3.81 \(\int e^{a x} \sin (b x) \, dx\)

Optimal. Leaf size=42 \[ \frac {a e^{a x} \sin (b x)}{a^2+b^2}-\frac {b e^{a x} \cos (b x)}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4432} \[ \frac {a e^{a x} \sin (b x)}{a^2+b^2}-\frac {b e^{a x} \cos (b x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a*x)*Sin[b*x],x]

[Out]

-((b*E^(a*x)*Cos[b*x])/(a^2 + b^2)) + (a*E^(a*x)*Sin[b*x])/(a^2 + 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^{a x} \sin (b x) \, dx &=-\frac {b e^{a x} \cos (b x)}{a^2+b^2}+\frac {a e^{a x} \sin (b x)}{a^2+b^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.69 \[ \frac {e^{a x} (a \sin (b x)-b \cos (b x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a*x)*Sin[b*x],x]

[Out]

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

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fricas [A]  time = 0.41, size = 33, normalized size = 0.79 \[ -\frac {b \cos \left (b x\right ) e^{\left (a x\right )} - a e^{\left (a x\right )} \sin \left (b x\right )}{a^{2} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.01, size = 38, normalized size = 0.90 \[ -{\left (\frac {b \cos \left (b x\right )}{a^{2} + b^{2}} - \frac {a \sin \left (b x\right )}{a^{2} + b^{2}}\right )} e^{\left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 41, normalized size = 0.98 \[ \frac {a \,{\mathrm e}^{a x} \sin \left (b x \right )}{a^{2}+b^{2}}-\frac {b \cos \left (b x \right ) {\mathrm e}^{a x}}{a^{2}+b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 29, normalized size = 0.69 \[ -\frac {{\left (b \cos \left (b x\right ) - a \sin \left (b x\right )\right )} e^{\left (a x\right )}}{a^{2} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.02, size = 29, normalized size = 0.69 \[ -\frac {{\mathrm {e}}^{a\,x}\,\left (b\,\cos \left (b\,x\right )-a\,\sin \left (b\,x\right )\right )}{a^2+b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.10, size = 139, normalized size = 3.31 \[ \begin {cases} 0 & \text {for}\: a = 0 \wedge b = 0 \\\frac {x e^{- i b x} \sin {\left (b x \right )}}{2} - \frac {i x e^{- i b x} \cos {\left (b x \right )}}{2} + \frac {i e^{- i b x} \sin {\left (b x \right )}}{2 b} & \text {for}\: a = - i b \\\frac {x e^{i b x} \sin {\left (b x \right )}}{2} + \frac {i x e^{i b x} \cos {\left (b x \right )}}{2} - \frac {i e^{i b x} \sin {\left (b x \right )}}{2 b} & \text {for}\: a = i b \\\frac {a e^{a x} \sin {\left (b x \right )}}{a^{2} + b^{2}} - \frac {b e^{a x} \cos {\left (b x \right )}}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(a*x)*sin(b*x),x)

[Out]

Piecewise((0, Eq(a, 0) & Eq(b, 0)), (x*exp(-I*b*x)*sin(b*x)/2 - I*x*exp(-I*b*x)*cos(b*x)/2 + I*exp(-I*b*x)*sin
(b*x)/(2*b), Eq(a, -I*b)), (x*exp(I*b*x)*sin(b*x)/2 + I*x*exp(I*b*x)*cos(b*x)/2 - I*exp(I*b*x)*sin(b*x)/(2*b),
Eq(a, I*b)), (a*exp(a*x)*sin(b*x)/(a**2 + b**2) - b*exp(a*x)*cos(b*x)/(a**2 + b**2), True))

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