### 3.931 $$\int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=57 $\frac{e^{-4 a-4 b x}}{256 b}-\frac{3 e^{4 a+4 b x}}{256 b}+\frac{e^{8 a+8 b x}}{512 b}+\frac{3 x}{64}$

[Out]

E^(-4*a - 4*b*x)/(256*b) - (3*E^(4*a + 4*b*x))/(256*b) + E^(8*a + 8*b*x)/(512*b) + (3*x)/64

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Rubi [A]  time = 0.0624658, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {2282, 12, 266, 43} $\frac{e^{-4 a-4 b x}}{256 b}-\frac{3 e^{4 a+4 b x}}{256 b}+\frac{e^{8 a+8 b x}}{512 b}+\frac{3 x}{64}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

E^(-4*a - 4*b*x)/(256*b) - (3*E^(4*a + 4*b*x))/(256*b) + E^(8*a + 8*b*x)/(512*b) + (3*x)/64

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{2 (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^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^4\right )^3}{x^5} \, dx,x,e^{a+b x}\right )}{64 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-1+x)^3}{x^2} \, dx,x,e^{4 a+4 b x}\right )}{256 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-3-\frac{1}{x^2}+\frac{3}{x}+x\right ) \, dx,x,e^{4 a+4 b x}\right )}{256 b}\\ &=\frac{e^{-4 a-4 b x}}{256 b}-\frac{3 e^{4 a+4 b x}}{256 b}+\frac{e^{8 a+8 b x}}{512 b}+\frac{3 x}{64}\\ \end{align*}

Mathematica [A]  time = 0.0435625, size = 45, normalized size = 0.79 $\frac{e^{-4 (a+b x)}-3 e^{4 (a+b x)}+\frac{1}{2} e^{8 (a+b x)}+12 b x}{256 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

(E^(-4*(a + b*x)) - 3*E^(4*(a + b*x)) + E^(8*(a + b*x))/2 + 12*b*x)/(256*b)

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Maple [A]  time = 0.014, size = 61, normalized size = 1.1 \begin{align*}{\frac{3\,x}{64}}-{\frac{\sinh \left ( 4\,bx+4\,a \right ) }{64\,b}}+{\frac{\sinh \left ( 8\,bx+8\,a \right ) }{512\,b}}-{\frac{\cosh \left ( 4\,bx+4\,a \right ) }{128\,b}}+{\frac{\cosh \left ( 8\,bx+8\,a \right ) }{512\,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)^3*sinh(b*x+a)^3,x)

[Out]

3/64*x-1/64/b*sinh(4*b*x+4*a)+1/512/b*sinh(8*b*x+8*a)-1/128*cosh(4*b*x+4*a)/b+1/512*cosh(8*b*x+8*a)/b

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Maxima [A]  time = 1.14565, size = 70, normalized size = 1.23 \begin{align*} -\frac{{\left (6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (8 \, b x + 8 \, a\right )}}{512 \, b} + \frac{3 \,{\left (b x + a\right )}}{64 \, b} + \frac{e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, 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)^3*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/512*(6*e^(-4*b*x - 4*a) - 1)*e^(8*b*x + 8*a)/b + 3/64*(b*x + a)/b + 1/256*e^(-4*b*x - 4*a)/b

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Fricas [B]  time = 1.73983, size = 513, normalized size = 9. \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{6} - 20 \, \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} - 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \, \sinh \left (b x + a\right )^{6} + 6 \,{\left (4 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} + 3 \,{\left (15 \, \cosh \left (b x + a\right )^{4} + 8 \, b x - 2\right )} \sinh \left (b x + a\right )^{2} - 6 \,{\left (\cosh \left (b x + a\right )^{5} + 2 \,{\left (4 \, b x + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{512 \,{\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)^3*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/512*(3*cosh(b*x + a)^6 - 20*cosh(b*x + a)^3*sinh(b*x + a)^3 + 45*cosh(b*x + a)^2*sinh(b*x + a)^4 - 6*cosh(b*
x + a)*sinh(b*x + a)^5 + 3*sinh(b*x + a)^6 + 6*(4*b*x - 1)*cosh(b*x + a)^2 + 3*(15*cosh(b*x + a)^4 + 8*b*x - 2
)*sinh(b*x + a)^2 - 6*(cosh(b*x + a)^5 + 2*(4*b*x + 1)*cosh(b*x + a))*sinh(b*x + a))/(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 [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(2*b*x+2*a)*cosh(b*x+a)**3*sinh(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20434, size = 81, normalized size = 1.42 \begin{align*} \frac{24 \, b x - 2 \,{\left (3 \, e^{\left (4 \, b x + 4 \, a\right )} - 1\right )} e^{\left (-4 \, b x - 4 \, a\right )} +{\left (e^{\left (8 \, b x + 16 \, a\right )} - 6 \, e^{\left (4 \, b x + 12 \, a\right )}\right )} e^{\left (-8 \, a\right )}}{512 \, 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)^3*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

1/512*(24*b*x - 2*(3*e^(4*b*x + 4*a) - 1)*e^(-4*b*x - 4*a) + (e^(8*b*x + 16*a) - 6*e^(4*b*x + 12*a))*e^(-8*a))
/b