3.88.4 \(\int \frac {e^{4-e^4+e^{x/2}-x^2} (10 x+5 e^{x/2} x^2-20 x^3)}{x} \, dx\)

Optimal. Leaf size=24 \[ 10 e^{4-e^4+e^{x/2}-x^2} x \]

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Rubi [B]  time = 0.12, antiderivative size = 56, normalized size of antiderivative = 2.33, number of steps used = 1, number of rules used = 1, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2288} \begin {gather*} \frac {10 e^{-x^2+e^{x/2}-e^4+4} \left (e^{x/2} x^2-4 x^3\right )}{\left (e^{x/2}-4 x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4 - E^4 + E^(x/2) - x^2)*(10*x + 5*E^(x/2)*x^2 - 20*x^3))/x,x]

[Out]

(10*E^(4 - E^4 + E^(x/2) - x^2)*(E^(x/2)*x^2 - 4*x^3))/((E^(x/2) - 4*x)*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 {10 e^{4-e^4+e^{x/2}-x^2} \left (e^{x/2} x^2-4 x^3\right )}{\left (e^{x/2}-4 x\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 1.00 \begin {gather*} 10 e^{4-e^4+e^{x/2}-x^2} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 - E^4 + E^(x/2) - x^2)*(10*x + 5*E^(x/2)*x^2 - 20*x^3))/x,x]

[Out]

10*E^(4 - E^4 + E^(x/2) - x^2)*x

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fricas [A]  time = 0.71, size = 25, normalized size = 1.04 \begin {gather*} 10 \, x^{2} e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} - \log \relax (x) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(1/2*x)-20*x^3+10*x)/exp(exp(4)-4)/exp(log(x)-exp(1/2*x)+x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 19, normalized size = 0.79 \begin {gather*} 10 \, x e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(1/2*x)-20*x^3+10*x)/exp(exp(4)-4)/exp(log(x)-exp(1/2*x)+x^2),x, algorithm="giac")

[Out]

10*x*e^(-x^2 - e^4 + e^(1/2*x) + 4)

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maple [A]  time = 0.09, size = 20, normalized size = 0.83




method result size



risch \(10 x \,{\mathrm e}^{4-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{2}}-x^{2}}\) \(20\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2*exp(1/2*x)-20*x^3+10*x)/exp(exp(4)-4)/exp(ln(x)-exp(1/2*x)+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 19, normalized size = 0.79 \begin {gather*} 10 \, x e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(1/2*x)-20*x^3+10*x)/exp(exp(4)-4)/exp(log(x)-exp(1/2*x)+x^2),x, algorithm="maxima")

[Out]

10*x*e^(-x^2 - e^4 + e^(1/2*x) + 4)

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mupad [B]  time = 5.51, size = 21, normalized size = 0.88 \begin {gather*} 10\,x\,{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^4\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{x/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(x/2) - log(x) - x^2)*exp(4 - exp(4))*(10*x + 5*x^2*exp(x/2) - 20*x^3),x)

[Out]

10*x*exp(-exp(4))*exp(4)*exp(-x^2)*exp(exp(x/2))

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sympy [A]  time = 0.21, size = 20, normalized size = 0.83 \begin {gather*} \frac {10 x e^{4} e^{- x^{2} + e^{\frac {x}{2}}}}{e^{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2*exp(1/2*x)-20*x**3+10*x)/exp(exp(4)-4)/exp(ln(x)-exp(1/2*x)+x**2),x)

[Out]

10*x*exp(4)*exp(-x**2 + exp(x/2))*exp(-exp(4))

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