3.544 \(\int e^{m x} \cos ^2(x) \, dx\)

Optimal. Leaf size=54 \[ \frac{2 e^{m x}}{m \left (m^2+4\right )}+\frac{m e^{m x} \cos ^2(x)}{m^2+4}+\frac{2 e^{m x} \sin (x) \cos (x)}{m^2+4} \]

[Out]

(2*E^(m*x))/(m*(4 + m^2)) + (E^(m*x)*m*Cos[x]^2)/(4 + m^2) + (2*E^(m*x)*Cos[x]*Sin[x])/(4 + m^2)

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Rubi [A]  time = 0.0230822, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4435, 2194} \[ \frac{2 e^{m x}}{m \left (m^2+4\right )}+\frac{m e^{m x} \cos ^2(x)}{m^2+4}+\frac{2 e^{m x} \sin (x) \cos (x)}{m^2+4} \]

Antiderivative was successfully verified.

[In]

Int[E^(m*x)*Cos[x]^2,x]

[Out]

(2*E^(m*x))/(m*(4 + m^2)) + (E^(m*x)*m*Cos[x]^2)/(4 + m^2) + (2*E^(m*x)*Cos[x]*Sin[x])/(4 + m^2)

Rule 4435

Int[Cos[(d_.) + (e_.)*(x_)]^(m_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Cos[d + e*x]^m)/(e^2*m^2 + b^2*c^2*Log[F]^2), x] + (Dist[(m*(m - 1)*e^2)/(e^2*m^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Cos[d + e*x]^(m - 2), x], x] + Simp[(e*m*F^(c*(a + b*x))*Sin[d + e*x]*Cos[d + e*x]^(m - 1))/(
e^2*m^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*m^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
m, 1]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin{align*} \int e^{m x} \cos ^2(x) \, dx &=\frac{e^{m x} m \cos ^2(x)}{4+m^2}+\frac{2 e^{m x} \cos (x) \sin (x)}{4+m^2}+\frac{2 \int e^{m x} \, dx}{4+m^2}\\ &=\frac{2 e^{m x}}{m \left (4+m^2\right )}+\frac{e^{m x} m \cos ^2(x)}{4+m^2}+\frac{2 e^{m x} \cos (x) \sin (x)}{4+m^2}\\ \end{align*}

Mathematica [A]  time = 0.0351539, size = 39, normalized size = 0.72 \[ \frac{e^{m x} \left (m^2 \cos (2 x)+m^2+2 m \sin (2 x)+4\right )}{2 m \left (m^2+4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(m*x)*Cos[x]^2,x]

[Out]

(E^(m*x)*(4 + m^2 + m^2*Cos[2*x] + 2*m*Sin[2*x]))/(2*m*(4 + m^2))

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Maple [A]  time = 0.02, size = 45, normalized size = 0.8 \begin{align*}{\frac{{{\rm e}^{mx}}}{2\,m}}+{\frac{m{{\rm e}^{mx}}\cos \left ( 2\,x \right ) }{2\,{m}^{2}+8}}+{\frac{{{\rm e}^{mx}}\sin \left ( 2\,x \right ) }{{m}^{2}+4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(m*x)*cos(x)^2,x)

[Out]

1/2*exp(m*x)/m+1/2*m/(m^2+4)*exp(m*x)*cos(2*x)+1/(m^2+4)*exp(m*x)*sin(2*x)

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Maxima [A]  time = 0.942968, size = 61, normalized size = 1.13 \begin{align*} \frac{m^{2} \cos \left (2 \, x\right ) e^{\left (m x\right )} + 2 \, m e^{\left (m x\right )} \sin \left (2 \, x\right ) +{\left (m^{2} + 4\right )} e^{\left (m x\right )}}{2 \,{\left (m^{3} + 4 \, m\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*cos(x)^2,x, algorithm="maxima")

[Out]

1/2*(m^2*cos(2*x)*e^(m*x) + 2*m*e^(m*x)*sin(2*x) + (m^2 + 4)*e^(m*x))/(m^3 + 4*m)

