3.69.66 $\int \frac{e^{\frac{1+e^3}{4+e^x-x}}(e^x(-1-e^3)\log (4)+(1+e^3)\log (4))}{32+2e^{2x}+e^x(16-4x)-16x+2x^2}dx$

Optimal. Leaf size=24 $$\frac{1}{2}{e}^{\frac{1+e^3}{4+e^x-x}}\log (4)$$

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Rubi [A]  time = 1.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 69, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.043, Rules used = {6688, 12, 6706} $$\begin{array}{c}\frac{1}{2}{e}^{\frac{1+e^3}{-x+e^x+4}}\log (4)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^((1 + E^3)/(4 + E^x - x))*(E^x*(-1 - E^3)*Log[4] + (1 + E^3)*Log[4]))/(32 + 2*E^(2*x) + E^x*(16 - 4*x)
- 16*x + 2*x^2),x]

[Out]

(E^((1 + E^3)/(4 + E^x - x))*Log[4])/2

Rule 12

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{1+e^3}{4+e^x-x}}(1+e^3)(1-e^x)\log (4)}{2{(4+e^x-x)}^2}dx\\ & =\frac{1}{2}((1+e^3)\log (4))\int \frac{e^{\frac{1+e^3}{4+e^x-x}}(1-e^x)}{{(4+e^x-x)}^2}dx\\ & =\frac{1}{2}{e}^{\frac{1+e^3}{4+e^x-x}}\log (4)\end{array}\end{array}$$

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Mathematica [A]  time = 0.32, size = 24, normalized size = 1.00 $$\begin{array}{c}\frac{1}{2}{e}^{\frac{1+e^3}{4+e^x-x}}\log (4)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 + E^3)/(4 + E^x - x))*(E^x*(-1 - E^3)*Log[4] + (1 + E^3)*Log[4]))/(32 + 2*E^(2*x) + E^x*(16 -
4*x) - 16*x + 2*x^2),x]

[Out]

(E^((1 + E^3)/(4 + E^x - x))*Log[4])/2

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fricas [A]  time = 0.71, size = 19, normalized size = 0.79 $$\begin{array}{c}{e}^{(-\frac{e^3+1}{x-e^x-4})}\log (2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="fricas")

[Out]

e^(-(e^3 + 1)/(x - e^x - 4))*log(2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int -\frac{((e^3+1)e^x\log (2)-(e^3+1)\log (2))e^{(-\frac{e^3+1}{x-e^x-4})}}{x^2-2(x-4)e^x-8x+e^{\left(2x\right)}+16}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="giac")

[Out]

integrate(-((e^3 + 1)*e^x*log(2) - (e^3 + 1)*log(2))*e^(-(e^3 + 1)/(x - e^x - 4))/(x^2 - 2*(x - 4)*e^x - 8*x +
e^(2*x) + 16), x)

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maple [A]  time = 0.17, size = 19, normalized size = 0.79

method	result	size
risch	$\ln (2){\mathrm{e}}^{\frac{{\mathrm{e}}^3+1}{{\mathrm{e}}^x-x+4}}$	$19$
norman	$\frac{x\ln (2){\mathrm{e}}^{\frac{{\mathrm{e}}^3+1}{{\mathrm{e}}^x-x+4}}-4\ln (2){\mathrm{e}}^{\frac{{\mathrm{e}}^3+1}{{\mathrm{e}}^x-x+4}}-{\mathrm{e}}^x\ln (2){\mathrm{e}}^{\frac{{\mathrm{e}}^3+1}{{\mathrm{e}}^x-x+4}}}{x-{\mathrm{e}}^x-4}$	$71$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-exp(3)-1)*ln(2)*exp(x)+2*(exp(3)+1)*ln(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)*exp(x)+
2*x^2-16*x+32),x,method=_RETURNVERBOSE)

[Out]

ln(2)*exp((exp(3)+1)/(exp(x)-x+4))

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maxima [A]  time = 0.54, size = 29, normalized size = 1.21 $$\begin{array}{c}{e}^{(-\frac{e^3}{x-e^x-4}-\frac{1}{x-e^x-4})}\log (2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-exp(3)-1)*log(2)*exp(x)+2*(exp(3)+1)*log(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)^2+(-4*x+16)
*exp(x)+2*x^2-16*x+32),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.42, size = 18, normalized size = 0.75 $$\begin{array}{c}{\mathrm{e}}^{\frac{{\mathrm{e}}^3+1}{{\mathrm{e}}^x-x+4}}\ln (2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(3) + 1)/(exp(x) - x + 4))*(2*log(2)*(exp(3) + 1) - 2*exp(x)*log(2)*(exp(3) + 1)))/(2*exp(2*x) -
16*x - exp(x)*(4*x - 16) + 2*x^2 + 32),x)

[Out]

exp((exp(3) + 1)/(exp(x) - x + 4))*log(2)

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sympy [A]  time = 0.37, size = 15, normalized size = 0.62 $$\begin{array}{c}{e}^{\frac{1+e^3}{-x+e^x+4}}\log (2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-exp(3)-1)*ln(2)*exp(x)+2*(exp(3)+1)*ln(2))*exp((exp(3)+1)/(exp(x)-x+4))/(2*exp(x)**2+(-4*x+16)*
exp(x)+2*x**2-16*x+32),x)

[Out]

exp((1 + exp(3))/(-x + exp(x) + 4))*log(2)

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