\(\int e^{\cos (x) \sin (x)} \cos (2 x) \, dx\) [682]

Optimal result

Integrand size = 12, antiderivative size = 10 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\frac {1}{2} \sin (2 x)} \]

[Out]

exp(1/2*sin(2*x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4441, 2225} \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\frac {1}{2} \sin (2 x)} \]

[In]

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

[Out]

E^(Sin[2*x]/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 4441

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[d/(b
*c), Subst[Int[SubstFor[1, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a +
b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int e^{x/2} \, dx,x,\sin (2 x)\right ) \\ & = e^{\frac {1}{2} \sin (2 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\cos (x) \sin (x)} \]

[In]

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

[Out]

E^(Cos[x]*Sin[x])

Maple [A] (verified)

Time = 208.58 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\({\mathrm e}^{\cos \left (x \right ) \sin \left (x \right )}\)	\(7\)
default	\({\mathrm e}^{\cos \left (x \right ) \sin \left (x \right )}\)	\(7\)
risch	\({\mathrm e}^{\cos \left (x \right ) \sin \left (x \right )}\)	\(7\)

[In]

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

[Out]

exp(cos(x)*sin(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.60 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\left (\cos \left (x\right ) \sin \left (x\right )\right )} \]

[In]

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

[Out]

e^(cos(x)*sin(x))

Sympy [F(-1)]

Timed out. \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\left (\frac {1}{2} \, \sin \left (2 \, x\right )\right )} \]

[In]

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

[Out]

e^(1/2*sin(2*x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.20 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx=e^{\left (\frac {\tan \left (x\right )}{\tan \left (x\right )^{2} + 1}\right )} \]

[In]

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

[Out]

e^(tan(x)/(tan(x)^2 + 1))

Mupad [B] (verification not implemented)

Time = 26.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{\cos (x) \sin (x)} \cos (2 x) \, dx={\mathrm {e}}^{\frac {\sin \left (2\,x\right )}{2}} \]

[In]

int(cos(2*x)*exp(cos(x)*sin(x)),x)

[Out]

exp(sin(2*x)/2)