3.902 $$\int e^{a+b x} \cosh (a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=57 $-\frac{e^{-2 a-2 b x}}{16 b}-\frac{e^{2 a+2 b x}}{16 b}+\frac{e^{4 a+4 b x}}{32 b}-\frac{x}{8}$

[Out]

-E^(-2*a - 2*b*x)/(16*b) - E^(2*a + 2*b*x)/(16*b) + E^(4*a + 4*b*x)/(32*b) - x/8

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Rubi [A]  time = 0.052599, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {2282, 12, 446, 75} $-\frac{e^{-2 a-2 b x}}{16 b}-\frac{e^{2 a+2 b x}}{16 b}+\frac{e^{4 a+4 b x}}{32 b}-\frac{x}{8}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0661357, size = 45, normalized size = 0.79 $-\frac{2 e^{-2 (a+b x)}+2 e^{2 (a+b x)}-e^{4 (a+b x)}+4 b x}{32 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 71, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) }{4}}-{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{bx}{8}}-{\frac{a}{8}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/4*cosh(b*x+a)^2*sinh(b*x+a)^2-1/4*cosh(b*x+a)^2+1/4*cosh(b*x+a)^3*sinh(b*x+a)-1/8*cosh(b*x+a)*sinh(b*x+
a)-1/8*b*x-1/8*a)

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Maxima [A]  time = 1.03779, size = 68, normalized size = 1.19 \begin{align*} -\frac{1}{8} \, x - \frac{a}{8 \, b} + \frac{e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/8*x - 1/8*a/b + 1/32*(e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a))/b - 1/16*e^(-2*b*x - 2*a)/b

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Fricas [B]  time = 1.53401, size = 261, normalized size = 4.58 \begin{align*} -\frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 3 \, \sinh \left (b x + a\right )^{3} + 2 \,{\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) -{\left (4 \, b x + 9 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )}{32 \,{\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)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/32*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 - 3*sinh(b*x + a)^3 + 2*(2*b*x + 1)*cosh(b*x + a) - (
4*b*x + 9*cosh(b*x + a)^2 - 2)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

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Sympy [A]  time = 26.8278, size = 177, normalized size = 3.11 \begin{align*} \begin{cases} - \frac{x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8} + \frac{x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8} + \frac{x e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} - \frac{x e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8} + \frac{3 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8 b} - \frac{e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4 b} + \frac{e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8 b} & \text{for}\: b \neq 0 \\x e^{a} \sinh ^{2}{\left (a \right )} \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13002, size = 77, normalized size = 1.35 \begin{align*} -\frac{4 \, b x - 2 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 4 \, a - e^{\left (4 \, b x + 4 \, a\right )} + 2 \, 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(b*x+a)*cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

-1/32*(4*b*x - 2*(e^(2*b*x + 2*a) - 1)*e^(-2*b*x - 2*a) + 4*a - e^(4*b*x + 4*a) + 2*e^(2*b*x + 2*a))/b