3.913 \(\int e^{a+b x} \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=69 \[ \frac{e^{-5 a-5 b x}}{320 b}-\frac{3 e^{-a-b x}}{64 b}-\frac{e^{3 a+3 b x}}{64 b}+\frac{e^{7 a+7 b x}}{448 b} \]

[Out]

E^(-5*a - 5*b*x)/(320*b) - (3*E^(-a - b*x))/(64*b) - E^(3*a + 3*b*x)/(64*b) + E^(7*a + 7*b*x)/(448*b)

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Rubi [A]  time = 0.0589133, antiderivative size = 69, 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} \[ \frac{e^{-5 a-5 b x}}{320 b}-\frac{3 e^{-a-b x}}{64 b}-\frac{e^{3 a+3 b x}}{64 b}+\frac{e^{7 a+7 b x}}{448 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

E^(-5*a - 5*b*x)/(320*b) - (3*E^(-a - b*x))/(64*b) - E^(3*a + 3*b*x)/(64*b) + E^(7*a + 7*b*x)/(448*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 ^3(a+b x) \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^4\right )^3}{64 x^6} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^4\right )^3}{x^6} \, dx,x,e^{a+b x}\right )}{64 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^6}+\frac{3}{x^2}-3 x^2+x^6\right ) \, dx,x,e^{a+b x}\right )}{64 b}\\ &=\frac{e^{-5 a-5 b x}}{320 b}-\frac{3 e^{-a-b x}}{64 b}-\frac{e^{3 a+3 b x}}{64 b}+\frac{e^{7 a+7 b x}}{448 b}\\ \end{align*}

Mathematica [A]  time = 0.0375875, size = 51, normalized size = 0.74 \[ \frac{e^{-5 (a+b x)} \left (-105 e^{4 (a+b x)}-35 e^{8 (a+b x)}+5 e^{12 (a+b x)}+7\right )}{2240 b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

(7 - 105*E^(4*(a + b*x)) - 35*E^(8*(a + b*x)) + 5*E^(12*(a + b*x)))/(2240*b*E^(5*(a + b*x)))

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Maple [B]  time = 0.013, size = 120, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4} \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{7}}-{\frac{3\, \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{35}}+{\frac{3\,\sinh \left ( bx+a \right ) }{35} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) }+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{35}}-{\frac{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{35}}-{\frac{2\,\cosh \left ( bx+a \right ) }{35}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x)

[Out]

1/b*(1/7*cosh(b*x+a)^4*sinh(b*x+a)^3-3/35*cosh(b*x+a)^4*sinh(b*x+a)+3/35*(2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a)+1
/7*sinh(b*x+a)^2*cosh(b*x+a)^5-2/35*cosh(b*x+a)^3*sinh(b*x+a)^2-2/35*cosh(b*x+a)*sinh(b*x+a)^2-2/35*cosh(b*x+a
))

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Maxima [A]  time = 1.04281, size = 73, normalized size = 1.06 \begin{align*} -\frac{{\left (15 \, e^{\left (4 \, b x + 4 \, a\right )} - 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{320 \, b} + \frac{e^{\left (7 \, b x + 7 \, a\right )} - 7 \, e^{\left (3 \, b x + 3 \, a\right )}}{448 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/320*(15*e^(4*b*x + 4*a) - 1)*e^(-5*b*x - 5*a)/b + 1/448*(e^(7*b*x + 7*a) - 7*e^(3*b*x + 3*a))/b

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Fricas [B]  time = 1.53266, size = 420, normalized size = 6.09 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{6} - 10 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 45 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \, \sinh \left (b x + a\right )^{6} + 5 \,{\left (9 \, \cosh \left (b x + a\right )^{4} - 7\right )} \sinh \left (b x + a\right )^{2} - 35 \, \cosh \left (b x + a\right )^{2} -{\left (3 \, \cosh \left (b x + a\right )^{5} - 35 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{560 \,{\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)^3*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/560*(3*cosh(b*x + a)^6 - 10*cosh(b*x + a)^3*sinh(b*x + a)^3 + 45*cosh(b*x + a)^2*sinh(b*x + a)^4 - 3*cosh(b*
x + a)*sinh(b*x + a)^5 + 3*sinh(b*x + a)^6 + 5*(9*cosh(b*x + a)^4 - 7)*sinh(b*x + a)^2 - 35*cosh(b*x + a)^2 -
(3*cosh(b*x + a)^5 - 35*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)**3*sinh(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.18104, size = 70, normalized size = 1.01 \begin{align*} -\frac{7 \,{\left (15 \, e^{\left (4 \, b x + 4 \, a\right )} - 1\right )} e^{\left (-5 \, b x - 5 \, a\right )} - 5 \, e^{\left (7 \, b x + 7 \, a\right )} + 35 \, e^{\left (3 \, b x + 3 \, a\right )}}{2240 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

-1/2240*(7*(15*e^(4*b*x + 4*a) - 1)*e^(-5*b*x - 5*a) - 5*e^(7*b*x + 7*a) + 35*e^(3*b*x + 3*a))/b