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Fricas [A]  time = 2.08634, size = 95, normalized size = 1.76 \begin{align*} \frac{2 \, m \cos \left (x\right ) e^{\left (m x\right )} \sin \left (x\right ) +{\left (m^{2} \cos \left (x\right )^{2} + 2\right )} e^{\left (m x\right )}}{m^{3} + 4 \, m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*cos(x)^2,x, algorithm="fricas")

[Out]

(2*m*cos(x)*e^(m*x)*sin(x) + (m^2*cos(x)^2 + 2)*e^(m*x))/(m^3 + 4*m)

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Sympy [A]  time = 3.42764, size = 265, normalized size = 4.91 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (x \right )}}{2} + \frac{x \cos ^{2}{\left (x \right )}}{2} + \frac{\sin{\left (x \right )} \cos{\left (x \right )}}{2} & \text{for}\: m = 0 \\- \frac{x e^{- 2 i x} \sin ^{2}{\left (x \right )}}{4} + \frac{i x e^{- 2 i x} \sin{\left (x \right )} \cos{\left (x \right )}}{2} + \frac{x e^{- 2 i x} \cos ^{2}{\left (x \right )}}{4} - \frac{e^{- 2 i x} \sin{\left (x \right )} \cos{\left (x \right )}}{4} + \frac{i e^{- 2 i x} \cos ^{2}{\left (x \right )}}{2} & \text{for}\: m = - 2 i \\- \frac{x e^{2 i x} \sin ^{2}{\left (x \right )}}{4} - \frac{i x e^{2 i x} \sin{\left (x \right )} \cos{\left (x \right )}}{2} + \frac{x e^{2 i x} \cos ^{2}{\left (x \right )}}{4} - \frac{e^{2 i x} \sin{\left (x \right )} \cos{\left (x \right )}}{4} - \frac{i e^{2 i x} \cos ^{2}{\left (x \right )}}{2} & \text{for}\: m = 2 i \\\frac{m^{2} e^{m x} \cos ^{2}{\left (x \right )}}{m^{3} + 4 m} + \frac{2 m e^{m x} \sin{\left (x \right )} \cos{\left (x \right )}}{m^{3} + 4 m} + \frac{2 e^{m x} \sin ^{2}{\left (x \right )}}{m^{3} + 4 m} + \frac{2 e^{m x} \cos ^{2}{\left (x \right )}}{m^{3} + 4 m} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*cos(x)**2,x)

[Out]

Piecewise((x*sin(x)**2/2 + x*cos(x)**2/2 + sin(x)*cos(x)/2, Eq(m, 0)), (-x*exp(-2*I*x)*sin(x)**2/4 + I*x*exp(-
2*I*x)*sin(x)*cos(x)/2 + x*exp(-2*I*x)*cos(x)**2/4 - exp(-2*I*x)*sin(x)*cos(x)/4 + I*exp(-2*I*x)*cos(x)**2/2,
Eq(m, -2*I)), (-x*exp(2*I*x)*sin(x)**2/4 - I*x*exp(2*I*x)*sin(x)*cos(x)/2 + x*exp(2*I*x)*cos(x)**2/4 - exp(2*I
*x)*sin(x)*cos(x)/4 - I*exp(2*I*x)*cos(x)**2/2, Eq(m, 2*I)), (m**2*exp(m*x)*cos(x)**2/(m**3 + 4*m) + 2*m*exp(m
*x)*sin(x)*cos(x)/(m**3 + 4*m) + 2*exp(m*x)*sin(x)**2/(m**3 + 4*m) + 2*exp(m*x)*cos(x)**2/(m**3 + 4*m), True))

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Giac [A]  time = 1.09468, size = 58, normalized size = 1.07 \begin{align*} \frac{1}{2} \,{\left (\frac{m \cos \left (2 \, x\right )}{m^{2} + 4} + \frac{2 \, \sin \left (2 \, x\right )}{m^{2} + 4}\right )} e^{\left (m x\right )} + \frac{e^{\left (m x\right )}}{2 \, m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)*cos(x)^2,x, algorithm="giac")

[Out]

1/2*(m*cos(2*x)/(m^2 + 4) + 2*sin(2*x)/(m^2 + 4))*e^(m*x) + 1/2*e^(m*x)/m