3.6.46 \(\int \frac {\cos ^3(\frac {x}{3})}{\sqrt {e^x}} \, dx\) [546]

Optimal. Leaf size=79 \[ -\frac {48 \cos \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}-\frac {2 \cos ^3\left (\frac {x}{3}\right )}{5 \sqrt {e^x}}+\frac {32 \sin \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}+\frac {4 \cos ^2\left (\frac {x}{3}\right ) \sin \left (\frac {x}{3}\right )}{5 \sqrt {e^x}} \]

[Out]

-48/65*cos(1/3*x)/exp(x)^(1/2)-2/5*cos(1/3*x)^3/exp(x)^(1/2)+32/65*sin(1/3*x)/exp(x)^(1/2)+4/5*cos(1/3*x)^2*si
n(1/3*x)/exp(x)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2319, 4520, 4518} \begin {gather*} \frac {32 \sin \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}-\frac {2 \cos ^3\left (\frac {x}{3}\right )}{5 \sqrt {e^x}}-\frac {48 \cos \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}+\frac {4 \sin \left (\frac {x}{3}\right ) \cos ^2\left (\frac {x}{3}\right )}{5 \sqrt {e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x/3]^3/Sqrt[E^x],x]

[Out]

(-48*Cos[x/3])/(65*Sqrt[E^x]) - (2*Cos[x/3]^3)/(5*Sqrt[E^x]) + (32*Sin[x/3])/(65*Sqrt[E^x]) + (4*Cos[x/3]^2*Si
n[x/3])/(5*Sqrt[E^x])

Rule 2319

Int[(u_.)*((a_.)*(F_)^(v_))^(n_), x_Symbol] :> Dist[(a*F^v)^n/F^(n*v), Int[u*F^(n*v), x], x] /; FreeQ[{F, a, n
}, x] &&  !IntegerQ[n]

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 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 \frac {\cos ^3\left (\frac {x}{3}\right )}{\sqrt {e^x}} \, dx &=\frac {e^{x/2} \int e^{-x/2} \cos ^3\left (\frac {x}{3}\right ) \, dx}{\sqrt {e^x}}\\ &=-\frac {2 \cos ^3\left (\frac {x}{3}\right )}{5 \sqrt {e^x}}+\frac {4 \cos ^2\left (\frac {x}{3}\right ) \sin \left (\frac {x}{3}\right )}{5 \sqrt {e^x}}+\frac {\left (8 e^{x/2}\right ) \int e^{-x/2} \cos \left (\frac {x}{3}\right ) \, dx}{15 \sqrt {e^x}}\\ &=-\frac {48 \cos \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}-\frac {2 \cos ^3\left (\frac {x}{3}\right )}{5 \sqrt {e^x}}+\frac {32 \sin \left (\frac {x}{3}\right )}{65 \sqrt {e^x}}+\frac {4 \cos ^2\left (\frac {x}{3}\right ) \sin \left (\frac {x}{3}\right )}{5 \sqrt {e^x}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 36, normalized size = 0.46 \begin {gather*} \frac {-135 \cos \left (\frac {x}{3}\right )-13 \cos (x)+90 \sin \left (\frac {x}{3}\right )+26 \sin (x)}{130 \sqrt {e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x/3]^3/Sqrt[E^x],x]

[Out]

(-135*Cos[x/3] - 13*Cos[x] + 90*Sin[x/3] + 26*Sin[x])/(130*Sqrt[E^x])

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Maple [A]
time = 0.09, size = 38, normalized size = 0.48

method	result	size
default	\(-\frac {{\mathrm e}^{-\frac {x}{2}} \cos \left (x \right )}{10}+\frac {{\mathrm e}^{-\frac {x}{2}} \sin \left (x \right )}{5}-\frac {27 \,{\mathrm e}^{-\frac {x}{2}} \cos \left (\frac {x}{3}\right )}{26}+\frac {9 \,{\mathrm e}^{-\frac {x}{2}} \sin \left (\frac {x}{3}\right )}{13}\)	\(38\)
risch	\(\frac {\left (-\frac {1}{1300}-\frac {i}{650}\right ) \left (-52 i {\mathrm e}^{-i x}+65 \,{\mathrm e}^{i x}-39 \,{\mathrm e}^{-i x}+\left (270-540 i\right ) \cos \left (\frac {x}{3}\right )+\left (-180+360 i\right ) \sin \left (\frac {x}{3}\right )\right )}{\sqrt {{\mathrm e}^{x}}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(1/3*x)^3/exp(x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/10*exp(-1/2*x)*cos(x)+1/5*exp(-1/2*x)*sin(x)-27/26*exp(-1/2*x)*cos(1/3*x)+9/13*exp(-1/2*x)*sin(1/3*x)

