3.69.10 $\int \frac{-20-160x-315x^2-195x^3-25x^4+e^4(96+256x+190x^2+25x^3)+(-20x-45x^2-25x^3+e^4(16+40x+25x^2))\log (\frac{e^4(-4-5x)+5x+5x^2}{4+5x})}{-100x-245x^2-170x^3-25x^4+e^4(80+216x+165x^2+25x^3)+(-20x-45x^2-25x^3+e^4(16+40x+25x^2))\log (\frac{e^4(-4-5x)+5x+5x^2}{4+5x})}dx$

Optimal. Leaf size=23 $$x+\log (5+x+\log (-e^4+x+\frac{x}{4+5x}))$$

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Rubi [A]  time = 1.07, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, integrand size = 192, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.016, Rules used = {6688, 6728, 6684} $$\begin{array}{c}x+\log (x+\log (\frac{5x(x+1)}{5x+4}-e^4)+5)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-20 - 160*x - 315*x^2 - 195*x^3 - 25*x^4 + E^4*(96 + 256*x + 190*x^2 + 25*x^3) + (-20*x - 45*x^2 - 25*x^3
+ E^4*(16 + 40*x + 25*x^2))*Log[(E^4*(-4 - 5*x) + 5*x + 5*x^2)/(4 + 5*x)])/(-100*x - 245*x^2 - 170*x^3 - 25*x
^4 + E^4*(80 + 216*x + 165*x^2 + 25*x^3) + (-20*x - 45*x^2 - 25*x^3 + E^4*(16 + 40*x + 25*x^2))*Log[(E^4*(-4 -
5*x) + 5*x + 5*x^2)/(4 + 5*x)]),x]

[Out]

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^4(6+x)(4+5x{)}^2-5(4+32x+63x^2+39x^3+5x^4)+(e^4(4+5x{)}^2-5x(4+9x+5x^2))\log (-e^4+\frac{5x(1+x)}{4+5x})}{(4+5x)(4e^4-5(1-e^4)x-5x^2)(5+x+\log (-e^4+\frac{5x(1+x)}{4+5x}))}dx\\ & =\int (1+\frac{-4(5-4e^4)-20(3-2e^4)x-5(14-5e^4)x^2-25x^3}{(4+5x)(4e^4-5(1-e^4)x-5x^2)(5+x+\log (-e^4+\frac{5x(1+x)}{4+5x}))})dx\\ & =x+\int \frac{-4(5-4e^4)-20(3-2e^4)x-5(14-5e^4)x^2-25x^3}{(4+5x)(4e^4-5(1-e^4)x-5x^2)(5+x+\log (-e^4+\frac{5x(1+x)}{4+5x}))}dx\\ & =x+\log (5+x+\log (-e^4+\frac{5x(1+x)}{4+5x}))\end{array}\end{array}$$

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Mathematica [A]  time = 0.15, size = 26, normalized size = 1.13 $$\begin{array}{c}x+\log (5+x+\log (-e^4+\frac{5x(1+x)}{4+5x}))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 160*x - 315*x^2 - 195*x^3 - 25*x^4 + E^4*(96 + 256*x + 190*x^2 + 25*x^3) + (-20*x - 45*x^2 -
25*x^3 + E^4*(16 + 40*x + 25*x^2))*Log[(E^4*(-4 - 5*x) + 5*x + 5*x^2)/(4 + 5*x)])/(-100*x - 245*x^2 - 170*x^3
- 25*x^4 + E^4*(80 + 216*x + 165*x^2 + 25*x^3) + (-20*x - 45*x^2 - 25*x^3 + E^4*(16 + 40*x + 25*x^2))*Log[(E^4
*(-4 - 5*x) + 5*x + 5*x^2)/(4 + 5*x)]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 33, normalized size = 1.43 $$\begin{array}{c}x+\log (x+\log \left(\frac{5x^2-(5x+4)e^4+5x}{5x+4}\right)+5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+190*x
^2+256*x+96)*exp(4)-25*x^4-195*x^3-315*x^2-160*x-20)/(((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-
4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+165*x^2+216*x+80)*exp(4)-25*x^4-170*x^3-245*x^2-100*x),x, algorithm="fri
cas")

