3.7.17 \(\int \frac {-12 x^2-4 x^3-2 x^4+e^3 (5+12 x+4 x^2+2 x^3)+(-6 x^2-2 x^3-x^4+e^3 (6 x+2 x^2+x^3)) \log (\frac {-e^3+x}{x})+(4 x^2+4 x^3+e^3 (-4 x-4 x^2)+(2 x^2+2 x^3+e^3 (-2 x-2 x^2)) \log (\frac {-e^3+x}{x})) \log (8+4 \log (\frac {-e^3+x}{x}))+(2 e^3 x-2 x^2+(e^3 x-x^2) \log (\frac {-e^3+x}{x})) \log ^2(8+4 \log (\frac {-e^3+x}{x}))}{-2 x^2-4 x^3-2 x^4+e^3 (2 x+4 x^2+2 x^3)+(-x^2-2 x^3-x^4+e^3 (x+2 x^2+x^3)) \log (\frac {-e^3+x}{x})+(4 x^2+4 x^3+e^3 (-4 x-4 x^2)+(2 x^2+2 x^3+e^3 (-2 x-2 x^2)) \log (\frac {-e^3+x}{x})) \log (8+4 \log (\frac {-e^3+x}{x}))+(2 e^3 x-2 x^2+(e^3 x-x^2) \log (\frac {-e^3+x}{x})) \log ^2(8+4 \log (\frac {-e^3+x}{x}))} \, dx\)
Optimal. Leaf size=29 \[ -1+x+\frac {5}{-1-x+\log \left (4 \left (2+\log \left (\frac {-e^3+x}{x}\right )\right )\right )} \]
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Rubi [A] time = 7.02, antiderivative size = 27, normalized size of antiderivative = 0.93,
number of steps used = 4, number of rules used = 3, integrand size = 430, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used
= {6688, 6742, 6686} \begin {gather*} x-\frac {5}{x-\log \left (4 \left (\log \left (1-\frac {e^3}{x}\right )+2\right )\right )+1} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-12*x^2 - 4*x^3 - 2*x^4 + E^3*(5 + 12*x + 4*x^2 + 2*x^3) + (-6*x^2 - 2*x^3 - x^4 + E^3*(6*x + 2*x^2 + x^3
))*Log[(-E^3 + x)/x] + (4*x^2 + 4*x^3 + E^3*(-4*x - 4*x^2) + (2*x^2 + 2*x^3 + E^3*(-2*x - 2*x^2))*Log[(-E^3 +
x)/x])*Log[8 + 4*Log[(-E^3 + x)/x]] + (2*E^3*x - 2*x^2 + (E^3*x - x^2)*Log[(-E^3 + x)/x])*Log[8 + 4*Log[(-E^3
+ x)/x]]^2)/(-2*x^2 - 4*x^3 - 2*x^4 + E^3*(2*x + 4*x^2 + 2*x^3) + (-x^2 - 2*x^3 - x^4 + E^3*(x + 2*x^2 + x^3))
*Log[(-E^3 + x)/x] + (4*x^2 + 4*x^3 + E^3*(-4*x - 4*x^2) + (2*x^2 + 2*x^3 + E^3*(-2*x - 2*x^2))*Log[(-E^3 + x)
/x])*Log[8 + 4*Log[(-E^3 + x)/x]] + (2*E^3*x - 2*x^2 + (E^3*x - x^2)*Log[(-E^3 + x)/x])*Log[8 + 4*Log[(-E^3 +
x)/x]]^2),x]
[Out]
x - 5/(1 + x - Log[4*(2 + Log[1 - E^3/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 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 {5 e^3+12 e^3 x-12 \left (1-\frac {e^3}{3}\right ) x^2-4 \left (1-\frac {e^3}{2}\right ) x^3-2 x^4+4 x (1+x) \left (-e^3+x\right ) \log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )+2 \left (e^3-x\right ) x \log ^2\left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )+\left (e^3-x\right ) x \log \left (1-\frac {e^3}{x}\right ) \left (6+2 x+x^2-2 (1+x) \log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )+\log ^2\left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )\right )}{\left (e^3-x\right ) x \left (2+\log \left (1-\frac {e^3}{x}\right )\right ) \left (1+x-\log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )\right )^2} \, dx\\ &=\int \left (1+\frac {5 \left (e^3+2 e^3 x-2 x^2+e^3 x \log \left (1-\frac {e^3}{x}\right )-x^2 \log \left (1-\frac {e^3}{x}\right )\right )}{\left (e^3-x\right ) x \left (2+\log \left (1-\frac {e^3}{x}\right )\right ) \left (1+x-\log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )\right )^2}\right ) \, dx\\ &=x+5 \int \frac {e^3+2 e^3 x-2 x^2+e^3 x \log \left (1-\frac {e^3}{x}\right )-x^2 \log \left (1-\frac {e^3}{x}\right )}{\left (e^3-x\right ) x \left (2+\log \left (1-\frac {e^3}{x}\right )\right ) \left (1+x-\log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )\right )^2} \, dx\\ &=x-\frac {5}{1+x-\log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 27, normalized size = 0.93 \begin {gather*} x+\frac {5}{-1-x+\log \left (4 \left (2+\log \left (1-\frac {e^3}{x}\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-12*x^2 - 4*x^3 - 2*x^4 + E^3*(5 + 12*x + 4*x^2 + 2*x^3) + (-6*x^2 - 2*x^3 - x^4 + E^3*(6*x + 2*x^2
+ x^3))*Log[(-E^3 + x)/x] + (4*x^2 + 4*x^3 + E^3*(-4*x - 4*x^2) + (2*x^2 + 2*x^3 + E^3*(-2*x - 2*x^2))*Log[(-
E^3 + x)/x])*Log[8 + 4*Log[(-E^3 + x)/x]] + (2*E^3*x - 2*x^2 + (E^3*x - x^2)*Log[(-E^3 + x)/x])*Log[8 + 4*Log[
(-E^3 + x)/x]]^2)/(-2*x^2 - 4*x^3 - 2*x^4 + E^3*(2*x + 4*x^2 + 2*x^3) + (-x^2 - 2*x^3 - x^4 + E^3*(x + 2*x^2 +
x^3))*Log[(-E^3 + x)/x] + (4*x^2 + 4*x^3 + E^3*(-4*x - 4*x^2) + (2*x^2 + 2*x^3 + E^3*(-2*x - 2*x^2))*Log[(-E^
3 + x)/x])*Log[8 + 4*Log[(-E^3 + x)/x]] + (2*E^3*x - 2*x^2 + (E^3*x - x^2)*Log[(-E^3 + x)/x])*Log[8 + 4*Log[(-
E^3 + x)/x]]^2),x]
[Out]
x + 5/(-1 - x + Log[4*(2 + Log[1 - E^3/x])])
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fricas [A] time = 0.76, size = 49, normalized size = 1.69 \begin {gather*} \frac {x^{2} - x \log \left (4 \, \log \left (\frac {x - e^{3}}{x}\right ) + 8\right ) + x - 5}{x - \log \left (4 \, \log \left (\frac {x - e^{3}}{x}\right ) + 8\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((x*exp(3)-x^2)*log((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp
(3)+2*x^3+2*x^2)*log((x-exp(3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+6*x
)*exp(3)-x^4-2*x^3-6*x^2)*log((x-exp(3))/x)+(2*x^3+4*x^2+12*x+5)*exp(3)-2*x^4-4*x^3-12*x^2)/(((x*exp(3)-x^2)*l
og((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3+2*x^2)*log((x-exp(
3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+x)*exp(3)-x^4-2*x^3-x^2)*log((x
-exp(3))/x)+(2*x^3+4*x^2+2*x)*exp(3)-2*x^4-4*x^3-2*x^2),x, algorithm="fricas")
[Out]
(x^2 - x*log(4*log((x - e^3)/x) + 8) + x - 5)/(x - log(4*log((x - e^3)/x) + 8) + 1)
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giac [B] time = 16.18, size = 1737, normalized size = 59.90 result too large to
display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((x*exp(3)-x^2)*log((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp
(3)+2*x^3+2*x^2)*log((x-exp(3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+6*x
)*exp(3)-x^4-2*x^3-6*x^2)*log((x-exp(3))/x)+(2*x^3+4*x^2+12*x+5)*exp(3)-2*x^4-4*x^3-12*x^2)/(((x*exp(3)-x^2)*l
og((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3+2*x^2)*log((x-exp(
3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+x)*exp(3)-x^4-2*x^3-x^2)*log((x
