3.19.9 \(\int \frac {e^{-e^{-5+e^5-x}} (20 x-4 x^2-x^3) (-20+8 x+3 x^2+e^{-5+e^5-x} (-20 x+4 x^2+x^3))}{-20 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=30 \[ e^{-e^{-5+e^5-x}} x \left (4+4 (4-x)-x^2\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1586, 2288} \begin {gather*} e^{-e^{-x+e^5-5}} \left (-x^3-4 x^2+20 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((20*x - 4*x^2 - x^3)*(-20 + 8*x + 3*x^2 + E^(-5 + E^5 - x)*(-20*x + 4*x^2 + x^3)))/(E^E^(-5 + E^5 - x)*(-
20*x + 4*x^2 + x^3)),x]

[Out]

(20*x - 4*x^2 - x^3)/E^E^(-5 + E^5 - x)

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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} &=-\int e^{-e^{-5+e^5-x}} \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right ) \, dx\\ &=e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 25, normalized size = 0.83 \begin {gather*} -e^{-e^{-5+e^5-x}} x \left (-20+4 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((20*x - 4*x^2 - x^3)*(-20 + 8*x + 3*x^2 + E^(-5 + E^5 - x)*(-20*x + 4*x^2 + x^3)))/(E^E^(-5 + E^5 -
x)*(-20*x + 4*x^2 + x^3)),x]

[Out]

-((x*(-20 + 4*x + x^2))/E^E^(-5 + E^5 - x))

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fricas [A]  time = 0.82, size = 27, normalized size = 0.90 \begin {gather*} e^{\left (-e^{\left (-x + e^{5} - 5\right )} + \log \left (-x^{3} - 4 \, x^{2} + 20 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2
-20*x),x, algorithm="fricas")

[Out]

e^(-e^(-x + e^5 - 5) + log(-x^3 - 4*x^2 + 20*x))

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giac [B]  time = 0.28, size = 77, normalized size = 2.57 \begin {gather*} -{\left (x^{3} e^{\left (-x + e^{5} - e^{\left (-x + e^{5} - 5\right )} - 5\right )} + 4 \, x^{2} e^{\left (-x + e^{5} - e^{\left (-x + e^{5} - 5\right )} - 5\right )} - 20 \, x e^{\left (-x + e^{5} - e^{\left (-x + e^{5} - 5\right )} - 5\right )}\right )} e^{\left (x - e^{5} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2
-20*x),x, algorithm="giac")

[Out]

-(x^3*e^(-x + e^5 - e^(-x + e^5 - 5) - 5) + 4*x^2*e^(-x + e^5 - e^(-x + e^5 - 5) - 5) - 20*x*e^(-x + e^5 - e^(
-x + e^5 - 5) - 5))*e^(x - e^5 + 5)

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maple [C]  time = 0.12, size = 151, normalized size = 5.03

method	result	size
risch	\(-x \left (x^{2}+4 x -20\right ) {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )^{2} \mathrm {csgn}\left (i x \right )}{2}+\frac {i \pi \mathrm {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+4 x -20\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x^{2}+4 x -20\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x^{2}+4 x -20\right )\right )}{2}-i \pi \mathrm {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )^{2}-{\mathrm e}^{{\mathrm e}^{5}-x -5}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(ln(-x^3-4*x^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),
x,method=_RETURNVERBOSE)

[Out]

-x*(x^2+4*x-20)*exp(1/2*I*Pi*csgn(I*x*(x^2+4*x-20))^3+1/2*I*Pi*csgn(I*x*(x^2+4*x-20))^2*csgn(I*x)+1/2*I*Pi*csg
n(I*x*(x^2+4*x-20))^2*csgn(I*(x^2+4*x-20))-1/2*I*Pi*csgn(I*x*(x^2+4*x-20))*csgn(I*x)*csgn(I*(x^2+4*x-20))-I*Pi
*csgn(I*x*(x^2+4*x-20))^2-exp(exp(5)-x-5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -{\left (3 \, x^{2} + {\left (x^{3} + 4 \, x^{2} - 20 \, x\right )} e^{\left (-x + e^{5} - 5\right )} + 8 \, x - 20\right )} e^{\left (-e^{\left (-x + e^{5} - 5\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2
-20*x),x, algorithm="maxima")

[Out]

integrate(-(3*x^2 + (x^3 + 4*x^2 - 20*x)*e^(-x + e^5 - 5) + 8*x - 20)*e^(-e^(-x + e^5 - 5)), x)

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mupad [B]  time = 1.43, size = 23, normalized size = 0.77 \begin {gather*} -x\,{\mathrm {e}}^{-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^5}}\,\left (x^2+4\,x-20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(20*x - 4*x^2 - x^3) - exp(exp(5) - x - 5))*(8*x + exp(exp(5) - x - 5)*(4*x^2 - 20*x + x^3) + 3*x^
2 - 20))/(4*x^2 - 20*x + x^3),x)

[Out]

-x*exp(-exp(-x)*exp(-5)*exp(exp(5)))*(4*x + x^2 - 20)

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sympy [A]  time = 0.22, size = 20, normalized size = 0.67 \begin {gather*} \left (- x^{3} - 4 x^{2} + 20 x\right ) e^{- e^{- x - 5 + e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+4*x**2-20*x)*exp(exp(5)-x-5)+3*x**2+8*x-20)*exp(ln(-x**3-4*x**2+20*x)-exp(exp(5)-x-5))/(x**3+
4*x**2-20*x),x)

[Out]

(-x**3 - 4*x**2 + 20*x)*exp(-exp(-x - 5 + exp(5)))

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