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

Optimal. Leaf size=57 \[ \frac{e^{-2 a-2 b x}}{16 b}-\frac{3 e^{2 a+2 b x}}{16 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{3 x}{8} \]

[Out]

E^(-2*a - 2*b*x)/(16*b) - (3*E^(2*a + 2*b*x))/(16*b) + E^(4*a + 4*b*x)/(32*b) + (3*x)/8

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Rubi [A]  time = 0.0384831, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 12, 266, 43} \[ \frac{e^{-2 a-2 b x}}{16 b}-\frac{3 e^{2 a+2 b x}}{16 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sinh[a + b*x]^3,x]

[Out]

E^(-2*a - 2*b*x)/(16*b) - (3*E^(2*a + 2*b*x))/(16*b) + E^(4*a + 4*b*x)/(32*b) + (3*x)/8

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{a+b x} \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{8 x^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{x^3} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-1+x)^3}{x^2} \, dx,x,e^{2 a+2 b x}\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-3-\frac{1}{x^2}+\frac{3}{x}+x\right ) \, dx,x,e^{2 a+2 b x}\right )}{16 b}\\ &=\frac{e^{-2 a-2 b x}}{16 b}-\frac{3 e^{2 a+2 b x}}{16 b}+\frac{e^{4 a+4 b x}}{32 b}+\frac{3 x}{8}\\ \end{align*}

Mathematica [A]  time = 0.0336144, size = 45, normalized size = 0.79 \[ \frac{e^{-2 (a+b x)}-3 e^{2 (a+b x)}+\frac{1}{2} e^{4 (a+b x)}+6 b x}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-2*(a + b*x)) - 3*E^(2*(a + b*x)) + E^(4*(a + b*x))/2 + 6*b*x)/(16*b)

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Maple [A]  time = 0.006, size = 67, normalized size = 1.2 \begin{align*}{\frac{1}{b} \left ( \left ({\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( bx+a \right ) }{8}} \right ) \cosh \left ( bx+a \right ) +{\frac{3\,bx}{8}}+{\frac{3\,a}{8}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*((1/4*sinh(b*x+a)^3-3/8*sinh(b*x+a))*cosh(b*x+a)+3/8*b*x+3/8*a+1/4*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*cosh(b*
x+a)^2)

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Maxima [A]  time = 1.11444, size = 72, normalized size = 1.26 \begin{align*} \frac{3 \,{\left (b x + a\right )}}{8 \, b} + \frac{e^{\left (4 \, b x + 4 \, a\right )}}{32 \, b} - \frac{3 \, e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b} + \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(b*x + a)/b + 1/32*e^(4*b*x + 4*a)/b - 3/16*e^(2*b*x + 2*a)/b + 1/16*e^(-2*b*x - 2*a)/b

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Fricas [B]  time = 2.05265, size = 259, normalized size = 4.54 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - \sinh \left (b x + a\right )^{3} + 6 \,{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right ) - 3 \,{\left (4 \, b x + \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(3*cosh(b*x + a)^3 + 9*cosh(b*x + a)*sinh(b*x + a)^2 - sinh(b*x + a)^3 + 6*(2*b*x - 1)*cosh(b*x + a) - 3*
(4*b*x + cosh(b*x + a)^2 + 2)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

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Sympy [A]  time = 27.0309, size = 187, normalized size = 3.28 \begin{align*} \begin{cases} \frac{3 x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8} - \frac{3 x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8} - \frac{3 x e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} + \frac{3 x e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8} + \frac{5 e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} - \frac{e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8 b} - \frac{e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\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)*sinh(b*x+a)**3,x)

[Out]

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

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Giac [A]  time = 1.13775, size = 77, normalized size = 1.35 \begin{align*} \frac{12 \, b x - 2 \,{\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 12 \, a + e^{\left (4 \, b x + 4 \, a\right )} - 6 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*b*x - 2*(3*e^(2*b*x + 2*a) - 1)*e^(-2*b*x - 2*a) + 12*a + e^(4*b*x + 4*a) - 6*e^(2*b*x + 2*a))/b