-exp(3))/x)+(2*x^3+4*x^2+2*x)*exp(3)-2*x^4-4*x^3-2*x^2),x, algorithm="giac")
[Out]
(x^4*log(x - e^3)*log((x - e^3)/x) - x^3*e^3*log(x - e^3)*log((x - e^3)/x) - x^4*log(x)*log((x - e^3)/x) + x^3
*e^3*log(x)*log((x - e^3)/x) - x^3*log(x - e^3)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + x^2*e^3*log(x -
e^3)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + x^3*log(x)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) -
x^2*e^3*log(x)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + 2*x^4*log(x - e^3) - 2*x^3*e^3*log(x - e^3) - 2
*x^4*log(x) + 2*x^3*e^3*log(x) + 2*x^4*log((x - e^3)/x) - 2*x^3*e^3*log((x - e^3)/x) + x^3*log(x - e^3)*log((x
- e^3)/x) - x^2*e^3*log(x - e^3)*log((x - e^3)/x) - x^3*log(x)*log((x - e^3)/x) + x^2*e^3*log(x)*log((x - e^3
)/x) - 2*x^3*log(x - e^3)*log(4*log((x - e^3)/x) + 8) + 2*x^2*e^3*log(x - e^3)*log(4*log((x - e^3)/x) + 8) + 2
*x^3*log(x)*log(4*log((x - e^3)/x) + 8) - 2*x^2*e^3*log(x)*log(4*log((x - e^3)/x) + 8) - 2*x^3*log((x - e^3)/x
)*log(4*log((x - e^3)/x) + 8) + 2*x^2*e^3*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + 4*x^4 - 4*x^3*e^3 + 2
*x^3*log(x - e^3) - 3*x^2*e^3*log(x - e^3) - 2*x^3*log(x) + 3*x^2*e^3*log(x) + 2*x^3*log((x - e^3)/x) - 2*x^2*
e^3*log((x - e^3)/x) - 5*x^2*log(x - e^3)*log((x - e^3)/x) + 5*x*e^3*log(x - e^3)*log((x - e^3)/x) + 5*x^2*log
(x)*log((x - e^3)/x) - 5*x*e^3*log(x)*log((x - e^3)/x) - 4*x^3*log(4*log((x - e^3)/x) + 8) + 4*x^2*e^3*log(4*l
og((x - e^3)/x) + 8) + x*e^3*log(x - e^3)*log(4*log((x - e^3)/x) + 8) - x*e^3*log(x)*log(4*log((x - e^3)/x) +
8) + 4*x^3 - 6*x^2*e^3 - 10*x^2*log(x - e^3) + 9*x*e^3*log(x - e^3) + 10*x^2*log(x) - 9*x*e^3*log(x) - 10*x^2*
log((x - e^3)/x) + 10*x*e^3*log((x - e^3)/x) + 2*x*e^3*log(4*log((x - e^3)/x) + 8) - 20*x^2 + 18*x*e^3 + 5*e^3
*log((x - e^3)/x) + 10*e^3)/(x^3*log(x - e^3)*log((x - e^3)/x) - x^2*e^3*log(x - e^3)*log((x - e^3)/x) - x^3*l
og(x)*log((x - e^3)/x) + x^2*e^3*log(x)*log((x - e^3)/x) - x^2*log(x - e^3)*log((x - e^3)/x)*log(4*log((x - e^
3)/x) + 8) + x*e^3*log(x - e^3)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + x^2*log(x)*log((x - e^3)/x)*log
(4*log((x - e^3)/x) + 8) - x*e^3*log(x)*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + 2*x^3*log(x - e^3) - 2*
x^2*e^3*log(x - e^3) - 2*x^3*log(x) + 2*x^2*e^3*log(x) + 2*x^3*log((x - e^3)/x) - 2*x^2*e^3*log((x - e^3)/x) +
x^2*log(x - e^3)*log((x - e^3)/x) - x*e^3*log(x - e^3)*log((x - e^3)/x) - x^2*log(x)*log((x - e^3)/x) + x*e^3
*log(x)*log((x - e^3)/x) - 2*x^2*log(x - e^3)*log(4*log((x - e^3)/x) + 8) + 2*x*e^3*log(x - e^3)*log(4*log((x
- e^3)/x) + 8) + 2*x^2*log(x)*log(4*log((x - e^3)/x) + 8) - 2*x*e^3*log(x)*log(4*log((x - e^3)/x) + 8) - 2*x^2
*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + 2*x*e^3*log((x - e^3)/x)*log(4*log((x - e^3)/x) + 8) + 4*x^3 -
4*x^2*e^3 + 2*x^2*log(x - e^3) - 3*x*e^3*log(x - e^3) - 2*x^2*log(x) + 3*x*e^3*log(x) + 2*x^2*log((x - e^3)/x
