3.2.8 \(\int \frac {-x+x^2+e^5 (-2+2 x)+(-x+x^2+e^5 (-2+2 x)) \log (x^2-2 x^3+x^4)+e^5 (-4+8 x) \log (\frac {1}{3} (1+\log (x^2-2 x^3+x^4)))}{-x+x^2+(-x+x^2) \log (x^2-2 x^3+x^4)} \, dx\)

Optimal. Leaf size=30 \[ x+e^5 \left (\log \left (x^2\right )+\log ^2\left (\frac {1}{3} \left (1+\log \left (\left (-x+x^2\right )^2\right )\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.80, antiderivative size = 34, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6741, 6742, 43, 6684, 6686} \begin {gather*} e^5 \log ^2\left (\frac {1}{3} \left (\log \left ((1-x)^2 x^2\right )+1\right )\right )+x+2 e^5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + x^2 + E^5*(-2 + 2*x) + (-x + x^2 + E^5*(-2 + 2*x))*Log[x^2 - 2*x^3 + x^4] + E^5*(-4 + 8*x)*Log[(1 +
Log[x^2 - 2*x^3 + x^4])/3])/(-x + x^2 + (-x + x^2)*Log[x^2 - 2*x^3 + x^4]),x]

[Out]

x + 2*E^5*Log[x] + E^5*Log[(1 + Log[(1 - x)^2*x^2])/3]^2

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6684

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

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 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 {x-x^2-e^5 (-2+2 x)-\left (-x+x^2+e^5 (-2+2 x)\right ) \log \left (x^2-2 x^3+x^4\right )-e^5 (-4+8 x) \log \left (\frac {1}{3} \left (1+\log \left (x^2-2 x^3+x^4\right )\right )\right )}{(1-x) x \left (1+\log \left ((-1+x)^2 x^2\right )\right )} \, dx\\ &=\int \left (\frac {2 e^5+x}{x}+\frac {4 e^5 (-1+2 x) \log \left (\frac {1}{3} \left (1+\log \left ((-1+x)^2 x^2\right )\right )\right )}{(-1+x) x \left (1+\log \left ((-1+x)^2 x^2\right )\right )}\right ) \, dx\\ &=\left (4 e^5\right ) \int \frac {(-1+2 x) \log \left (\frac {1}{3} \left (1+\log \left ((-1+x)^2 x^2\right )\right )\right )}{(-1+x) x \left (1+\log \left ((-1+x)^2 x^2\right )\right )} \, dx+\int \frac {2 e^5+x}{x} \, dx\\ &=e^5 \log ^2\left (\frac {1}{3} \left (1+\log \left ((1-x)^2 x^2\right )\right )\right )+\int \left (1+\frac {2 e^5}{x}\right ) \, dx\\ &=x+2 e^5 \log (x)+e^5 \log ^2\left (\frac {1}{3} \left (1+\log \left ((1-x)^2 x^2\right )\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 32, normalized size = 1.07 \begin {gather*} x+2 e^5 \log (x)+e^5 \log ^2\left (\frac {1}{3} \left (1+\log \left ((-1+x)^2 x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^2 + E^5*(-2 + 2*x) + (-x + x^2 + E^5*(-2 + 2*x))*Log[x^2 - 2*x^3 + x^4] + E^5*(-4 + 8*x)*Log
[(1 + Log[x^2 - 2*x^3 + x^4])/3])/(-x + x^2 + (-x + x^2)*Log[x^2 - 2*x^3 + x^4]),x]

[Out]

x + 2*E^5*Log[x] + E^5*Log[(1 + Log[(-1 + x)^2*x^2])/3]^2

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 31, normalized size = 1.03 \begin {gather*} e^{5} \log \left (\frac {1}{3} \, \log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + \frac {1}{3}\right )^{2} + 2 \, e^{5} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*ex
p(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2)+x^2-x),x, algorithm="fricas")

[Out]

e^5*log(1/3*log(x^4 - 2*x^3 + x^2) + 1/3)^2 + 2*e^5*log(x) + x

________________________________________________________________________________________

giac [A]  time = 0.78, size = 51, normalized size = 1.70 \begin {gather*} -2 \, e^{5} \log \relax (3) \log \left (\log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + 1\right ) + e^{5} \log \left (\log \left (x^{4} - 2 \, x^{3} + x^{2}\right ) + 1\right )^{2} + 2 \, e^{5} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*ex
p(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2)+x^2-x),x, algorithm="giac")

