3.73 \(\int e^x \sin (a+b x) \, dx\)

Optimal. Leaf size=37 \[ \frac{e^x \sin (a+b x)}{b^2+1}-\frac{b e^x \cos (a+b x)}{b^2+1} \]

[Out]

-((b*E^x*Cos[a + b*x])/(1 + b^2)) + (E^x*Sin[a + b*x])/(1 + b^2)

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Rubi [A]  time = 0.0129877, antiderivative size = 37, 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 = {4432} \[ \frac{e^x \sin (a+b x)}{b^2+1}-\frac{b e^x \cos (a+b x)}{b^2+1} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sin[a + b*x],x]

[Out]

-((b*E^x*Cos[a + b*x])/(1 + b^2)) + (E^x*Sin[a + b*x])/(1 + b^2)

Rule 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[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 \sin (a+b x) \, dx &=-\frac{b e^x \cos (a+b x)}{1+b^2}+\frac{e^x \sin (a+b x)}{1+b^2}\\ \end{align*}

Mathematica [A]  time = 0.0579994, size = 27, normalized size = 0.73 \[ \frac{e^x (\sin (a+b x)-b \cos (a+b x))}{b^2+1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sin[a + b*x],x]

[Out]

(E^x*(-(b*Cos[a + b*x]) + Sin[a + b*x]))/(1 + b^2)

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Maple [A]  time = 0.003, size = 36, normalized size = 1. \begin{align*} -{\frac{{{\rm e}^{x}}b\cos \left ( bx+a \right ) }{{b}^{2}+1}}+{\frac{{{\rm e}^{x}}\sin \left ( bx+a \right ) }{{b}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*exp(x)*cos(b*x+a)/(b^2+1)+exp(x)*sin(b*x+a)/(b^2+1)

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Maxima [A]  time = 0.987557, size = 38, normalized size = 1.03 \begin{align*} -\frac{{\left (b \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} e^{x}}{b^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.462817, size = 70, normalized size = 1.89 \begin{align*} -\frac{b \cos \left (b x + a\right ) e^{x} - e^{x} \sin \left (b x + a\right )}{b^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.49205, size = 114, normalized size = 3.08 \begin{align*} \begin{cases} \frac{x e^{x} \sin{\left (a - i x \right )}}{2} + \frac{i x e^{x} \cos{\left (a - i x \right )}}{2} - \frac{i e^{x} \cos{\left (a - i x \right )}}{2} & \text{for}\: b = - i \\\frac{x e^{x} \sin{\left (a + i x \right )}}{2} - \frac{i x e^{x} \cos{\left (a + i x \right )}}{2} + \frac{e^{x} \sin{\left (a + i x \right )}}{2} & \text{for}\: b = i \\- \frac{b e^{x} \cos{\left (a + b x \right )}}{b^{2} + 1} + \frac{e^{x} \sin{\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)*sin(b*x+a),x)

[Out]

Piecewise((x*exp(x)*sin(a - I*x)/2 + I*x*exp(x)*cos(a - I*x)/2 - I*exp(x)*cos(a - I*x)/2, Eq(b, -I)), (x*exp(x
)*sin(a + I*x)/2 - I*x*exp(x)*cos(a + I*x)/2 + exp(x)*sin(a + I*x)/2, Eq(b, I)), (-b*exp(x)*cos(a + b*x)/(b**2
+ 1) + exp(x)*sin(a + b*x)/(b**2 + 1), True))

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Giac [A]  time = 1.28129, size = 47, normalized size = 1.27 \begin{align*} -{\left (\frac{b \cos \left (b x + a\right )}{b^{2} + 1} - \frac{\sin \left (b x + a\right )}{b^{2} + 1}\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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