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

Optimal. Leaf size=24 \[ x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 1.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 6741, 6688, 6742, 6686} \begin {gather*} \log ^8\left (x-\log \left (\frac {x}{x+2}\right )\right )+x \left (e^2+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[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 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]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 24, normalized size = 1.00 \begin {gather*} x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 24, normalized size = 1.00 \begin {gather*} \log \left (x - \log \left (\frac {x}{x + 2}\right )\right )^{8} + x e^{2} + x \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (-8 x^{2}-16 x +16\right ) \ln \left (-\ln \left (\frac {x}{2+x}\right )+x \right )^{7}+\left (\left (x^{2}+2 x \right ) \ln \relax (5)+\left (x^{2}+2 x \right ) {\mathrm e}^{2}\right ) \ln \left (\frac {x}{2+x}\right )+\left (-x^{3}-2 x^{2}\right ) \ln \relax (5)+\left (-x^{3}-2 x^{2}\right ) {\mathrm e}^{2}}{\left (x^{2}+2 x \right ) \ln \left (\frac {x}{2+x}\right )-x^{3}-2 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 21, normalized size = 0.88 \begin {gather*} \log \left (x + \log \left (x + 2\right ) - \log \relax (x)\right )^{8} + x {\left (e^{2} + \log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 4.53, size = 23, normalized size = 0.96 \begin {gather*} {\ln \left (x-\ln \left (\frac {x}{x+2}\right )\right )}^8+x\,\left ({\mathrm {e}}^2+\ln \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.46, size = 19, normalized size = 0.79 \begin {gather*} x \left (\log {\relax (5 )} + e^{2}\right ) + \log {\left (x - \log {\left (\frac {x}{x + 2} \right )} \right )}^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________