\(\int \frac {e^2 (-2 x^2-x^3)+(-2 x^2-x^3) \log (5)+(e^2 (2 x+x^2)+(2 x+x^2) \log (5)) \log (\frac {x}{2+x})+(16-16 x-8 x^2) \log ^7(x-\log (\frac {x}{2+x}))}{-2 x^2-x^3+(2 x+x^2) \log (\frac {x}{2+x})} \, dx\) [6402]

Optimal result

Integrand size = 117, antiderivative size = 24 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \]

[Out]

x*(ln(5)+exp(2))+ln(-ln(x/(2+x))+x)^8

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 6873, 6820, 6874, 6818} \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\log ^8\left (x-\log \left (\frac {x}{x+2}\right )\right )+x \left (e^2+\log (5)\right ) \]

[In]

Int[(E^2*(-2*x^2 - x^3) + (-2*x^2 - x^3)*Log[5] + (E^2*(2*x + x^2) + (2*x + x^2)*Log[5])*Log[x/(2 + x)] + (16
- 16*x - 8*x^2)*Log[x - Log[x/(2 + x)]]^7)/(-2*x^2 - x^3 + (2*x + x^2)*Log[x/(2 + x)]),x]

[Out]

x*(E^2 + Log[5]) + Log[x - Log[x/(2 + x)]]^8

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 6818

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 6820

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

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-2 x^2-x^3\right ) \left (e^2+\log (5)\right )+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx \\ & = \int \frac {-\left (\left (-2 x^2-x^3\right ) \left (e^2+\log (5)\right )\right )-\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )-\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{x (2+x) \left (x-\log \left (\frac {x}{2+x}\right )\right )} \, dx \\ & = \int \frac {x^2 (2+x) \left (e^2+\log (5)\right )-x (2+x) \left (e^2+\log (5)\right ) \log \left (\frac {x}{2+x}\right )+8 \left (-2+2 x+x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{x (2+x) \left (x-\log \left (\frac {x}{2+x}\right )\right )} \, dx \\ & = \int \left (e^2 \left (1+\frac {\log (5)}{e^2}\right )+\frac {8 \left (-2+2 x+x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{x (2+x) \left (x-\log \left (\frac {x}{2+x}\right )\right )}\right ) \, dx \\ & = x \left (e^2+\log (5)\right )+8 \int \frac {\left (-2+2 x+x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{x (2+x) \left (x-\log \left (\frac {x}{2+x}\right )\right )} \, dx \\ & = x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \]

[In]

Integrate[(E^2*(-2*x^2 - x^3) + (-2*x^2 - x^3)*Log[5] + (E^2*(2*x + x^2) + (2*x + x^2)*Log[5])*Log[x/(2 + x)]
+ (16 - 16*x - 8*x^2)*Log[x - Log[x/(2 + x)]]^7)/(-2*x^2 - x^3 + (2*x + x^2)*Log[x/(2 + x)]),x]

[Out]

x*(E^2 + Log[5]) + Log[x - Log[x/(2 + x)]]^8

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(59\) vs. \(2(23)=46\).

Time = 2.96 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50

method	result	size
default	\(\ln \left (-\frac {\frac {\ln \left (\frac {x}{2+x}\right ) x}{2+x}+\frac {2 x}{2+x}-\ln \left (\frac {x}{2+x}\right )}{\frac {x}{2+x}-1}\right )^{8}+{\mathrm e}^{2} x +x \ln \left (5\right )\)	\(60\)

[In]

int(((-8*x^2-16*x+16)*ln(-ln(x/(2+x))+x)^7+((x^2+2*x)*ln(5)+(x^2+2*x)*exp(2))*ln(x/(2+x))+(-x^3-2*x^2)*ln(5)+(
-x^3-2*x^2)*exp(2))/((x^2+2*x)*ln(x/(2+x))-x^3-2*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(-(ln(x/(2+x))/(2+x)*x+2*x/(2+x)-ln(x/(2+x)))/(x/(2+x)-1))^8+exp(2)*x+x*ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\log \left (x - \log \left (\frac {x}{x + 2}\right )\right )^{8} + x e^{2} + x \log \left (5\right ) \]

[In]

integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^
2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2*x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="fricas")

[Out]

log(x - log(x/(x + 2)))^8 + x*e^2 + x*log(5)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (\log {\left (5 \right )} + e^{2}\right ) + \log {\left (x - \log {\left (\frac {x}{x + 2} \right )} \right )}^{8} \]

[In]

integrate(((-8*x**2-16*x+16)*ln(-ln(x/(2+x))+x)**7+((x**2+2*x)*ln(5)+(x**2+2*x)*exp(2))*ln(x/(2+x))+(-x**3-2*x
**2)*ln(5)+(-x**3-2*x**2)*exp(2))/((x**2+2*x)*ln(x/(2+x))-x**3-2*x**2),x)

[Out]

x*(log(5) + exp(2)) + log(x - log(x/(x + 2)))**8

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\log \left (x + \log \left (x + 2\right ) - \log \left (x\right )\right )^{8} + x {\left (e^{2} + \log \left (5\right )\right )} \]

[In]

integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^
2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2*x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="maxima")

[Out]

log(x + log(x + 2) - log(x))^8 + x*(e^2 + log(5))

Giac [F(-1)]

Timed out. \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\text {Timed out} \]

[In]

integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^
2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2*x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 12.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx={\ln \left (x-\ln \left (\frac {x}{x+2}\right )\right )}^8+x\,\left ({\mathrm {e}}^2+\ln \left (5\right )\right ) \]

[In]

int((exp(2)*(2*x^2 + x^3) + log(5)*(2*x^2 + x^3) + log(x - log(x/(x + 2)))^7*(16*x + 8*x^2 - 16) - log(x/(x +
2))*(exp(2)*(2*x + x^2) + log(5)*(2*x + x^2)))/(2*x^2 + x^3 - log(x/(x + 2))*(2*x + x^2)),x)

[Out]

x*(exp(2) + log(5)) + log(x - log(x/(x + 2)))^8