3.104 \(\int e^x \cos (a+b x) \, dx\)

Optimal. Leaf size=36 \[ \frac{b e^x \sin (a+b x)}{b^2+1}+\frac{e^x \cos (a+b x)}{b^2+1} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0123241, antiderivative size = 36, 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} \[ \frac{b e^x \sin (a+b x)}{b^2+1}+\frac{e^x \cos (a+b x)}{b^2+1} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Cos[a + b*x],x]

[Out]

(E^x*Cos[a + b*x])/(1 + b^2) + (b*E^x*Sin[a + b*x])/(1 + 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^x \cos (a+b x) \, dx &=\frac{e^x \cos (a+b x)}{1+b^2}+\frac{b e^x \sin (a+b x)}{1+b^2}\\ \end{align*}

Mathematica [A]  time = 0.0504032, size = 26, normalized size = 0.72 \[ \frac{e^x (b \sin (a+b x)+\cos (a+b x))}{b^2+1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Cos[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 35, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{x}}\cos \left ( bx+a \right ) }{{b}^{2}+1}}+{\frac{{{\rm e}^{x}}b\sin \left ( bx+a \right ) }{{b}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.09355, size = 34, normalized size = 0.94 \begin{align*} \frac{{\left (b \sin \left (b x + a\right ) + \cos \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)*cos(b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.461251, size = 69, normalized size = 1.92 \begin{align*} \frac{b e^{x} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) e^{x}}{b^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 2.47465, size = 114, normalized size = 3.17 \begin{align*} \begin{cases} - \frac{i x e^{x} \sin{\left (a - i x \right )}}{2} + \frac{x e^{x} \cos{\left (a - i x \right )}}{2} + \frac{e^{x} \cos{\left (a - i x \right )}}{2} & \text{for}\: b = - i \\\frac{i x e^{x} \sin{\left (a + i x \right )}}{2} + \frac{x e^{x} \cos{\left (a + i x \right )}}{2} - \frac{i e^{x} \sin{\left (a + i x \right )}}{2} & \text{for}\: b = i \\\frac{b e^{x} \sin{\left (a + b x \right )}}{b^{2} + 1} + \frac{e^{x} \cos{\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)*cos(b*x+a),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.13738, size = 45, normalized size = 1.25 \begin{align*}{\left (\frac{b \sin \left (b x + a\right )}{b^{2} + 1} + \frac{\cos \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)*cos(b*x+a),x, algorithm="giac")

[Out]

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