### 3.933 $$\int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=57 $\frac{e^{-2 a-2 b x}}{32 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{e^{6 a+6 b x}}{96 b}-\frac{x}{8}$

[Out]

E^(-2*a - 2*b*x)/(32*b) + E^(4*a + 4*b*x)/(32*b) + E^(6*a + 6*b*x)/(96*b) - x/8

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Rubi [A]  time = 0.0522093, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {2282, 12, 446, 75} $\frac{e^{-2 a-2 b x}}{32 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{e^{6 a+6 b x}}{96 b}-\frac{x}{8}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

E^(-2*a - 2*b*x)/(32*b) + E^(4*a + 4*b*x)/(32*b) + E^(6*a + 6*b*x)/(96*b) - x/8

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (1+x^2\right )^3}{16 x^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (1+x^2\right )^3}{x^3} \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-1+x) (1+x)^3}{x^2} \, dx,x,e^{2 a+2 b x}\right )}{32 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^2}-\frac{2}{x}+2 x+x^2\right ) \, dx,x,e^{2 a+2 b x}\right )}{32 b}\\ &=\frac{e^{-2 a-2 b x}}{32 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{e^{6 a+6 b x}}{96 b}-\frac{x}{8}\\ \end{align*}

Mathematica [A]  time = 0.0579636, size = 43, normalized size = 0.75 $\frac{3 e^{-2 (a+b x)}+3 e^{4 (a+b x)}+e^{6 (a+b x)}-12 b x}{96 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3/E^(2*(a + b*x)) + 3*E^(4*(a + b*x)) + E^(6*(a + b*x)) - 12*b*x)/(96*b)

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Maple [A]  time = 0.007, size = 89, normalized size = 1.6 \begin{align*} -{\frac{x}{8}}-{\frac{\sinh \left ( 2\,bx+2\,a \right ) }{32\,b}}+{\frac{\sinh \left ( 4\,bx+4\,a \right ) }{32\,b}}+{\frac{\sinh \left ( 6\,bx+6\,a \right ) }{96\,b}}+{\frac{\cosh \left ( 2\,bx+2\,a \right ) }{32\,b}}+{\frac{\cosh \left ( 4\,bx+4\,a \right ) }{32\,b}}+{\frac{\cosh \left ( 6\,bx+6\,a \right ) }{96\,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),x)

[Out]

-1/8*x-1/32*sinh(2*b*x+2*a)/b+1/32/b*sinh(4*b*x+4*a)+1/96/b*sinh(6*b*x+6*a)+1/32*cosh(2*b*x+2*a)/b+1/32*cosh(4
*b*x+4*a)/b+1/96*cosh(6*b*x+6*a)/b

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Maxima [A]  time = 1.06912, size = 70, normalized size = 1.23 \begin{align*} \frac{{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{96 \, b} - \frac{b x + a}{8 \, b} + \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, 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),x, algorithm="maxima")

[Out]

1/96*(3*e^(-2*b*x - 2*a) + 1)*e^(6*b*x + 6*a)/b - 1/8*(b*x + a)/b + 1/32*e^(-2*b*x - 2*a)/b

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Fricas [B]  time = 1.88129, size = 413, normalized size = 7.25 \begin{align*} \frac{4 \, \cosh \left (b x + a\right )^{4} - 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 4 \, \sinh \left (b x + a\right )^{4} - 3 \,{\left (4 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} - 3 \,{\left (4 \, b x - 8 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \,{\left (4 \, \cosh \left (b x + a\right )^{3} - 3 \,{\left (4 \, b x + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{96 \,{\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),x, algorithm="fricas")

[Out]

1/96*(4*cosh(b*x + a)^4 - 8*cosh(b*x + a)*sinh(b*x + a)^3 + 4*sinh(b*x + a)^4 - 3*(4*b*x - 1)*cosh(b*x + a)^2
- 3*(4*b*x - 8*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*(4*cosh(b*x + a)^3 - 3*(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 [A]  time = 103.65, size = 240, normalized size = 4.21 \begin{align*} \begin{cases} \frac{x e^{2 a} e^{2 b x} \sinh ^{4}{\left (a + b x \right )}}{8} - \frac{x e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4} + \frac{x e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{4} - \frac{x e^{2 a} e^{2 b x} \cosh ^{4}{\left (a + b x \right )}}{8} - \frac{e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} + \frac{e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac{e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{24 b} + \frac{e^{2 a} e^{2 b x} \cosh ^{4}{\left (a + b x \right )}}{12 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh{\left (a \right )} \cosh ^{3}{\left (a \right )} & \text{otherwise} \end{cases} \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),x)

[Out]

Piecewise((x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**4/8 - x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3*cosh(a + b*x)/4 +
x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**3/4 - x*exp(2*a)*exp(2*b*x)*cosh(a + b*x)**4/8 - exp(2*a)*
exp(2*b*x)*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + exp(2*a)*exp(2*b*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(4*b)
- exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**3/(24*b) + exp(2*a)*exp(2*b*x)*cosh(a + b*x)**4/(12*b), Ne(
b, 0)), (x*exp(2*a)*sinh(a)*cosh(a)**3, True))

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Giac [A]  time = 1.17915, size = 82, normalized size = 1.44 \begin{align*} -\frac{12 \, b x - 3 \,{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} -{\left (e^{\left (6 \, b x + 12 \, a\right )} + 3 \, e^{\left (4 \, b x + 10 \, a\right )}\right )} e^{\left (-6 \, a\right )}}{96 \, 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),x, algorithm="giac")

[Out]

-1/96*(12*b*x - 3*(2*e^(2*b*x + 2*a) + 1)*e^(-2*b*x - 2*a) - (e^(6*b*x + 12*a) + 3*e^(4*b*x + 10*a))*e^(-6*a))
/b