3.9.72 \(\int e^{2 (a+b x)} \sinh ^3(a+b x) \, dx\) [872]

Optimal. Leaf size=66 \[ \frac {e^{-a-b x}}{8 b}+\frac {3 e^{a+b x}}{8 b}-\frac {e^{3 a+3 b x}}{8 b}+\frac {e^{5 a+5 b x}}{40 b} \]

[Out]

1/8*exp(-b*x-a)/b+3/8*exp(b*x+a)/b-1/8*exp(3*b*x+3*a)/b+1/40*exp(5*b*x+5*a)/b

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Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2320, 12, 276} \begin {gather*} \frac {e^{-a-b x}}{8 b}+\frac {3 e^{a+b x}}{8 b}-\frac {e^{3 a+3 b x}}{8 b}+\frac {e^{5 a+5 b x}}{40 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Sinh[a + b*x]^3,x]

[Out]

E^(-a - b*x)/(8*b) + (3*E^(a + b*x))/(8*b) - E^(3*a + 3*b*x)/(8*b) + E^(5*a + 5*b*x)/(40*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 276

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 2320

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^{2 (a+b x)} \sinh ^3(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{8 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{x^2} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac {\text {Subst}\left (\int \left (3-\frac {1}{x^2}-3 x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac {e^{-a-b x}}{8 b}+\frac {3 e^{a+b x}}{8 b}-\frac {e^{3 a+3 b x}}{8 b}+\frac {e^{5 a+5 b x}}{40 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 50, normalized size = 0.76 \begin {gather*} \frac {e^{-a-b x} \left (5+15 e^{2 (a+b x)}-5 e^{4 (a+b x)}+e^{6 (a+b x)}\right )}{40 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*(a + b*x))*Sinh[a + b*x]^3,x]

[Out]

(E^(-a - b*x)*(5 + 15*E^(2*(a + b*x)) - 5*E^(4*(a + b*x)) + E^(6*(a + b*x))))/(40*b)

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Maple [A]
time = 1.23, size = 80, normalized size = 1.21

method	result	size
risch	\(\frac {{\mathrm e}^{-b x -a}}{8 b}+\frac {3 \,{\mathrm e}^{b x +a}}{8 b}-\frac {{\mathrm e}^{3 b x +3 a}}{8 b}+\frac {{\mathrm e}^{5 b x +5 a}}{40 b}\)	\(55\)
default	\(\frac {\sinh \left (b x +a \right )}{4 b}-\frac {\sinh \left (3 b x +3 a \right )}{8 b}+\frac {\sinh \left (5 b x +5 a \right )}{40 b}+\frac {\cosh \left (b x +a \right )}{2 b}-\frac {\cosh \left (3 b x +3 a \right )}{8 b}+\frac {\cosh \left (5 b x +5 a \right )}{40 b}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*sinh(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/4*sinh(b*x+a)/b-1/8/b*sinh(3*b*x+3*a)+1/40/b*sinh(5*b*x+5*a)+1/2*cosh(b*x+a)/b-1/8*cosh(3*b*x+3*a)/b+1/40*co
sh(5*b*x+5*a)/b

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Maxima [A]
time = 0.27, size = 53, normalized size = 0.80 \begin {gather*} -\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 15 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{40 \, b} + \frac {e^{\left (-b x - a\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/40*(5*e^(-2*b*x - 2*a) - 15*e^(-4*b*x - 4*a) - 1)*e^(5*b*x + 5*a)/b + 1/8*e^(-b*x - a)/b

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Fricas [A]
time = 0.37, size = 105, normalized size = 1.59 \begin {gather*} \frac {3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 2 \, \sinh \left (b x + a\right )^{3} - 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 5 \, \cosh \left (b x + a\right )}{20 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/20*(3*cosh(b*x + a)^3 + 9*cosh(b*x + a)*sinh(b*x + a)^2 - 2*sinh(b*x + a)^3 - 2*(3*cosh(b*x + a)^2 + 5)*sinh
(b*x + a) + 5*cosh(b*x + a))/(b*cosh(b*x + a)^2 - 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (49) = 98\).
time = 1.38, size = 124, normalized size = 1.88 \begin {gather*} \begin {cases} \frac {2 e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )}}{5 b} + \frac {e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{5 b} - \frac {4 e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac {2 e^{2 a} e^{2 b x} \cosh ^{3}{\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a)**3,x)

[Out]

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

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Giac [A]
time = 0.42, size = 54, normalized size = 0.82 \begin {gather*} \frac {e^{\left (5 \, b x + 5 \, a\right )}}{40 \, b} - \frac {e^{\left (3 \, b x + 3 \, a\right )}}{8 \, b} + \frac {3 \, e^{\left (b x + a\right )}}{8 \, b} + \frac {e^{\left (-b x - a\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

1/40*e^(5*b*x + 5*a)/b - 1/8*e^(3*b*x + 3*a)/b + 3/8*e^(b*x + a)/b + 1/8*e^(-b*x - a)/b

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Mupad [B]
time = 0.28, size = 45, normalized size = 0.68 \begin {gather*} \frac {15\,{\mathrm {e}}^{a+b\,x}+5\,{\mathrm {e}}^{-a-b\,x}-5\,{\mathrm {e}}^{3\,a+3\,b\,x}+{\mathrm {e}}^{5\,a+5\,b\,x}}{40\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*a + 2*b*x)*sinh(a + b*x)^3,x)

[Out]

(15*exp(a + b*x) + 5*exp(- a - b*x) - 5*exp(3*a + 3*b*x) + exp(5*a + 5*b*x))/(40*b)

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