\(\int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx\) [547]

Optimal result

Integrand size = 14, antiderivative size = 36 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=\frac {e^{2 x}}{16}-\frac {1}{80} e^{2 x} \cos (4 x)-\frac {1}{40} e^{2 x} \sin (4 x) \]

[Out]

1/16*exp(2*x)-1/80*exp(2*x)*cos(4*x)-1/40*exp(2*x)*sin(4*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4557, 2225, 4518} \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=\frac {e^{2 x}}{16}-\frac {1}{40} e^{2 x} \sin (4 x)-\frac {1}{80} e^{2 x} \cos (4 x) \]

[In]

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

[Out]

E^(2*x)/16 - (E^(2*x)*Cos[4*x])/80 - (E^(2*x)*Sin[4*x])/40

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 4518

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]

Rule 4557

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{2 x}}{8}-\frac {1}{8} e^{2 x} \cos (4 x)\right ) \, dx \\ & = \frac {1}{8} \int e^{2 x} \, dx-\frac {1}{8} \int e^{2 x} \cos (4 x) \, dx \\ & = \frac {e^{2 x}}{16}-\frac {1}{80} e^{2 x} \cos (4 x)-\frac {1}{40} e^{2 x} \sin (4 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.58 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=-\frac {1}{80} e^{2 x} (-5+\cos (4 x)+2 \sin (4 x)) \]

[In]

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

[Out]

-1/80*(E^(2*x)*(-5 + Cos[4*x] + 2*Sin[4*x]))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.53

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{2 x} \left (-5+\cos \left (4 x \right )+2 \sin \left (4 x \right )\right )}{80}\)	\(19\)
default	\(\frac {{\mathrm e}^{2 x}}{16}-\frac {{\mathrm e}^{2 x} \cos \left (4 x \right )}{80}-\frac {{\mathrm e}^{2 x} \sin \left (4 x \right )}{40}\)	\(28\)
risch	\(\frac {{\mathrm e}^{2 x}}{16}-\frac {{\mathrm e}^{\left (2+4 i\right ) x}}{160}+\frac {i {\mathrm e}^{\left (2+4 i\right ) x}}{80}-\frac {{\mathrm e}^{\left (2-4 i\right ) x}}{160}-\frac {i {\mathrm e}^{\left (2-4 i\right ) x}}{80}\)	\(42\)
norman	\(\frac {-\frac {{\mathrm e}^{2 x} \tan \left (\frac {x}{2}\right )}{5}+\frac {3 \,{\mathrm e}^{2 x} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{5}+\frac {7 \,{\mathrm e}^{2 x} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{5}-\frac {{\mathrm e}^{2 x} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{2}-\frac {7 \,{\mathrm e}^{2 x} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{5}+\frac {3 \,{\mathrm e}^{2 x} \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{5}+\frac {{\mathrm e}^{2 x} \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{5}+\frac {{\mathrm e}^{2 x} \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{20}+\frac {{\mathrm e}^{2 x}}{20}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{4}}\)	\(113\)

[In]

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

[Out]

-1/80*exp(2*x)*(-5+cos(4*x)+2*sin(4*x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=-\frac {1}{10} \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} e^{\left (2 \, x\right )} \sin \left (x\right ) - \frac {1}{20} \, {\left (2 \, \cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} - 1\right )} e^{\left (2 \, x\right )} \]

[In]

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

[Out]

-1/10*(2*cos(x)^3 - cos(x))*e^(2*x)*sin(x) - 1/20*(2*cos(x)^4 - 2*cos(x)^2 - 1)*e^(2*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (29) = 58\).

Time = 0.52 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.94 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=\frac {e^{2 x} \sin ^{4}{\left (x \right )}}{20} + \frac {e^{2 x} \sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{10} + \frac {e^{2 x} \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{5} - \frac {e^{2 x} \sin {\left (x \right )} \cos ^{3}{\left (x \right )}}{10} + \frac {e^{2 x} \cos ^{4}{\left (x \right )}}{20} \]

[In]

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

[Out]

exp(2*x)*sin(x)**4/20 + exp(2*x)*sin(x)**3*cos(x)/10 + exp(2*x)*sin(x)**2*cos(x)**2/5 - exp(2*x)*sin(x)*cos(x)
**3/10 + exp(2*x)*cos(x)**4/20

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=-\frac {1}{80} \, \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} - \frac {1}{40} \, e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) + \frac {1}{16} \, e^{\left (2 \, x\right )} \]

[In]

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

[Out]

-1/80*cos(4*x)*e^(2*x) - 1/40*e^(2*x)*sin(4*x) + 1/16*e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=-\frac {1}{80} \, {\left (\cos \left (4 \, x\right ) + 2 \, \sin \left (4 \, x\right )\right )} e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (2 \, x\right )} \]

[In]

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

[Out]

-1/80*(cos(4*x) + 2*sin(4*x))*e^(2*x) + 1/16*e^(2*x)

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.50 \[ \int e^{2 x} \cos ^2(x) \sin ^2(x) \, dx=-\frac {{\mathrm {e}}^{2\,x}\,\left (\cos \left (4\,x\right )+2\,\sin \left (4\,x\right )-5\right )}{80} \]

[In]

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

[Out]

-(exp(2*x)*(cos(4*x) + 2*sin(4*x) - 5))/80