[Out]

-2*e^5*log(3)*log(log(x^4 - 2*x^3 + x^2) + 1) + e^5*log(log(x^4 - 2*x^3 + x^2) + 1)^2 + 2*e^5*log(x) + x

________________________________________________________________________________________

maple [A]  time = 0.29, size = 32, normalized size = 1.07

method	result	size
norman	\(x +{\mathrm e}^{5} \ln \left (\frac {\ln \left (x^{4}-2 x^{3}+x^{2}\right )}{3}+\frac {1}{3}\right )^{2}+2 \,{\mathrm e}^{5} \ln \relax (x )\)	\(32\)
default	\(x +2 \,{\mathrm e}^{5} \ln \relax (x )-2 \,{\mathrm e}^{5} \ln \relax (3) \ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )+{\mathrm e}^{5} \ln \left (\ln \left (x^{4}-2 x^{3}+x^{2}\right )+1\right )^{2}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x-4)*exp(5)*ln(1/3*ln(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+x^2-x)*ln(x^4-2*x^3+x^2)+(2*x-2)*exp(5)+x^2-
x)/((x^2-x)*ln(x^4-2*x^3+x^2)+x^2-x),x,method=_RETURNVERBOSE)

[Out]

x+exp(5)*ln(1/3*ln(x^4-2*x^3+x^2)+1/3)^2+2*exp(5)*ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.93, size = 45, normalized size = 1.50 \begin {gather*} -2 \, e^{5} \log \relax (3) \log \left (2 \, \log \left (x - 1\right ) + 2 \, \log \relax (x) + 1\right ) + e^{5} \log \left (2 \, \log \left (x - 1\right ) + 2 \, \log \relax (x) + 1\right )^{2} + 2 \, e^{5} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-4)*exp(5)*log(1/3*log(x^4-2*x^3+x^2)+1/3)+((2*x-2)*exp(5)+x^2-x)*log(x^4-2*x^3+x^2)+(2*x-2)*ex
p(5)+x^2-x)/((x^2-x)*log(x^4-2*x^3+x^2)+x^2-x),x, algorithm="maxima")

[Out]

-2*e^5*log(3)*log(2*log(x - 1) + 2*log(x) + 1) + e^5*log(2*log(x - 1) + 2*log(x) + 1)^2 + 2*e^5*log(x) + x

________________________________________________________________________________________

mupad [B]  time = 0.80, size = 31, normalized size = 1.03 \begin {gather*} {\mathrm {e}}^5\,{\ln \left (\frac {\ln \left (x^4-2\,x^3+x^2\right )}{3}+\frac {1}{3}\right )}^2+x+2\,{\mathrm {e}}^5\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2 - 2*x^3 + x^4)*(x^2 - x + exp(5)*(2*x - 2)) - x + x^2 + exp(5)*(2*x - 2) + exp(5)*log(log(x^2 -
2*x^3 + x^4)/3 + 1/3)*(8*x - 4))/(x + log(x^2 - 2*x^3 + x^4)*(x - x^2) - x^2),x)

[Out]

x + exp(5)*log(log(x^2 - 2*x^3 + x^4)/3 + 1/3)^2 + 2*exp(5)*log(x)

________________________________________________________________________________________

sympy [A]  time = 0.59, size = 34, normalized size = 1.13 \begin {gather*} x + 2 e^{5} \log {\relax (x )} + e^{5} \log {\left (\frac {\log {\left (x^{4} - 2 x^{3} + x^{2} \right )}}{3} + \frac {1}{3} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-4)*exp(5)*ln(1/3*ln(x**4-2*x**3+x**2)+1/3)+((2*x-2)*exp(5)+x**2-x)*ln(x**4-2*x**3+x**2)+(2*x-2
)*exp(5)+x**2-x)/((x**2-x)*ln(x**4-2*x**3+x**2)+x**2-x),x)

[Out]

x + 2*exp(5)*log(x) + exp(5)*log(log(x**4 - 2*x**3 + x**2)/3 + 1/3)**2

________________________________________________________________________________________