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

Optimal. Leaf size=69 \[ \frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{-a-b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{80 b} \]

[Out]

E^(-3*a - 3*b*x)/(48*b) - E^(-a - b*x)/(8*b) - E^(3*a + 3*b*x)/(24*b) + E^(5*a + 5*b*x)/(80*b)

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Rubi [A]  time = 0.0510697, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2282, 12, 448} \[ \frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{-a-b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3*a - 3*b*x)/(48*b) - E^(-a - b*x)/(8*b) - E^(3*a + 3*b*x)/(24*b) + E^(5*a + 5*b*x)/(80*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 448

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

Rubi steps

\begin{align*} \int e^{a+b x} \cosh (a+b x) \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1-x^2\right ) \left (1-x^2\right )^3}{16 x^4} \, 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^4} \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^4}+\frac{2}{x^2}-2 x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac{e^{-3 a-3 b x}}{48 b}-\frac{e^{-a-b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{80 b}\\ \end{align*}

Mathematica [A]  time = 0.0618103, size = 51, normalized size = 0.74 \[ \frac{e^{-3 (a+b x)} \left (-30 e^{2 (a+b x)}-10 e^{6 (a+b x)}+3 e^{8 (a+b x)}+5\right )}{240 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5 - 30*E^(2*(a + b*x)) - 10*E^(6*(a + b*x)) + 3*E^(8*(a + b*x)))/(240*b*E^(3*(a + b*x)))

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Maple [A]  time = 0.01, size = 90, 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 ) ^{3}}{5}}-{\frac{\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{5}}+{\frac{\sinh \left ( bx+a \right ) }{5}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{5}}-{\frac{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{2\,\cosh \left ( bx+a \right ) }{15}} \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)^3,x)

[Out]

1/b*(1/5*cosh(b*x+a)^2*sinh(b*x+a)^3-1/5*sinh(b*x+a)*cosh(b*x+a)^2+1/5*sinh(b*x+a)+1/5*cosh(b*x+a)^3*sinh(b*x+
a)^2-2/15*cosh(b*x+a)*sinh(b*x+a)^2-2/15*cosh(b*x+a))

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Maxima [A]  time = 0.994325, size = 76, normalized size = 1.1 \begin{align*} -\frac{{\left (6 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} + \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} - 10 \, e^{\left (3 \, b x + 3 \, a\right )}}{240 \, 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)^3,x, algorithm="maxima")

[Out]

-1/48*(6*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a)/b + 1/240*(3*e^(5*b*x + 5*a) - 10*e^(3*b*x + 3*a))/b

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Fricas [A]  time = 1.56103, size = 302, normalized size = 4.38 \begin{align*} \frac{\cosh \left (b x + a\right )^{4} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} +{\left (6 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )^{2} - 5 \, \cosh \left (b x + a\right )^{2} -{\left (\cosh \left (b x + a\right )^{3} - 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{30 \,{\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)^3,x, algorithm="fricas")

[Out]

1/30*(cosh(b*x + a)^4 - cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + (6*cosh(b*x + a)^2 - 5)*sinh(b*x + a
)^2 - 5*cosh(b*x + a)^2 - (cosh(b*x + a)^3 - 5*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a
))

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Sympy [A]  time = 92.6981, size = 139, normalized size = 2.01 \begin{align*} \begin{cases} \frac{e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{5 b} - \frac{e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{5 b} + \frac{e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac{2 e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b} - \frac{2 e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\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)**3,x)

[Out]

Piecewise((exp(a)*exp(b*x)*sinh(a + b*x)**4/(5*b) - exp(a)*exp(b*x)*sinh(a + b*x)**3*cosh(a + b*x)/(5*b) + exp
(a)*exp(b*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(5*b) + 2*exp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)**3/(15*b)
- 2*exp(a)*exp(b*x)*cosh(a + b*x)**4/(15*b), Ne(b, 0)), (x*exp(a)*sinh(a)**3*cosh(a), True))

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Giac [A]  time = 1.14918, size = 70, normalized size = 1.01 \begin{align*} -\frac{5 \,{\left (6 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} + 10 \, e^{\left (3 \, b x + 3 \, a\right )}}{240 \, 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)^3,x, algorithm="giac")

[Out]

-1/240*(5*(6*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) - 3*e^(5*b*x + 5*a) + 10*e^(3*b*x + 3*a))/b