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

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 41, 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 = {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 = 0.03, 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)

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 31, normalized size = 0.76 \[ \frac {a \cos \left (b x\right ) e^{\left (a x\right )} + b e^{\left (a x\right )} \sin \left (b x\right )}{a^{2} + b^{2}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.01, size = 36, normalized size = 0.88 \[ {\left (\frac {a \cos \left (b x\right )}{a^{2} + b^{2}} + \frac {b \sin \left (b x\right )}{a^{2} + b^{2}}\right )} e^{\left (a x\right )} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 40, normalized size = 0.98 \[ \frac {a \cos \left (b x \right ) {\mathrm e}^{a x}}{a^{2}+b^{2}}+\frac {b \,{\mathrm e}^{a x} \sin \left (b x \right )}{a^{2}+b^{2}} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 27, normalized size = 0.66 \[ \frac {{\left (a \cos \left (b x\right ) + b \sin \left (b x\right )\right )} e^{\left (a x\right )}}{a^{2} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.03, size = 27, normalized size = 0.66 \[ \frac {{\mathrm {e}}^{a\,x}\,\left (a\,\cos \left (b\,x\right )+b\,\sin \left (b\,x\right )\right )}{a^2+b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.11, size = 136, normalized size = 3.32 \[ \begin {cases} x & \text {for}\: a = 0 \wedge b = 0 \\\frac {i x e^{- i b x} \sin {\left (b x \right )}}{2} + \frac {x e^{- i b x} \cos {\left (b x \right )}}{2} + \frac {e^{- i b x} \sin {\left (b x \right )}}{2 b} & \text {for}\: a = - i b \\- \frac {i x e^{i b x} \sin {\left (b x \right )}}{2} + \frac {x e^{i b x} \cos {\left (b x \right )}}{2} + \frac {e^{i b x} \sin {\left (b x \right )}}{2 b} & \text {for}\: a = i b \\\frac {a e^{a x} \cos {\left (b x \right )}}{a^{2} + b^{2}} + \frac {b e^{a x} \sin {\left (b x \right )}}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \]

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 + exp(-I*b*x)*sin(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 + exp(I*b*x)*sin(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))

________________________________________________________________________________________