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

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

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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 {gather*} x+\log \left (x+\log \left (\frac {5 x (x+1)}{5 x+4}-e^4\right )+5\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 (6+x) (4+5 x)^2-5 \left (4+32 x+63 x^2+39 x^3+5 x^4\right )+\left (e^4 (4+5 x)^2-5 x \left (4+9 x+5 x^2\right )\right ) \log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )} \, dx\\ &=\int \left (1+\frac {-4 \left (5-4 e^4\right )-20 \left (3-2 e^4\right ) x-5 \left (14-5 e^4\right ) x^2-25 x^3}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )}\right ) \, dx\\ &=x+\int \frac {-4 \left (5-4 e^4\right )-20 \left (3-2 e^4\right ) x-5 \left (14-5 e^4\right ) x^2-25 x^3}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )} \, dx\\ &=x+\log \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )\\ \end {aligned} \end {gather*}

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

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 {gather*} x + \log \left (x + \log \left (\frac {5 \, x^{2} - {\left (5 \, x + 4\right )} e^{4} + 5 \, x}{5 \, x + 4}\right ) + 5\right ) \end {gather*}

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 {gather*} x + \log \left (x + \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x - 4 \, e^{4}}{5 \, x + 4}\right ) + 5\right ) \end {gather*}

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

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 {gather*} x + \log \left (x + \log \left (5 \, x^{2} - 5 \, x {\left (e^{4} - 1\right )} - 4 \, e^{4}\right ) - \log \left (5 \, x + 4\right ) + 5\right ) \end {gather*}

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 {gather*} \text {Hanged} \end {gather*}

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 {gather*} x + \log {\left (x + \log {\left (\frac {5 x^{2} + 5 x + \left (- 5 x - 4\right ) e^{4}}{5 x + 4} \right )} + 5 \right )} \end {gather*}

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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