3.42.33 \(\int \frac {-11 x+e^{\sqrt [4]{e}} (11 x+12 x^2)+(2 x+e^{\sqrt [4]{e}} (-2 x-2 x^2)) \log (x+e^{\sqrt [4]{e}} (-x-x^2))}{-25-5 x^2+e^{\sqrt [4]{e}} (25+25 x+5 x^2+5 x^3)+(10+x^2+e^{\sqrt [4]{e}} (-10-10 x-x^2-x^3)) \log (x+e^{\sqrt [4]{e}} (-x-x^2))+(-1+e^{\sqrt [4]{e}} (1+x)) \log ^2(x+e^{\sqrt [4]{e}} (-x-x^2))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (-1+\frac {x^2}{-5+\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )}\right ) \]

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Rubi [A]  time = 4.69, antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 5, number of rules used = 3, integrand size = 181, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 6742, 6684} \begin {gather*} \log \left (x^2-\log \left (x-e^{\sqrt [4]{e}} x (x+1)\right )+5\right )-\log \left (5-\log \left (x-e^{\sqrt [4]{e}} x (x+1)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11*x + E^E^(1/4)*(11*x + 12*x^2) + (2*x + E^E^(1/4)*(-2*x - 2*x^2))*Log[x + E^E^(1/4)*(-x - x^2)])/(-25
- 5*x^2 + E^E^(1/4)*(25 + 25*x + 5*x^2 + 5*x^3) + (10 + x^2 + E^E^(1/4)*(-10 - 10*x - x^2 - x^3))*Log[x + E^E^
(1/4)*(-x - x^2)] + (-1 + E^E^(1/4)*(1 + x))*Log[x + E^E^(1/4)*(-x - x^2)]^2),x]

[Out]

-Log[5 - Log[x - E^E^(1/4)*x*(1 + x)]] + Log[5 + x^2 - Log[x - E^E^(1/4)*x*(1 + 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 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 \left (11-e^{\sqrt [4]{e}} (11+12 x)+2 \left (-1+e^{\sqrt [4]{e}} (1+x)\right ) \log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )}{\left (1-e^{\sqrt [4]{e}}-e^{\sqrt [4]{e}} x\right ) \left (5-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right ) \left (5+x^2-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )} \, dx\\ &=\int \left (\frac {-1+e^{\sqrt [4]{e}}+2 e^{\sqrt [4]{e}} x+2 \left (1-e^{\sqrt [4]{e}}\right ) x^2-2 e^{\sqrt [4]{e}} x^3}{x \left (1-e^{\sqrt [4]{e}}-e^{\sqrt [4]{e}} x\right ) \left (5+x^2-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )}-\frac {-1+e^{\sqrt [4]{e}}+2 e^{\sqrt [4]{e}} x}{x \left (-1+e^{\sqrt [4]{e}}+e^{\sqrt [4]{e}} x\right ) \left (-5+\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )}\right ) \, dx\\ &=\int \frac {-1+e^{\sqrt [4]{e}}+2 e^{\sqrt [4]{e}} x+2 \left (1-e^{\sqrt [4]{e}}\right ) x^2-2 e^{\sqrt [4]{e}} x^3}{x \left (1-e^{\sqrt [4]{e}}-e^{\sqrt [4]{e}} x\right ) \left (5+x^2-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )} \, dx-\int \frac {-1+e^{\sqrt [4]{e}}+2 e^{\sqrt [4]{e}} x}{x \left (-1+e^{\sqrt [4]{e}}+e^{\sqrt [4]{e}} x\right ) \left (-5+\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )} \, dx\\ &=-\log \left (5-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )+\log \left (5+x^2-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 48, normalized size = 1.78 \begin {gather*} -\log \left (5-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right )+\log \left (5+x^2-\log \left (x-e^{\sqrt [4]{e}} x (1+x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11*x + E^E^(1/4)*(11*x + 12*x^2) + (2*x + E^E^(1/4)*(-2*x - 2*x^2))*Log[x + E^E^(1/4)*(-x - x^2)])
/(-25 - 5*x^2 + E^E^(1/4)*(25 + 25*x + 5*x^2 + 5*x^3) + (10 + x^2 + E^E^(1/4)*(-10 - 10*x - x^2 - x^3))*Log[x
+ E^E^(1/4)*(-x - x^2)] + (-1 + E^E^(1/4)*(1 + x))*Log[x + E^E^(1/4)*(-x - x^2)]^2),x]