[Out]

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

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giac [A]  time = 0.30, size = 33, normalized size = 1.43 $$\begin{array}{c}x+\log (x+\log \left(\frac{5x^2-5x{e}^4+5x-4e^4}{5x+4}\right)+5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+190*x
^2+256*x+96)*exp(4)-25*x^4-195*x^3-315*x^2-160*x-20)/(((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-
4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+165*x^2+216*x+80)*exp(4)-25*x^4-170*x^3-245*x^2-100*x),x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.18, size = 33, normalized size = 1.43

method	result	size
norman	$x+\ln (x+\ln \left(\frac{(-5x-4){\mathrm{e}}^4+5x^2+5x}{4+5x}\right)+5)$	$33$
risch	$x+\ln (x+\ln \left(\frac{(-5x-4){\mathrm{e}}^4+5x^2+5x}{4+5x}\right)+5)$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*ln(((-5*x-4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+190*x^2+256*
x+96)*exp(4)-25*x^4-195*x^3-315*x^2-160*x-20)/(((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*ln(((-5*x-4)*exp(4
)+5*x^2+5*x)/(4+5*x))+(25*x^3+165*x^2+216*x+80)*exp(4)-25*x^4-170*x^3-245*x^2-100*x),x,method=_RETURNVERBOSE)

[Out]

x+ln(x+ln(((-5*x-4)*exp(4)+5*x^2+5*x)/(4+5*x))+5)

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maxima [A]  time = 0.45, size = 32, normalized size = 1.39 $$\begin{array}{c}x+\log (x+\log (5x^2-5x(e^4-1)-4e^4)-\log (5x+4)+5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+190*x
^2+256*x+96)*exp(4)-25*x^4-195*x^3-315*x^2-160*x-20)/(((25*x^2+40*x+16)*exp(4)-25*x^3-45*x^2-20*x)*log(((-5*x-
4)*exp(4)+5*x^2+5*x)/(4+5*x))+(25*x^3+165*x^2+216*x+80)*exp(4)-25*x^4-170*x^3-245*x^2-100*x),x, algorithm="max
ima")

[Out]

x + log(x + log(5*x^2 - 5*x*(e^4 - 1) - 4*e^4) - log(5*x + 4) + 5)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\text{Hanged}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((160*x + log((5*x + 5*x^2 - exp(4)*(5*x + 4))/(5*x + 4))*(20*x - exp(4)*(40*x + 25*x^2 + 16) + 45*x^2 + 25
*x^3) - exp(4)*(256*x + 190*x^2 + 25*x^3 + 96) + 315*x^2 + 195*x^3 + 25*x^4 + 20)/(100*x + log((5*x + 5*x^2 -
exp(4)*(5*x + 4))/(5*x + 4))*(20*x - exp(4)*(40*x + 25*x^2 + 16) + 45*x^2 + 25*x^3) - exp(4)*(216*x + 165*x^2
+ 25*x^3 + 80) + 245*x^2 + 170*x^3 + 25*x^4),x)

[Out]

\text{Hanged}

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sympy [A]  time = 0.70, size = 31, normalized size = 1.35 $$\begin{array}{c}x+\log (x+\log \left(\frac{5x^2+5x+(-5x-4)e^4}{5x+4}\right)+5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((25*x**2+40*x+16)*exp(4)-25*x**3-45*x**2-20*x)*ln(((-5*x-4)*exp(4)+5*x**2+5*x)/(4+5*x))+(25*x**3+1
90*x**2+256*x+96)*exp(4)-25*x**4-195*x**3-315*x**2-160*x-20)/(((25*x**2+40*x+16)*exp(4)-25*x**3-45*x**2-20*x)*
ln(((-5*x-4)*exp(4)+5*x**2+5*x)/(4+5*x))+(25*x**3+165*x**2+216*x+80)*exp(4)-25*x**4-170*x**3-245*x**2-100*x),x
)

[Out]

x + log(x + log((5*x**2 + 5*x + (-5*x - 4)*exp(4))/(5*x + 4)) + 5)

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