3.336 \(\int e^x \sinh (a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac{e^x \sinh (a+b x)}{1-b^2}-\frac{b e^x \cosh (a+b x)}{1-b^2} \]

[Out]

-((b*E^x*Cosh[a + b*x])/(1 - b^2)) + (E^x*Sinh[a + b*x])/(1 - b^2)

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Rubi [A]  time = 0.0132719, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {5474} \[ \frac{e^x \sinh (a+b x)}{1-b^2}-\frac{b e^x \cosh (a+b x)}{1-b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sinh[a + b*x],x]

[Out]

-((b*E^x*Cosh[a + b*x])/(1 - b^2)) + (E^x*Sinh[a + b*x])/(1 - b^2)

Rule 5474

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*} \int e^x \sinh (a+b x) \, dx &=-\frac{b e^x \cosh (a+b x)}{1-b^2}+\frac{e^x \sinh (a+b x)}{1-b^2}\\ \end{align*}

Mathematica [A]  time = 0.0479623, size = 28, normalized size = 0.68 \[ \frac{e^x (b \cosh (a+b x)-\sinh (a+b x))}{b^2-1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sinh[a + b*x],x]

[Out]

(E^x*(b*Cosh[a + b*x] - Sinh[a + b*x]))/(-1 + b^2)

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Maple [A]  time = 0.016, size = 62, normalized size = 1.5 \begin{align*} -{\frac{\sinh \left ( \left ( b-1 \right ) x+a \right ) }{2\,b-2}}+{\frac{\sinh \left ( \left ( 1+b \right ) x+a \right ) }{2+2\,b}}+{\frac{\cosh \left ( \left ( b-1 \right ) x+a \right ) }{2\,b-2}}+{\frac{\cosh \left ( \left ( 1+b \right ) x+a \right ) }{2+2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b-1)*sinh((b-1)*x+a)+1/2/(1+b)*sinh((1+b)*x+a)+1/2*cosh((b-1)*x+a)/(b-1)+1/2*cosh((1+b)*x+a)/(1+b)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74592, size = 134, normalized size = 3.27 \begin{align*} \frac{b \cosh \left (b x + a\right ) \cosh \left (x\right ) + b \cosh \left (b x + a\right ) \sinh \left (x\right ) -{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \sinh \left (b x + a\right )}{b^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.842607, size = 99, normalized size = 2.41 \begin{align*} \begin{cases} \frac{x e^{x} \sinh{\left (a - x \right )}}{2} + \frac{x e^{x} \cosh{\left (a - x \right )}}{2} + \frac{e^{x} \sinh{\left (a - x \right )}}{2} & \text{for}\: b = -1 \\\frac{x e^{x} \sinh{\left (a + x \right )}}{2} - \frac{x e^{x} \cosh{\left (a + x \right )}}{2} + \frac{e^{x} \cosh{\left (a + x \right )}}{2} & \text{for}\: b = 1 \\\frac{b e^{x} \cosh{\left (a + b x \right )}}{b^{2} - 1} - \frac{e^{x} \sinh{\left (a + b x \right )}}{b^{2} - 1} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(x)*sinh(a - x)/2 + x*exp(x)*cosh(a - x)/2 + exp(x)*sinh(a - x)/2, Eq(b, -1)), (x*exp(x)*sinh(
a + x)/2 - x*exp(x)*cosh(a + x)/2 + exp(x)*cosh(a + x)/2, Eq(b, 1)), (b*exp(x)*cosh(a + b*x)/(b**2 - 1) - exp(
x)*sinh(a + b*x)/(b**2 - 1), True))

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Giac [A]  time = 1.16365, size = 43, normalized size = 1.05 \begin{align*} \frac{e^{\left (b x + a + x\right )}}{2 \,{\left (b + 1\right )}} + \frac{e^{\left (-b x - a + x\right )}}{2 \,{\left (b - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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