3.6.44 \(\int e^{m x} \cos ^2(x) \, dx\) [544]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4520, 2225} \begin {gather*} \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} \end {gather*}

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 2225

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]

Rule 4520

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]

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 = 0.02, size = 39, normalized size = 0.72 \begin {gather*} \frac {e^{m x} \left (4+m^2+m^2 \cos (2 x)+2 m \sin (2 x)\right )}{2 m \left (4+m^2\right )} \end {gather*}

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))

________________________________________________________________________________________

Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.47, size = 178, normalized size = 3.30 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {x}{2}+\frac {\text {Sin}\left [2 x\right ]}{4},m\text {==}0\right \},\left \{\frac {\left (2 I+2 I x \text {Sin}\left [2 x\right ]+2 x \text {Cos}\left [2 x\right ]-2 I \text {Cos}\left [2 x\right ]+3 \text {Sin}\left [2 x\right ]\right ) E^{-2 I x}}{8},m\text {==}-2 I\right \},\left \{\frac {\left (-2 I-2 I x \text {Sin}\left [2 x\right ]+2 x \text {Cos}\left [2 x\right ]+2 I \text {Cos}\left [2 x\right ]+3 \text {Sin}\left [2 x\right ]\right ) E^{2 I x}}{8},m\text {==}2 I\right \}\right \},\frac {2 m \text {Cos}\left [x\right ] E^{m x} \text {Sin}\left [x\right ]}{4 m+m^3}+\frac {m^2 E^{m x} \text {Cos}\left [x\right ]^2}{4 m+m^3}+\frac {2 E^{m x} \text {Cos}\left [x\right ]^2}{4 m+m^3}+\frac {2 E^{m x} \text {Sin}\left [x\right ]^2}{4 m+m^3}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Piecewise[{{x / 2 + Sin[2 x] / 4, m == 0}, {(2 I + 2 I x Sin[2 x] + 2 x Cos[2 x] - 2 I Cos[2 x] + 3 Sin[2 x])
E ^ (-2 I x) / 8, m == -2 I}, {(-2 I - 2 I x Sin[2 x] + 2 x Cos[2 x] + 2 I Cos[2 x] + 3 Sin[2 x]) E ^ (2 I x)
/ 8, m == 2 I}}, 2 m Cos[x] E ^ (m x) Sin[x] / (4 m + m ^ 3) + m ^ 2 E ^ (m x) Cos[x] ^ 2 / (4 m + m ^ 3) + 2
E ^ (m x) Cos[x] ^ 2 / (4 m + m ^ 3) + 2 E ^ (m x) Sin[x] ^ 2 / (4 m + m ^ 3)]

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 45, normalized size = 0.83

method	result	size
risch	\(\frac {{\mathrm e}^{m x}}{2 m}+\frac {{\mathrm e}^{\left (2 i+m \right ) x}}{8 i+4 m}+\frac {{\mathrm e}^{x \left (m -2 i\right )}}{4 m -8 i}\)	\(41\)
default	\(\frac {{\mathrm e}^{m x}}{2 m}+\frac {m \,{\mathrm e}^{m x} \cos \left (2 x \right )}{2 m^{2}+8}+\frac {{\mathrm e}^{m x} \sin \left (2 x \right )}{m^{2}+4}\)	\(45\)
norman	\(\frac {\frac {\left (m^{2}+2\right ) {\mathrm e}^{m x}}{m \left (m^{2}+4\right )}+\frac {\left (m^{2}+2\right ) {\mathrm e}^{m x} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{m \left (m^{2}+4\right )}+\frac {4 \,{\mathrm e}^{m x} \tan \left (\frac {x}{2}\right )}{m^{2}+4}-\frac {4 \,{\mathrm e}^{m x} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{m^{2}+4}-\frac {2 \left (m^{2}-2\right ) {\mathrm e}^{m x} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{m \left (m^{2}+4\right )}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(m*x)*cos(x)^2,x,method=_RETURNVERBOSE)

[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)

________________________________________________________________________________________

Maxima [A]
time = 0.24, size = 45, normalized size = 0.83 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.32, size = 37, normalized size = 0.69 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Sympy [A]
time = 0.44, size = 265, normalized size = 4.91 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________

Giac [A]
time = 0.00, size = 45, normalized size = 0.83 \begin {gather*} \frac {\mathrm {e}^{m x}}{2 m}+\mathrm {e}^{m x} \left (\frac {2 m \cos \left (2 x\right )}{\left (2 m\right )^{2}+16}+\frac {4 \sin \left (2 x\right )}{\left (2 m\right )^{2}+16}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.05, size = 37, normalized size = 0.69 \begin {gather*} \frac {{\mathrm {e}}^{m\,x}}{2\,m}+\frac {{\mathrm {e}}^{m\,x}\,\left (2\,\sin \left (2\,x\right )+m\,\cos \left (2\,x\right )\right )}{2\,\left (m^2+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________