3.264 \(\int e^{a+b x} \cosh ^4(a+b x) \, dx\)

Optimal. Leaf size=83 \[ -\frac {e^{-3 a-3 b x}}{48 b}-\frac {e^{-a-b x}}{4 b}+\frac {3 e^{a+b x}}{8 b}+\frac {e^{3 a+3 b x}}{12 b}+\frac {e^{5 a+5 b x}}{80 b} \]

[Out]

-1/48*exp(-3*b*x-3*a)/b-1/4*exp(-b*x-a)/b+3/8*exp(b*x+a)/b+1/12*exp(3*b*x+3*a)/b+1/80*exp(5*b*x+5*a)/b

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Rubi [A]  time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2282, 12, 270} \[ -\frac {e^{-3 a-3 b x}}{48 b}-\frac {e^{-a-b x}}{4 b}+\frac {3 e^{a+b x}}{8 b}+\frac {e^{3 a+3 b x}}{12 b}+\frac {e^{5 a+5 b x}}{80 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[a + b*x]^4,x]

[Out]

-E^(-3*a - 3*b*x)/(48*b) - E^(-a - b*x)/(4*b) + (3*E^(a + b*x))/(8*b) + E^(3*a + 3*b*x)/(12*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 ^4(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^4}{16 x^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^4}{x^4} \, dx,x,e^{a+b x}\right )}{16 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (6+\frac {1}{x^4}+\frac {4}{x^2}+4 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}}{4 b}+\frac {3 e^{a+b x}}{8 b}+\frac {e^{3 a+3 b x}}{12 b}+\frac {e^{5 a+5 b x}}{80 b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 62, normalized size = 0.75 \[ \frac {e^{-3 (a+b x)} \left (-60 e^{2 (a+b x)}+90 e^{4 (a+b x)}+20 e^{6 (a+b x)}+3 e^{8 (a+b x)}-5\right )}{240 b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cosh[a + b*x]^4,x]

[Out]

(-5 - 60*E^(2*(a + b*x)) + 90*E^(4*(a + b*x)) + 20*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 = 0.52, size = 113, normalized size = 1.36 \[ -\frac {\cosh \left (b x + a\right )^{4} - 16 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )^{2} + 20 \, \cosh \left (b x + a\right )^{2} - 16 \, {\left (\cosh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 45}{120 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/120*(cosh(b*x + a)^4 - 16*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 10)*sinh
(b*x + a)^2 + 20*cosh(b*x + a)^2 - 16*(cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a) - 45)/(b*cosh(b*x + a)
- b*sinh(b*x + a))

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giac [A]  time = 0.12, size = 60, normalized size = 0.72 \[ -\frac {5 \, {\left (12 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 20 \, e^{\left (3 \, b x + 3 \, a\right )} - 90 \, e^{\left (b x + a\right )}}{240 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^4,x, algorithm="giac")

[Out]

-1/240*(5*(12*e^(2*b*x + 2*a) + 1)*e^(-3*b*x - 3*a) - 3*e^(5*b*x + 5*a) - 20*e^(3*b*x + 3*a) - 90*e^(b*x + a))
/b

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maple [A]  time = 0.22, size = 45, normalized size = 0.54 \[ \frac {\frac {\left (\cosh ^{5}\left (b x +a \right )\right )}{5}+\left (\frac {8}{15}+\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{5}+\frac {4 \left (\cosh ^{2}\left (b x +a \right )\right )}{15}\right ) \sinh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)^4,x)

[Out]

1/b*(1/5*cosh(b*x+a)^5+(8/15+1/5*cosh(b*x+a)^4+4/15*cosh(b*x+a)^2)*sinh(b*x+a))

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maxima [A]  time = 0.32, size = 68, normalized size = 0.82 \[ \frac {e^{\left (5 \, b x + 5 \, a\right )}}{80 \, b} + \frac {e^{\left (3 \, b x + 3 \, a\right )}}{12 \, b} + \frac {3 \, e^{\left (b x + a\right )}}{8 \, b} - \frac {e^{\left (-b x - a\right )}}{4 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^4,x, algorithm="maxima")

[Out]

1/80*e^(5*b*x + 5*a)/b + 1/12*e^(3*b*x + 3*a)/b + 3/8*e^(b*x + a)/b - 1/4*e^(-b*x - a)/b - 1/48*e^(-3*b*x - 3*
a)/b

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mupad [B]  time = 0.50, size = 58, normalized size = 0.70 \[ \frac {90\,{\mathrm {e}}^{a+b\,x}-60\,{\mathrm {e}}^{-a-b\,x}-5\,{\mathrm {e}}^{-3\,a-3\,b\,x}+20\,{\mathrm {e}}^{3\,a+3\,b\,x}+3\,{\mathrm {e}}^{5\,a+5\,b\,x}}{240\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^4*exp(a + b*x),x)

[Out]

(90*exp(a + b*x) - 60*exp(- a - b*x) - 5*exp(- 3*a - 3*b*x) + 20*exp(3*a + 3*b*x) + 3*exp(5*a + 5*b*x))/(240*b
)

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sympy [A]  time = 54.78, size = 139, normalized size = 1.67 \[ \begin {cases} \frac {8 e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{15 b} - \frac {8 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{15 b} - \frac {4 e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac {4 e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{5 b} + \frac {e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x e^{a} \cosh ^{4}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)**4,x)

[Out]

Piecewise((8*exp(a)*exp(b*x)*sinh(a + b*x)**4/(15*b) - 8*exp(a)*exp(b*x)*sinh(a + b*x)**3*cosh(a + b*x)/(15*b)
- 4*exp(a)*exp(b*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(5*b) + 4*exp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)**
3/(5*b) + exp(a)*exp(b*x)*cosh(a + b*x)**4/(5*b), Ne(b, 0)), (x*exp(a)*cosh(a)**4, True))

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