3.23 \(\int e^x \cos ^2(e^x) \sin (e^x) \, dx\)

Optimal. Leaf size=10 \[ -\frac {1}{3} \cos ^3\left (e^x\right ) \]

[Out]

-1/3*cos(exp(x))^3

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Rubi [A]  time = 0.03, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2282, 2565, 30} \[ -\frac {1}{3} \cos ^3\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[E^x]^3/3

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int e^x \cos ^2\left (e^x\right ) \sin \left (e^x\right ) \, dx &=\operatorname {Subst}\left (\int \cos ^2(x) \sin (x) \, dx,x,e^x\right )\\ &=-\operatorname {Subst}\left (\int x^2 \, dx,x,\cos \left (e^x\right )\right )\\ &=-\frac {1}{3} \cos ^3\left (e^x\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.90 \[ -\frac {1}{4} \cos \left (e^x\right )-\frac {1}{12} \cos \left (3 e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Cos[E^x] - Cos[3*E^x]/12

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fricas [A]  time = 0.44, size = 7, normalized size = 0.70 \[ -\frac {1}{3} \, \cos \left (e^{x}\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*cos(e^x)^3

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giac [A]  time = 1.19, size = 7, normalized size = 0.70 \[ -\frac {1}{3} \, \cos \left (e^{x}\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*cos(e^x)^3

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maple [A]  time = 0.01, size = 8, normalized size = 0.80 \[ -\frac {\left (\cos ^{3}\left ({\mathrm e}^{x}\right )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*cos(exp(x))^3

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maxima [A]  time = 0.45, size = 7, normalized size = 0.70 \[ -\frac {1}{3} \, \cos \left (e^{x}\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*cos(e^x)^3

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mupad [B]  time = 0.19, size = 7, normalized size = 0.70 \[ -\frac {{\cos \left ({\mathrm {e}}^x\right )}^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(exp(x))^3/3

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sympy [A]  time = 1.96, size = 8, normalized size = 0.80 \[ - \frac {\cos ^{3}{\left (e^{x} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(exp(x))**3/3

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