3.33.56 $\int \frac{e^{\frac{x}{4+e^2}}\log (2)+(-4-e^2)\log (2)+(e^{\frac{x}{4+e^2}}(8+2e^2)-8x-2e^{2}x){\log }^2(e^{\frac{x}{4+e^2}}-x)}{(e^{\frac{x}{4+e^2}}(4+e^2)-4x-e^{2}x){\log }^2(e^{\frac{x}{4+e^2}}-x)}dx$

Optimal. Leaf size=34 $$e^{\frac{5}{\log (2)}}+2x-\frac{\log (2)}{\log (e^{\frac{x}{4+e^2}}-x)}$$

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Rubi [A]  time = 0.56, antiderivative size = 26, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, integrand size = 121, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.025, Rules used = {6, 6688, 6686} $$\begin{array}{c}2x-\frac{\log (2)}{\log (e^{\frac{x}{4+e^2}}-x)}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{x}{4+e^2}}\log (2)+(-4-e^2)\log (2)+(e^{\frac{x}{4+e^2}}(8+2e^2)-8x-2e^{2}x){\log }^2(e^{\frac{x}{4+e^2}}-x)}{(e^{\frac{x}{4+e^2}}(4+e^2)+(-4-e^2)x){\log }^2(e^{\frac{x}{4+e^2}}-x)}dx\\ & =\int (2+\frac{(e^{\frac{x}{4+e^2}}-4(1+\frac{e^2}{4}))\log (2)}{(4+e^2)(e^{\frac{x}{4+e^2}}-x){\log }^2(e^{\frac{x}{4+e^2}}-x)})dx\\ & =2x+\frac{\log (2)\int \frac{e^{\frac{x}{4+e^2}}-4(1+\frac{e^2}{4})}{(e^{\frac{x}{4+e^2}}-x){\log }^2(e^{\frac{x}{4+e^2}}-x)}dx}{4+e^2}\\ & =2x-\frac{\log (2)}{\log (e^{\frac{x}{4+e^2}}-x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 26, normalized size = 0.76 $$\begin{array}{c}2x-\frac{\log (2)}{\log (e^{\frac{x}{4+e^2}}-x)}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

2*x - Log[2]/Log[E^(x/(4 + E^2)) - x]

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fricas [A]  time = 0.54, size = 39, normalized size = 1.15 $$\begin{array}{c}\frac{2x\log (-x+e^{\left(\frac{x}{e^2+4}\right)})-\log (2)}{\log (-x+e^{\left(\frac{x}{e^2+4}\right)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+exp(2)))-x)^2+log(2)*exp(x/(4+exp(2)))
+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x/(4+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="fric
as")

[Out]

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

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giac [A]  time = 1.07, size = 39, normalized size = 1.15 $$\begin{array}{c}\frac{2x\log (-x+e^{\left(\frac{x}{e^2+4}\right)})-\log (2)}{\log (-x+e^{\left(\frac{x}{e^2+4}\right)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 25, normalized size = 0.74

method	result	size
risch	$2x-\frac{\ln (2)}{\ln ({\mathrm{e}}^{\frac{x}{4+{\mathrm{e}}^2}}-x)}$	$25$
norman	$\frac{2x\ln ({\mathrm{e}}^{\frac{x}{4+{\mathrm{e}}^2}}-x)-\ln (2)}{\ln ({\mathrm{e}}^{\frac{x}{4+{\mathrm{e}}^2}}-x)}$	$40$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.81, size = 39, normalized size = 1.15 $$\begin{array}{c}\frac{2x\log (-x+e^{\left(\frac{x}{e^2+4}\right)})-\log (2)}{\log (-x+e^{\left(\frac{x}{e^2+4}\right)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(2)+8)*exp(x/(4+exp(2)))-2*exp(2)*x-8*x)*log(exp(x/(4+exp(2)))-x)^2+log(2)*exp(x/(4+exp(2)))
+(-exp(2)-4)*log(2))/((4+exp(2))*exp(x/(4+exp(2)))-exp(2)*x-4*x)/log(exp(x/(4+exp(2)))-x)^2,x, algorithm="maxi
ma")

[Out]

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

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mupad [B]  time = 2.24, size = 24, normalized size = 0.71 $$\begin{array}{c}2x-\frac{\ln (2)}{\ln ({\mathrm{e}}^{\frac{x}{{\mathrm{e}}^2+4}}-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.28, size = 17, normalized size = 0.50 $$\begin{array}{c}2x-\frac{\log (2)}{\log (-x+e^{\frac{x}{4+e^2}})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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