3.65.82 \(\int \frac {1}{30} e^{\frac {1}{10} (2 e^{x/4}-x-10 \log (\frac {2}{x}))} (-40+2 x-e^{x/4} x) \, dx\)

Optimal. Leaf size=41 \[ \frac {2}{3 \left (1-\frac {x}{-\frac {2 e^{\frac {1}{5} \left (-e^{x/4}+\frac {x}{2}\right )}}{x}+x}\right )} \]

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Rubi [A]  time = 0.12, antiderivative size = 51, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 2274, 2288} \begin {gather*} -\frac {e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (2 x-e^{x/4} x\right )}{3 \left (2-e^{x/4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*E^(x/4) - x - 10*Log[2/x])/10)*(-40 + 2*x - E^(x/4)*x))/30,x]

[Out]

-1/3*(E^((2*E^(x/4) - x)/10)*x*(2*x - E^(x/4)*x))/(2 - E^(x/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{30} \int e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx\\ &=\frac {1}{30} \int \frac {1}{2} e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (-40+2 x-e^{x/4} x\right ) \, dx\\ &=\frac {1}{60} \int e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (-40+2 x-e^{x/4} x\right ) \, dx\\ &=-\frac {e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (2 x-e^{x/4} x\right )}{3 \left (2-e^{x/4}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.63 \begin {gather*} -\frac {1}{3} e^{\frac {e^{x/4}}{5}-\frac {x}{10}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*E^(x/4) - x - 10*Log[2/x])/10)*(-40 + 2*x - E^(x/4)*x))/30,x]

[Out]

-1/3*(E^(E^(x/4)/5 - x/10)*x^2)

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fricas [A]  time = 0.85, size = 22, normalized size = 0.54 \begin {gather*} -\frac {2}{3} \, x e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )} - \log \left (\frac {2}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-x*exp(1/4*x)+2*x-40)/exp(log(2/x)-1/5*exp(1/4*x)+1/10*x),x, algorithm="fricas")

[Out]

-2/3*x*e^(-1/10*x + 1/5*e^(1/4*x) - log(2/x))

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giac [A]  time = 0.18, size = 16, normalized size = 0.39 \begin {gather*} -\frac {1}{3} \, x^{2} e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-x*exp(1/4*x)+2*x-40)/exp(log(2/x)-1/5*exp(1/4*x)+1/10*x),x, algorithm="giac")

[Out]

-1/3*x^2*e^(-1/10*x + 1/5*e^(1/4*x))

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maple [A]  time = 0.28, size = 17, normalized size = 0.41

method	result	size
risch	\(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\)	\(17\)
default	\(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\)	\(23\)
norman	\(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\)	\(23\)
meijerg	\(-\frac {4 \left (1-{\mathrm e}^{-x \ln \relax (2)}\right )}{3 \ln \relax (2)}-\frac {8 \left (1-\frac {\left (2-\frac {x \left (-4 \ln \relax (2)+1\right )}{2}\right ) {\mathrm e}^{\frac {x \left (-4 \ln \relax (2)+1\right )}{4}}}{2}\right )}{15 \left (-4 \ln \relax (2)+1\right )^{2}}+\frac {1-\frac {\left (2+2 x \ln \relax (2)\right ) {\mathrm e}^{-x \ln \relax (2)}}{2}}{15 \ln \relax (2)^{2}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/30*(-x*exp(1/4*x)+2*x-40)/exp(ln(2/x)-1/5*exp(1/4*x)+1/10*x),x,method=_RETURNVERBOSE)

[Out]

-1/3*x^2*exp(1/5*exp(1/4*x)-1/10*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{60} \, \int {\left (x e^{\left (\frac {1}{4} \, x\right )} - 2 \, x + 40\right )} x e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-x*exp(1/4*x)+2*x-40)/exp(log(2/x)-1/5*exp(1/4*x)+1/10*x),x, algorithm="maxima")

[Out]

-1/60*integrate((x*e^(1/4*x) - 2*x + 40)*x*e^(-1/10*x + 1/5*e^(1/4*x)), x)

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mupad [B]  time = 4.14, size = 16, normalized size = 0.39 \begin {gather*} -\frac {x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x/4}}{5}-\frac {x}{10}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(exp(x/4)/5 - x/10 - log(2/x))*((x*exp(x/4))/30 - x/15 + 4/3),x)

[Out]

-(x^2*exp(exp(x/4)/5 - x/10))/3

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sympy [A]  time = 0.17, size = 17, normalized size = 0.41 \begin {gather*} - \frac {x^{2} e^{- \frac {x}{10} + \frac {e^{\frac {x}{4}}}{5}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/30*(-x*exp(1/4*x)+2*x-40)/exp(ln(2/x)-1/5*exp(1/4*x)+1/10*x),x)

[Out]

-x**2*exp(-x/10 + exp(x/4)/5)/3

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