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]

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

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Rubi [A]  time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

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 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^{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*}

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Mathematica [A]  time = 0.06, 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]

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

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fricas [B]  time = 0.43, size = 95, normalized size = 1.67 \[ -\frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 3 \, \sinh \left (b x + a\right )^{3} + 2 \, {\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) - {\left (4 \, b x + 9 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )}{32 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

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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giac [A]  time = 0.12, size = 57, normalized size = 1.00 \[ -\frac {4 \, b x - 2 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 4 \, a - e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} \]

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

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maple [A]  time = 0.13, size = 53, normalized size = 0.93 \[ \frac {\frac {\left (\sinh ^{4}\left (b x +a \right )\right )}{4}+\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{4}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}-\frac {b x}{8}-\frac {a}{8}}{b} \]

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*sinh(b*x+a)^4+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 = 0.34, size = 50, normalized size = 0.88 \[ -\frac {1}{8} \, x - \frac {a}{8 \, b} + \frac {e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} \]

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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mupad [B]  time = 0.28, size = 43, normalized size = 0.75 \[ -\frac {x}{8}-\frac {\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}}{16}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{16}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{32}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 18.39, size = 177, normalized size = 3.11 \[ \begin {cases} - \frac {x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8} + \frac {x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8} + \frac {x e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} - \frac {x e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8} + \frac {3 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8 b} - \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b} + \frac {e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{2}{\relax (a )} \cosh {\relax (a )} & \text {otherwise} \end {cases} \]

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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