3.50.79 \(\int e^{-e^3-e^x x+2 x^3-2 x^2 \log (x)} (-1-10 x+32 x^2-6 x^3+e^x (-5-4 x+x^2)+(-20 x+4 x^2) \log (x)) \, dx\)

Optimal. Leaf size=32 \[ 8-e^{-x \left (e^x+\frac {e^3}{x}+2 x (-x+\log (x))\right )} (-5+x) \]

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Rubi [B]  time = 1.36, antiderivative size = 92, normalized size of antiderivative = 2.88, number of steps used = 1, number of rules used = 1, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2288} \begin {gather*} \frac {e^{2 x^3-e^x x-e^3} x^{-2 x^2} \left (6 x^3-32 x^2+e^x \left (-x^2+4 x+5\right )+4 \left (5 x-x^2\right ) \log (x)+10 x\right )}{-6 x^2+e^x x+2 x+e^x+4 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-E^3 - E^x*x + 2*x^3 - 2*x^2*Log[x])*(-1 - 10*x + 32*x^2 - 6*x^3 + E^x*(-5 - 4*x + x^2) + (-20*x + 4*x^
2)*Log[x]),x]

[Out]

(E^(-E^3 - E^x*x + 2*x^3)*(10*x - 32*x^2 + 6*x^3 + E^x*(5 + 4*x - x^2) + 4*(5*x - x^2)*Log[x]))/(x^(2*x^2)*(E^
x + 2*x + E^x*x - 6*x^2 + 4*x*Log[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 {e^{-e^3-e^x x+2 x^3} x^{-2 x^2} \left (10 x-32 x^2+6 x^3+e^x \left (5+4 x-x^2\right )+4 \left (5 x-x^2\right ) \log (x)\right )}{e^x+2 x+e^x x-6 x^2+4 x \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.12, size = 31, normalized size = 0.97 \begin {gather*} -e^{-e^3-e^x x+2 x^3} (-5+x) x^{-2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-E^3 - E^x*x + 2*x^3 - 2*x^2*Log[x])*(-1 - 10*x + 32*x^2 - 6*x^3 + E^x*(-5 - 4*x + x^2) + (-20*x
+ 4*x^2)*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.69, size = 28, normalized size = 0.88 \begin {gather*} -{\left (x - 5\right )} e^{\left (2 \, x^{3} - 2 \, x^{2} \log \relax (x) - x e^{x} - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-20*x)*log(x)+(x^2-4*x-5)*exp(x)-6*x^3+32*x^2-10*x-1)/exp(2*x^2*log(x)+exp(x)*x+exp(3)-2*x^3)
,x, algorithm="fricas")

[Out]

-(x - 5)*e^(2*x^3 - 2*x^2*log(x) - x*e^x - e^3)

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giac [A]  time = 0.24, size = 52, normalized size = 1.62 \begin {gather*} -x e^{\left (2 \, x^{3} - 2 \, x^{2} \log \relax (x) - x e^{x} - e^{3}\right )} + 5 \, e^{\left (2 \, x^{3} - 2 \, x^{2} \log \relax (x) - x e^{x} - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-20*x)*log(x)+(x^2-4*x-5)*exp(x)-6*x^3+32*x^2-10*x-1)/exp(2*x^2*log(x)+exp(x)*x+exp(3)-2*x^3)
,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 32, normalized size = 1.00

method	result	size
risch	\(\left (5-x \right ) x^{-2 x^{2}} {\mathrm e}^{-{\mathrm e}^{x} x -{\mathrm e}^{3}+2 x^{3}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2-20*x)*ln(x)+(x^2-4*x-5)*exp(x)-6*x^3+32*x^2-10*x-1)/exp(2*x^2*ln(x)+exp(x)*x+exp(3)-2*x^3),x,metho
d=_RETURNVERBOSE)

[Out]

(5-x)/(x^(2*x^2))*exp(-exp(x)*x-exp(3)+2*x^3)

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maxima [A]  time = 0.45, size = 28, normalized size = 0.88 \begin {gather*} -{\left (x - 5\right )} e^{\left (2 \, x^{3} - 2 \, x^{2} \log \relax (x) - x e^{x} - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-20*x)*log(x)+(x^2-4*x-5)*exp(x)-6*x^3+32*x^2-10*x-1)/exp(2*x^2*log(x)+exp(x)*x+exp(3)-2*x^3)
,x, algorithm="maxima")

[Out]

-(x - 5)*e^(2*x^3 - 2*x^2*log(x) - x*e^x - e^3)

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mupad [B]  time = 3.36, size = 31, normalized size = 0.97 \begin {gather*} -\frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^{2\,x^3}\,\left (x-5\right )}{x^{2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(2*x^3 - 2*x^2*log(x) - x*exp(x) - exp(3))*(10*x + exp(x)*(4*x - x^2 + 5) + log(x)*(20*x - 4*x^2) - 32
*x^2 + 6*x^3 + 1),x)

[Out]

-(exp(-x*exp(x))*exp(-exp(3))*exp(2*x^3)*(x - 5))/x^(2*x^2)

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sympy [A]  time = 0.45, size = 26, normalized size = 0.81 \begin {gather*} \left (5 - x\right ) e^{2 x^{3} - 2 x^{2} \log {\relax (x )} - x e^{x} - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2-20*x)*ln(x)+(x**2-4*x-5)*exp(x)-6*x**3+32*x**2-10*x-1)/exp(2*x**2*ln(x)+exp(x)*x+exp(3)-2*x
**3),x)

[Out]

(5 - x)*exp(2*x**3 - 2*x**2*log(x) - x*exp(x) - exp(3))

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