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]

-1/32*exp(-2*b*x-2*a)/b-1/16*exp(2*b*x+2*a)/b+1/96*exp(6*b*x+6*a)/b

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Rubi [A]  time = 0.06, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 0.47, size = 108, normalized size = 2.08 \[ -\frac {\cosh \left (b x + a\right )^{4} - 8 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 3}{48 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]

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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giac [A]  time = 0.14, size = 43, normalized size = 0.83 \[ \frac {{\left (e^{\left (6 \, b x + 12 \, a\right )} - 6 \, e^{\left (2 \, b x + 8 \, a\right )}\right )} e^{\left (-6 \, a\right )} - 3 \, e^{\left (-2 \, b x - 2 \, a\right )}}{96 \, b} \]

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

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maple [A]  time = 0.26, size = 58, normalized size = 1.12 \[ -\frac {\sinh \left (2 b x +2 a \right )}{32 b}+\frac {\sinh \left (6 b x +6 a \right )}{96 b}-\frac {3 \cosh \left (2 b x +2 a \right )}{32 b}+\frac {\cosh \left (6 b x +6 a \right )}{96 b} \]

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 = 0.32, size = 42, normalized size = 0.81 \[ -\frac {{\left (6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{96 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b} \]

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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mupad [B]  time = 0.53, size = 39, normalized size = 0.75 \[ -\frac {3\,{\mathrm {e}}^{-2\,a-2\,b\,x}+6\,{\mathrm {e}}^{2\,a+2\,b\,x}-{\mathrm {e}}^{6\,a+6\,b\,x}}{96\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*exp(- 2*a - 2*b*x) + 6*exp(2*a + 2*b*x) - exp(6*a + 6*b*x))/(96*b)

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sympy [A]  time = 57.10, size = 128, normalized size = 2.46 \[ \begin {cases} - \frac {5 e^{2 a} e^{2 b x} \sinh ^{4}{\left (a + b x \right )}}{48 b} + \frac {5 e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{24 b} + \frac {e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} - \frac {e^{2 a} e^{2 b x} \cosh ^{4}{\left (a + b x \right )}}{16 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\relax (a )} \cosh ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**4/(48*b) + 5*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3*cosh(a + b*
x)/(24*b) + exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) - exp(2*a)*exp(2*b*x)*cosh(a + b*x)**4/(1
6*b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**2*cosh(a)**2, True))

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