3.69.1 $\int \frac{e^5(5-2x)-e^5\log (4)+e^{\frac{e^e-2x^2}{e^5}}(-e^5+4x^2+4x\log (4))}{e^5}dx$

Optimal. Leaf size=27 $$(5-e^{\frac{e^e-2x^2}{e^5}}-x)(x+\log (4))$$

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Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.67, number of steps used = 3, number of rules used = 2, integrand size = 53, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.038, Rules used = {12, 2288} $$\begin{array}{c}-\frac{e^{\frac{e^e-2x^2}{e^5}}(x^2+x\log (4))}{x}-\frac{1}{4}(5-2x{)}^2-x\log (4)\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(5 - 2*x)^2 - x*Log[4] - (E^((E^E - 2*x^2)/E^5)*(x^2 + x*Log[4]))/x

Rule 12

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int (e^5(5-2x)-e^5\log (4)+e^{\frac{e^e-2x^2}{e^5}}(-e^5+4x^2+4x\log (4)))dx}{e^5}\\ & =-\frac{1}{4}(5-2x{)}^2-x\log (4)+\frac{\int e^{\frac{e^e-2x^2}{e^5}}(-e^5+4x^2+4x\log (4))dx}{e^5}\\ & =-\frac{1}{4}(5-2x{)}^2-x\log (4)-\frac{e^{\frac{e^e-2x^2}{e^5}}(x^2+x\log (4))}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 34, normalized size = 1.26 $$\begin{array}{c}-x(-5+x+\log (4))-\frac{1}{4}{e}^{\frac{e^e-2x^2}{e^5}}(4x+\log (256))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(-5 + x + Log[4])) - (E^((E^E - 2*x^2)/E^5)*(4*x + Log[256]))/4

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fricas [A]  time = 0.51, size = 38, normalized size = 1.41 $$\begin{array}{c}-x^2-(x+2\log (2))e^{(-(2x^2-e^e)e^{(-5)})}-2x\log (2)+5x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(-2*x+5)*exp(5))/exp(5),x
, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 50, normalized size = 1.85 $$\begin{array}{c}-(2x{e}^5\log (2)+(x^2-5x)e^5+(x{e}^5+2e^5\log (2))e^{(-(2x^2-e^e)e^{(-5)})})e^{(-5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(-2*x+5)*exp(5))/exp(5),x
, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 44, normalized size = 1.63

method	result	size
risch	$-2x\ln (2)-x^2+5x+(-2{\mathrm{e}}^5\ln (2)-x{\mathrm{e}}^5){\mathrm{e}}^{-2x^2{\mathrm{e}}^{-5}+{\mathrm{e}}^{\mathrm{e}}{\mathrm{e}}^{-5}-5}$	$44$
norman	$(-2\ln (2)+5)x-x^2-x{\mathrm{e}}^{({\mathrm{e}}^{\mathrm{e}}-2x^2){\mathrm{e}}^{-5}}-2\ln (2){\mathrm{e}}^{({\mathrm{e}}^{\mathrm{e}}-2x^2){\mathrm{e}}^{-5}}$	$52$
default	${\mathrm{e}}^{-5}({\mathrm{e}}^5(-x^2+5x)-{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{e}}{\mathrm{e}}^{-5}}{\mathrm{e}}^{5}x{\mathrm{e}}^{-2x^2{\mathrm{e}}^{-5}}-2{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{e}}{\mathrm{e}}^{-5}}\ln (2){\mathrm{e}}^{-2x^2{\mathrm{e}}^{-5}}{\mathrm{e}}^5-2x{\mathrm{e}}^5\ln (2))$	$76$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x*ln(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*ln(2)+(-2*x+5)*exp(5))/exp(5),x,method=
_RETURNVERBOSE)

[Out]

-2*x*ln(2)-x^2+5*x+(-2*exp(5)*ln(2)-x*exp(5))*exp(-2*x^2*exp(-5)+exp(exp(1))*exp(-5)-5)

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maxima [A]  time = 0.45, size = 54, normalized size = 2.00 $$\begin{array}{c}-(2x{e}^5\log (2)+(x^2-5x)e^5+(x{e}^{(e^{(e-5)}+5)}+2e^{(e^{(e-5)}+5)}\log (2))e^{(-2x^2{e}^{(-5)})})e^{(-5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(-2*x+5)*exp(5))/exp(5),x
, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.18, size = 51, normalized size = 1.89 $$\begin{array}{c}5x-2x\ln (2)-2{\mathrm{e}}^{{\mathrm{e}}^{-5}{\mathrm{e}}^{\mathrm{e}}-2x^2{\mathrm{e}}^{-5}}\ln (2)-x{\mathrm{e}}^{{\mathrm{e}}^{-5}{\mathrm{e}}^{\mathrm{e}}-2x^2{\mathrm{e}}^{-5}}-x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-5)*(2*exp(5)*log(2) - exp(exp(-5)*(exp(exp(1)) - 2*x^2))*(8*x*log(2) - exp(5) + 4*x^2) + exp(5)*(2*x
- 5)),x)

[Out]

5*x - 2*x*log(2) - 2*exp(exp(-5)*exp(exp(1)) - 2*x^2*exp(-5))*log(2) - x*exp(exp(-5)*exp(exp(1)) - 2*x^2*exp(-
5)) - x^2

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sympy [A]  time = 0.14, size = 34, normalized size = 1.26 $$\begin{array}{c}-x^2+x(5-2\log (2))+(-x-2\log (2))e^{\frac{-2x^2+e^e}{e^5}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*ln(2)-exp(5)+4*x**2)*exp((exp(exp(1))-2*x**2)/exp(5))-2*exp(5)*ln(2)+(-2*x+5)*exp(5))/exp(5),x
)

[Out]

-x**2 + x*(5 - 2*log(2)) + (-x - 2*log(2))*exp((-2*x**2 + exp(E))*exp(-5))

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