3.80 \(\int e^{a x} \cos (b x) \, dx\)

Optimal. Leaf size=41 \[ \frac{b e^{a x} \sin (b x)}{a^2+b^2}+\frac{a e^{a x} \cos (b x)}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.011306, 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 = {4433} \[ \frac{b e^{a x} \sin (b x)}{a^2+b^2}+\frac{a e^{a x} \cos (b x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4433

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

Mathematica [A]  time = 0.025097, size = 28, normalized size = 0.68 \[ \frac{e^{a x} (a \cos (b x)+b \sin (b x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 40, normalized size = 1. \begin{align*}{\frac{a{{\rm e}^{ax}}\cos \left ( bx \right ) }{{a}^{2}+{b}^{2}}}+{\frac{{{\rm e}^{ax}}b\sin \left ( bx \right ) }{{a}^{2}+{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.966976, size = 36, normalized size = 0.88 \begin{align*} \frac{{\left (a \cos \left (b x\right ) + b \sin \left (b x\right )\right )} e^{\left (a x\right )}}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.559013, size = 74, normalized size = 1.8 \begin{align*} \frac{a \cos \left (b x\right ) e^{\left (a x\right )} + b e^{\left (a x\right )} \sin \left (b x\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.32268, size = 139, normalized size = 3.39 \begin{align*} \begin{cases} x & \text{for}\: a = 0 \wedge b = 0 \\\frac{i x e^{- i b x} \sin{\left (b x \right )}}{2} + \frac{x e^{- i b x} \cos{\left (b x \right )}}{2} + \frac{i e^{- i b x} \cos{\left (b x \right )}}{2 b} & \text{for}\: a = - i b \\- \frac{i x e^{i b x} \sin{\left (b x \right )}}{2} + \frac{x e^{i b x} \cos{\left (b x \right )}}{2} - \frac{i e^{i b x} \cos{\left (b x \right )}}{2 b} & \text{for}\: a = i b \\\frac{a e^{a x} \cos{\left (b x \right )}}{a^{2} + b^{2}} + \frac{b e^{a x} \sin{\left (b x \right )}}{a^{2} + b^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08904, size = 49, normalized size = 1.2 \begin{align*}{\left (\frac{a \cos \left (b x\right )}{a^{2} + b^{2}} + \frac{b \sin \left (b x\right )}{a^{2} + b^{2}}\right )} e^{\left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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