosh(a)**2, True))

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Giac [A]  time = 1.14842, size = 49, normalized size = 1. \begin{align*} \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} - 30 \, e^{\left (b x + a\right )} - 5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{240 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*e^(5*b*x + 5*a) - 30*e^(b*x + a) - 5*e^(-3*b*x - 3*a))/b