[Out]

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

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fricas [A]  time = 0.68, size = 40, normalized size = 1.48 \begin {gather*} \log \left (-x^{2} + \log \left (-{\left (x^{2} + x\right )} e^{\left (e^{\frac {1}{4}}\right )} + x\right ) - 5\right ) - \log \left (\log \left (-{\left (x^{2} + x\right )} e^{\left (e^{\frac {1}{4}}\right )} + x\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(exp(1/4))+2*x)*log((-x^2-x)*exp(exp(1/4))+x)+(12*x^2+11*x)*exp(exp(1/4))-11*x)/((
(x+1)*exp(exp(1/4))-1)*log((-x^2-x)*exp(exp(1/4))+x)^2+((-x^3-x^2-10*x-10)*exp(exp(1/4))+x^2+10)*log((-x^2-x)*
exp(exp(1/4))+x)+(5*x^3+5*x^2+25*x+25)*exp(exp(1/4))-5*x^2-25),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.44, size = 48, normalized size = 1.78 \begin {gather*} \log \left (-x^{2} + \log \left (-x^{2} e^{\left (e^{\frac {1}{4}}\right )} - x e^{\left (e^{\frac {1}{4}}\right )} + x\right ) - 5\right ) - \log \left (\log \left (-x^{2} e^{\left (e^{\frac {1}{4}}\right )} - x e^{\left (e^{\frac {1}{4}}\right )} + x\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(exp(1/4))+2*x)*log((-x^2-x)*exp(exp(1/4))+x)+(12*x^2+11*x)*exp(exp(1/4))-11*x)/((
(x+1)*exp(exp(1/4))-1)*log((-x^2-x)*exp(exp(1/4))+x)^2+((-x^3-x^2-10*x-10)*exp(exp(1/4))+x^2+10)*log((-x^2-x)*
exp(exp(1/4))+x)+(5*x^3+5*x^2+25*x+25)*exp(exp(1/4))-5*x^2-25),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 47, normalized size = 1.74

