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]

1/48*exp(-3*b*x-3*a)/b-1/8*exp(-b*x-a)/b-1/24*exp(3*b*x+3*a)/b+1/80*exp(5*b*x+5*a)/b

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 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]

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 ^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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.48, size = 111, normalized size = 1.61 \[ \frac {\cosh \left (b x + a\right )^{4} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + {\left (6 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )^{2} - 5 \, \cosh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{3} - 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{30 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

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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giac [A]  time = 0.12, size = 52, normalized size = 0.75 \[ -\frac {5 \, {\left (6 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} + 10 \, e^{\left (3 \, b x + 3 \, a\right )}}{240 \, b} \]

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

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maple [A]  time = 0.16, size = 44, normalized size = 0.64 \[ \frac {\frac {\left (\sinh ^{5}\left (b x +a \right )\right )}{5}+\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cosh ^{3}\left (b x +a \right )\right )}{15}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)*sinh(b*x+a)^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 56, normalized size = 0.81 \[ -\frac {{\left (6 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} + \frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 10 \, e^{\left (3 \, b x + 3 \, a\right )}}{240 \, b} \]

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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mupad [B]  time = 0.55, size = 50, normalized size = 0.72 \[ -\frac {30\,{\mathrm {e}}^{-a-b\,x}-5\,{\mathrm {e}}^{-3\,a-3\,b\,x}+10\,{\mathrm {e}}^{3\,a+3\,b\,x}-3\,{\mathrm {e}}^{5\,a+5\,b\,x}}{240\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*exp(a + b*x)*sinh(a + b*x)^3,x)

[Out]

-(30*exp(- a - b*x) - 5*exp(- 3*a - 3*b*x) + 10*exp(3*a + 3*b*x) - 3*exp(5*a + 5*b*x))/(240*b)

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sympy [A]  time = 60.11, size = 139, normalized size = 2.01 \[ \begin {cases} \frac {e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{5 b} - \frac {e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{5 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac {2 e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b} - \frac {2 e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{15 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\relax (a )} \cosh {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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