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]

-1/48*exp(-3*b*x-3*a)/b-1/8*exp(b*x+a)/b+1/80*exp(5*b*x+5*a)/b

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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 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]

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^{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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 0.44, size = 90, normalized size = 1.84 \[ -\frac {\cosh \left (b x + a\right )^{4} - 16 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - 16 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 15}{120 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

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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giac [A]  time = 0.13, size = 36, normalized size = 0.73 \[ \frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 30 \, e^{\left (b x + a\right )} - 5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{240 \, b} \]

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

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maple [A]  time = 0.36, size = 70, normalized size = 1.43 \[ \frac {\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cosh ^{3}\left (b x +a \right )\right )}{15}+\frac {\sinh \left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{5}-\frac {\left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{3}\right ) \sinh \left (b x +a \right )}{5}}{b} \]

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

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maxima [A]  time = 0.34, size = 38, normalized size = 0.78 \[ \frac {e^{\left (5 \, b x + 5 \, a\right )} - 10 \, e^{\left (b x + a\right )}}{80 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} \]

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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mupad [B]  time = 0.52, size = 36, normalized size = 0.73 \[ -\frac {30\,{\mathrm {e}}^{a+b\,x}+5\,{\mathrm {e}}^{-3\,a-3\,b\,x}-3\,{\mathrm {e}}^{5\,a+5\,b\,x}}{240\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(30*exp(a + b*x) + 5*exp(- 3*a - 3*b*x) - 3*exp(5*a + 5*b*x))/(240*b)

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sympy [A]  time = 62.53, size = 144, normalized size = 2.94 \[ \begin {cases} - \frac {2 e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{15 b} + \frac {2 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{15 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac {2 e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b} - \frac {2 e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{15 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{2}{\relax (a )} \cosh ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

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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