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Maxima [A]
time = 1.95, size = 27, normalized size = 0.34 \begin {gather*} -\frac {1}{130} \, {\left (135 \, \cos \left (\frac {1}{3} \, x\right ) + 13 \, \cos \left (x\right ) - 90 \, \sin \left (\frac {1}{3} \, x\right ) - 26 \, \sin \left (x\right )\right )} e^{\left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/3*x)^3/exp(x)^(1/2),x, algorithm="maxima")

[Out]

-1/130*(135*cos(1/3*x) + 13*cos(x) - 90*sin(1/3*x) - 26*sin(x))*e^(-1/2*x)

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Fricas [A]
time = 0.90, size = 42, normalized size = 0.53 \begin {gather*} \frac {4}{65} \, {\left (13 \, \cos \left (\frac {1}{3} \, x\right )^{2} + 8\right )} e^{\left (-\frac {1}{2} \, x\right )} \sin \left (\frac {1}{3} \, x\right ) - \frac {2}{65} \, {\left (13 \, \cos \left (\frac {1}{3} \, x\right )^{3} + 24 \, \cos \left (\frac {1}{3} \, x\right )\right )} e^{\left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/3*x)^3/exp(x)^(1/2),x, algorithm="fricas")

[Out]

4/65*(13*cos(1/3*x)^2 + 8)*e^(-1/2*x)*sin(1/3*x) - 2/65*(13*cos(1/3*x)^3 + 24*cos(1/3*x))*e^(-1/2*x)

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Sympy [A]
time = 0.50, size = 76, normalized size = 0.96 \begin {gather*} \frac {32 \sin ^{3}{\left (\frac {x}{3} \right )}}{65 \sqrt {e^{x}}} - \frac {48 \sin ^{2}{\left (\frac {x}{3} \right )} \cos {\left (\frac {x}{3} \right )}}{65 \sqrt {e^{x}}} + \frac {84 \sin {\left (\frac {x}{3} \right )} \cos ^{2}{\left (\frac {x}{3} \right )}}{65 \sqrt {e^{x}}} - \frac {74 \cos ^{3}{\left (\frac {x}{3} \right )}}{65 \sqrt {e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/3*x)**3/exp(x)**(1/2),x)

[Out]

32*sin(x/3)**3/(65*sqrt(exp(x))) - 48*sin(x/3)**2*cos(x/3)/(65*sqrt(exp(x))) + 84*sin(x/3)*cos(x/3)**2/(65*sqr
t(exp(x))) - 74*cos(x/3)**3/(65*sqrt(exp(x)))

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Giac [A]
time = 0.64, size = 33, normalized size = 0.42 \begin {gather*} -\frac {9}{26} \, {\left (3 \, \cos \left (\frac {1}{3} \, x\right ) - 2 \, \sin \left (\frac {1}{3} \, x\right )\right )} e^{\left (-\frac {1}{2} \, x\right )} - \frac {1}{10} \, {\left (\cos \left (x\right ) - 2 \, \sin \left (x\right )\right )} e^{\left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/3*x)^3/exp(x)^(1/2),x, algorithm="giac")

[Out]

-9/26*(3*cos(1/3*x) - 2*sin(1/3*x))*e^(-1/2*x) - 1/10*(cos(x) - 2*sin(x))*e^(-1/2*x)

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Mupad [B]
time = 0.30, size = 39, normalized size = 0.49 \begin {gather*} -\frac {{\mathrm {e}}^{-\frac {x}{2}}\,\left (\frac {8\,{\cos \left (\frac {x}{3}\right )}^3}{5}-\frac {16\,\sin \left (\frac {x}{3}\right )\,{\cos \left (\frac {x}{3}\right )}^2}{5}+\frac {192\,\cos \left (\frac {x}{3}\right )}{65}-\frac {128\,\sin \left (\frac {x}{3}\right )}{65}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x/3)^3/exp(x)^(1/2),x)

[Out]

-(exp(-x/2)*((192*cos(x/3))/65 - (128*sin(x/3))/65 - (16*cos(x/3)^2*sin(x/3))/5 + (8*cos(x/3)^3)/5))/4

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