\(\int e^{a+b x} \sinh ^3(a+b x) \, dx\) [302]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {e^{-2 a-2 b x}}{16 b}-\frac {3 e^{2 a+2 b x}}{16 b}+\frac {e^{4 a+4 b x}}{32 b}+\frac {3 x}{8} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 12, 272, 45} \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {e^{-2 a-2 b x}}{16 b}-\frac {3 e^{2 a+2 b x}}{16 b}+\frac {e^{4 a+4 b x}}{32 b}+\frac {3 x}{8} \]

[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

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])

Rule 272

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 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{8 x^3} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{x^3} \, dx,x,e^{a+b x}\right )}{8 b} \\ & = \frac {\text {Subst}\left (\int \frac {(-1+x)^3}{x^2} \, dx,x,e^{2 a+2 b x}\right )}{16 b} \\ & = \frac {\text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {e^{-2 (a+b x)}-3 e^{2 (a+b x)}+\frac {1}{2} e^{4 (a+b x)}+6 b x}{16 b} \]

[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)

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {{\mathrm e}^{-2 b x -2 a}}{16 b}-\frac {3 \,{\mathrm e}^{2 b x +2 a}}{16 b}+\frac {{\mathrm e}^{4 b x +4 a}}{32 b}+\frac {3 x}{8}\)	\(47\)
derivativedivides	\(\frac {\left (\frac {\sinh \left (b x +a \right )^{3}}{4}-\frac {3 \sinh \left (b x +a \right )}{8}\right ) \cosh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}+\frac {\sinh \left (b x +a \right )^{4}}{4}}{b}\)	\(49\)
default	\(\frac {\left (\frac {\sinh \left (b x +a \right )^{3}}{4}-\frac {3 \sinh \left (b x +a \right )}{8}\right ) \cosh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}+\frac {\sinh \left (b x +a \right )^{4}}{4}}{b}\)	\(49\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (46) = 92\).

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.67 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - \sinh \left (b x + a\right )^{3} + 6 \, {\left (2 \, b x - 1\right )} \cosh \left (b x + a\right ) - 3 \, {\left (4 \, b x + \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )}{32 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (48) = 96\).

Time = 0.84 (sec) , antiderivative size = 207, normalized size of antiderivative = 3.63 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\begin {cases} \frac {3 x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8} - \frac {3 x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8} - \frac {3 x e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} + \frac {3 x e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8} - \frac {3 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b} + \frac {e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac {5 e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {3 \, {\left (b x + a\right )}}{8 \, b} + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{32 \, b} - \frac {3 \, e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {12 \, b x - 2 \, {\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 12 \, a + e^{\left (4 \, b x + 4 \, a\right )} - 6 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int e^{a+b x} \sinh ^3(a+b x) \, dx=\frac {3\,x}{8}+\frac {\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}}{16}-\frac {3\,{\mathrm {e}}^{2\,a+2\,b\,x}}{16}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{32}}{b} \]

[In]

int(exp(a + b*x)*sinh(a + b*x)^3,x)

[Out]

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