method	result	size
norman	\(-\ln \left (\ln \left (\left (-x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}+x \right )-5\right )+\ln \left (x^{2}-\ln \left (\left (-x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}+x \right )+5\right )\)	\(47\)
risch	\(\ln \left (-x^{2}+\ln \left (\left (-x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}+x \right )-5\right )-\ln \left (\ln \left (\left (-x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}+x \right )-5\right )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-2*x)*exp(exp(1/4))+2*x)*ln((-x^2-x)*exp(exp(1/4))+x)+(12*x^2+11*x)*exp(exp(1/4))-11*x)/(((x+1)*e
xp(exp(1/4))-1)*ln((-x^2-x)*exp(exp(1/4))+x)^2+((-x^3-x^2-10*x-10)*exp(exp(1/4))+x^2+10)*ln((-x^2-x)*exp(exp(1
/4))+x)+(5*x^3+5*x^2+25*x+25)*exp(exp(1/4))-5*x^2-25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 46, normalized size = 1.70 \begin {gather*} \log \left (-x^{2} + \log \left (-x e^{\left (e^{\frac {1}{4}}\right )} - e^{\left (e^{\frac {1}{4}}\right )} + 1\right ) + \log \relax (x) - 5\right ) - \log \left (\log \left (-x e^{\left (e^{\frac {1}{4}}\right )} - e^{\left (e^{\frac {1}{4}}\right )} + 1\right ) + \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(exp(1/4))+2*x)*log((-x^2-x)*exp(exp(1/4))+x)+(12*x^2+11*x)*exp(exp(1/4))-11*x)/((
(x+1)*exp(exp(1/4))-1)*log((-x^2-x)*exp(exp(1/4))+x)^2+((-x^3-x^2-10*x-10)*exp(exp(1/4))+x^2+10)*log((-x^2-x)*
exp(exp(1/4))+x)+(5*x^3+5*x^2+25*x+25)*exp(exp(1/4))-5*x^2-25),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.49, size = 250, normalized size = 9.26 \begin {gather*} \ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^{1/4}}}{2}-x-x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{1/4}}+x^2+x^3-\frac {1}{2}\right )+\ln \left (\left (x^2-\ln \left (-x\,\left ({\mathrm {e}}^{{\mathrm {e}}^{1/4}}+x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-1\right )\right )+5\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{1/4}}+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-1\right )\right )-\ln \left (20\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-2\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )-10\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}+20\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}+2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )+4\,x^2\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )-20\,x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-20\,x^2+4\,x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )-4\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )-4\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\,\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}-x\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\right )+10\right )-\ln \left (x-\frac {{\mathrm {e}}^{-{\mathrm {e}}^{1/4}}}{2}+\frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - exp(exp(1/4))*(x + x^2))*(2*x - exp(exp(1/4))*(2*x + 2*x^2)) - 11*x + exp(exp(1/4))*(11*x + 12*x^
2))/(log(x - exp(exp(1/4))*(x + x^2))^2*(exp(exp(1/4))*(x + 1) - 1) + log(x - exp(exp(1/4))*(x + x^2))*(x^2 -
exp(exp(1/4))*(10*x + x^2 + x^3 + 10) + 10) + exp(exp(1/4))*(25*x + 5*x^2 + 5*x^3 + 25) - 5*x^2 - 25),x)

[Out]

log(exp(-exp(1/4))/2 - x - x^2*exp(-exp(1/4)) + x^2 + x^3 - 1/2) + log((x^2 - log(-x*(exp(exp(1/4)) + x*exp(ex
p(1/4)) - 1)) + 5)*(exp(exp(1/4)) + 2*x*exp(exp(1/4)) - 1)) - log(20*x^2*exp(exp(1/4)) - 2*log(x - x^2*exp(exp
(1/4)) - x*exp(exp(1/4))) - 10*exp(exp(1/4)) + 20*x^3*exp(exp(1/4)) + 2*exp(exp(1/4))*log(x - x^2*exp(exp(1/4)
) - x*exp(exp(1/4))) + 4*x^2*log(x - x^2*exp(exp(1/4)) - x*exp(exp(1/4))) - 20*x*exp(exp(1/4)) - 20*x^2 + 4*x*
exp(exp(1/4))*log(x - x^2*exp(exp(1/4)) - x*exp(exp(1/4))) - 4*x^2*exp(exp(1/4))*log(x - x^2*exp(exp(1/4)) - x
*exp(exp(1/4))) - 4*x^3*exp(exp(1/4))*log(x - x^2*exp(exp(1/4)) - x*exp(exp(1/4))) + 10) - log(x - exp(-exp(1/
4))/2 + 1/2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-2*x)*exp(exp(1/4))+2*x)*ln((-x**2-x)*exp(exp(1/4))+x)+(12*x**2+11*x)*exp(exp(1/4))-11*x)/
(((x+1)*exp(exp(1/4))-1)*ln((-x**2-x)*exp(exp(1/4))+x)**2+((-x**3-x**2-10*x-10)*exp(exp(1/4))+x**2+10)*ln((-x*
*2-x)*exp(exp(1/4))+x)+(5*x**3+5*x**2+25*x+25)*exp(exp(1/4))-5*x**2-25),x)

[Out]

Exception raised: PolynomialError

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