\(\int e^{2 (a+b x)} \sinh (a+b x) \, dx\) [874]

Optimal result

Integrand size = 16, antiderivative size = 32 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=-\frac {e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b} \]

[Out]

-1/2*exp(b*x+a)/b+1/6*exp(3*b*x+3*a)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2320, 12} \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\frac {e^{3 a+3 b x}}{6 b}-\frac {e^{a+b x}}{2 b} \]

[In]

Int[E^(2*(a + b*x))*Sinh[a + b*x],x]

[Out]

-1/2*E^(a + b*x)/b + E^(3*a + 3*b*x)/(6*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 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 {1}{2} \left (-1+x^2\right ) \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,e^{a+b x}\right )}{2 b} \\ & = -\frac {e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\frac {e^{a+b x} \left (-3+e^{2 (a+b x)}\right )}{6 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {{\mathrm e}^{b x +a}}{2 b}+\frac {{\mathrm e}^{3 b x +3 a}}{6 b}\)	\(27\)
parallelrisch	\(-\frac {{\mathrm e}^{2 b x +2 a} \left (\cosh \left (b x +a \right )-2 \sinh \left (b x +a \right )\right )}{3 b}\)	\(30\)
default	\(-\frac {\sinh \left (b x +a \right )}{2 b}+\frac {\sinh \left (3 b x +3 a \right )}{6 b}-\frac {\cosh \left (b x +a \right )}{2 b}+\frac {\cosh \left (3 b x +3 a \right )}{6 b}\)	\(52\)

[In]

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

[Out]

-1/2*exp(b*x+a)/b+1/6*exp(3*b*x+3*a)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 3}{6 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

[In]

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

[Out]

1/6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 3)/(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. 54 vs. \(2 (22) = 44\).

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\begin {cases} \frac {2 e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )}}{3 b} - \frac {e^{2 a} e^{2 b x} \cosh {\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\frac {e^{\left (3 \, b x + 3 \, a\right )}}{6 \, b} - \frac {e^{\left (b x + a\right )}}{2 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=\frac {e^{\left (3 \, b x + 3 \, a\right )}}{6 \, b} - \frac {e^{\left (b x + a\right )}}{2 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int e^{2 (a+b x)} \sinh (a+b x) \, dx=-\frac {3\,{\mathrm {e}}^{a+b\,x}-{\mathrm {e}}^{3\,a+3\,b\,x}}{6\,b} \]

[In]

int(exp(2*a + 2*b*x)*sinh(a + b*x),x)

[Out]

-(3*exp(a + b*x) - exp(3*a + 3*b*x))/(6*b)