\(\int e^{a+b x} \sinh (c+d x) \, dx\) [880]

Optimal result

Integrand size = 14, antiderivative size = 54 \[ \int e^{a+b x} \sinh (c+d x) \, dx=-\frac {d e^{a+b x} \cosh (c+d x)}{b^2-d^2}+\frac {b e^{a+b x} \sinh (c+d x)}{b^2-d^2} \]

[Out]

-d*exp(b*x+a)*cosh(d*x+c)/(b^2-d^2)+b*exp(b*x+a)*sinh(d*x+c)/(b^2-d^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5582} \[ \int e^{a+b x} \sinh (c+d x) \, dx=\frac {b e^{a+b x} \sinh (c+d x)}{b^2-d^2}-\frac {d e^{a+b x} \cosh (c+d x)}{b^2-d^2} \]

[In]

Int[E^(a + b*x)*Sinh[c + d*x],x]

[Out]

-((d*E^(a + b*x)*Cosh[c + d*x])/(b^2 - d^2)) + (b*E^(a + b*x)*Sinh[c + d*x])/(b^2 - d^2)

Rule 5582

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x)
)*(Sinh[d + e*x]/(e^2 - b^2*c^2*Log[F]^2)), x] + Simp[e*F^(c*(a + b*x))*(Cosh[d + e*x]/(e^2 - b^2*c^2*Log[F]^2
)), x] /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 - b^2*c^2*Log[F]^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {d e^{a+b x} \cosh (c+d x)}{b^2-d^2}+\frac {b e^{a+b x} \sinh (c+d x)}{b^2-d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \sinh (c+d x) \, dx=\frac {e^{a+b x} (-d \cosh (c+d x)+b \sinh (c+d x))}{(b-d) (b+d)} \]

[In]

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

[Out]

(E^(a + b*x)*(-(d*Cosh[c + d*x]) + b*Sinh[c + d*x]))/((b - d)*(b + d))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{b x +a} \left (b \sinh \left (d x +c \right )-d \cosh \left (d x +c \right )\right )}{b^{2}-d^{2}}\)	\(37\)
risch	\(\frac {{\mathrm e}^{b x +d x +a +c}}{2 b +2 d}-\frac {{\mathrm e}^{b x -d x +a -c}}{2 \left (b -d \right )}\)	\(41\)
default	\(-\frac {\sinh \left (a -c +\left (b -d \right ) x \right )}{2 \left (b -d \right )}+\frac {\sinh \left (a +c +\left (b +d \right ) x \right )}{2 b +2 d}-\frac {\cosh \left (a -c +\left (b -d \right ) x \right )}{2 \left (b -d \right )}+\frac {\cosh \left (a +c +\left (b +d \right ) x \right )}{2 b +2 d}\)	\(78\)

[In]

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

[Out]

exp(b*x+a)/(b^2-d^2)*(b*sinh(d*x+c)-d*cosh(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.24 \[ \int e^{a+b x} \sinh (c+d x) \, dx=-\frac {d \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + d \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) - {\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{b^{2} - d^{2}} \]

[In]

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

[Out]

-(d*cosh(b*x + a)*cosh(d*x + c) + d*cosh(d*x + c)*sinh(b*x + a) - (b*cosh(b*x + a) + b*sinh(b*x + a))*sinh(d*x
+ c))/(b^2 - d^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (42) = 84\).

Time = 0.38 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.72 \[ \int e^{a+b x} \sinh (c+d x) \, dx=\begin {cases} x e^{a} \sinh {\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {x e^{a} e^{- d x} \sinh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{- d x} \cosh {\left (c + d x \right )}}{2} + \frac {e^{a} e^{- d x} \sinh {\left (c + d x \right )}}{2 d} + \frac {e^{a} e^{- d x} \cosh {\left (c + d x \right )}}{d} & \text {for}\: b = - d \\\frac {x e^{a} e^{d x} \sinh {\left (c + d x \right )}}{2} - \frac {x e^{a} e^{d x} \cosh {\left (c + d x \right )}}{2} - \frac {e^{a} e^{d x} \sinh {\left (c + d x \right )}}{2 d} + \frac {e^{a} e^{d x} \cosh {\left (c + d x \right )}}{d} & \text {for}\: b = d \\\frac {b e^{a} e^{b x} \sinh {\left (c + d x \right )}}{b^{2} - d^{2}} - \frac {d e^{a} e^{b x} \cosh {\left (c + d x \right )}}{b^{2} - d^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(exp(b*x+a)*sinh(d*x+c),x)

[Out]

Piecewise((x*exp(a)*sinh(c), Eq(b, 0) & Eq(d, 0)), (x*exp(a)*exp(-d*x)*sinh(c + d*x)/2 + x*exp(a)*exp(-d*x)*co
sh(c + d*x)/2 + exp(a)*exp(-d*x)*sinh(c + d*x)/(2*d) + exp(a)*exp(-d*x)*cosh(c + d*x)/d, Eq(b, -d)), (x*exp(a)
*exp(d*x)*sinh(c + d*x)/2 - x*exp(a)*exp(d*x)*cosh(c + d*x)/2 - exp(a)*exp(d*x)*sinh(c + d*x)/(2*d) + exp(a)*e
xp(d*x)*cosh(c + d*x)/d, Eq(b, d)), (b*exp(a)*exp(b*x)*sinh(c + d*x)/(b**2 - d**2) - d*exp(a)*exp(b*x)*cosh(c
+ d*x)/(b**2 - d**2), True))

Maxima [F(-2)]

Exception generated. \[ \int e^{a+b x} \sinh (c+d x) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-d/b>0)', see `assume?` for mo
re details)I

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.74 \[ \int e^{a+b x} \sinh (c+d x) \, dx=\frac {e^{\left (b x + d x + a + c\right )}}{2 \, {\left (b + d\right )}} - \frac {e^{\left (b x - d x + a - c\right )}}{2 \, {\left (b - d\right )}} \]

[In]

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

[Out]

1/2*e^(b*x + d*x + a + c)/(b + d) - 1/2*e^(b*x - d*x + a - c)/(b - d)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int e^{a+b x} \sinh (c+d x) \, dx=-\frac {{\mathrm {e}}^{a-c+b\,x-d\,x}\,\left (b+d-b\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{2\,\left (b^2-d^2\right )} \]

[In]

int(exp(a + b*x)*sinh(c + d*x),x)

[Out]

-(exp(a - c + b*x - d*x)*(b + d - b*exp(2*c + 2*d*x) + d*exp(2*c + 2*d*x)))/(2*(b^2 - d^2))