) - 2*x*e^3*log((x - e^3)/x) - 4*x^2*log(4*log((x - e^3)/x) + 8) + 4*x*e^3*log(4*log((x - e^3)/x) + 8) + e^3*l
og(x - e^3)*log(4*log((x - e^3)/x) + 8) - e^3*log(x)*log(4*log((x - e^3)/x) + 8) + 4*x^2 - 6*x*e^3 - e^3*log(x
- e^3) + e^3*log(x) + 2*e^3*log(4*log((x - e^3)/x) + 8) - 2*e^3)
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maple [F] time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (x \,{\mathrm e}^{3}-x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+2 x \,{\mathrm e}^{3}-2 x^{2}\right ) \ln \left (4 \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+8\right )^{2}+\left (\left (\left (-2 x^{2}-2 x \right ) {\mathrm e}^{3}+2 x^{3}+2 x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+\left (-4 x^{2}-4 x \right ) {\mathrm e}^{3}+4 x^{3}+4 x^{2}\right ) \ln \left (4 \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+8\right )+\left (\left (x^{3}+2 x^{2}+6 x \right ) {\mathrm e}^{3}-x^{4}-2 x^{3}-6 x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+\left (2 x^{3}+4 x^{2}+12 x +5\right ) {\mathrm e}^{3}-2 x^{4}-4 x^{3}-12 x^{2}}{\left (\left (x \,{\mathrm e}^{3}-x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+2 x \,{\mathrm e}^{3}-2 x^{2}\right ) \ln \left (4 \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+8\right )^{2}+\left (\left (\left (-2 x^{2}-2 x \right ) {\mathrm e}^{3}+2 x^{3}+2 x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+\left (-4 x^{2}-4 x \right ) {\mathrm e}^{3}+4 x^{3}+4 x^{2}\right ) \ln \left (4 \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+8\right )+\left (\left (x^{3}+2 x^{2}+x \right ) {\mathrm e}^{3}-x^{4}-2 x^{3}-x^{2}\right ) \ln \left (\frac {x -{\mathrm e}^{3}}{x}\right )+\left (2 x^{3}+4 x^{2}+2 x \right ) {\mathrm e}^{3}-2 x^{4}-4 x^{3}-2 x^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((((x*exp(3)-x^2)*ln((x-exp(3))/x)+2*x*exp(3)-2*x^2)*ln(4*ln((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3
+2*x^2)*ln((x-exp(3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*ln(4*ln((x-exp(3))/x)+8)+((x^3+2*x^2+6*x)*exp(3)-x^4
-2*x^3-6*x^2)*ln((x-exp(3))/x)+(2*x^3+4*x^2+12*x+5)*exp(3)-2*x^4-4*x^3-12*x^2)/(((x*exp(3)-x^2)*ln((x-exp(3))/
x)+2*x*exp(3)-2*x^2)*ln(4*ln((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3+2*x^2)*ln((x-exp(3))/x)+(-4*x^2-4*
x)*exp(3)+4*x^3+4*x^2)*ln(4*ln((x-exp(3))/x)+8)+((x^3+2*x^2+x)*exp(3)-x^4-2*x^3-x^2)*ln((x-exp(3))/x)+(2*x^3+4
*x^2+2*x)*exp(3)-2*x^4-4*x^3-2*x^2),x)
[Out]
int((((x*exp(3)-x^2)*ln((x-exp(3))/x)+2*x*exp(3)-2*x^2)*ln(4*ln((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3
+2*x^2)*ln((x-exp(3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*ln(4*ln((x-exp(3))/x)+8)+((x^3+2*x^2+6*x)*exp(3)-x^4
-2*x^3-6*x^2)*ln((x-exp(3))/x)+(2*x^3+4*x^2+12*x+5)*exp(3)-2*x^4-4*x^3-12*x^2)/(((x*exp(3)-x^2)*ln((x-exp(3))/
x)+2*x*exp(3)-2*x^2)*ln(4*ln((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3+2*x^2)*ln((x-exp(3))/x)+(-4*x^2-4*
x)*exp(3)+4*x^3+4*x^2)*ln(4*ln((x-exp(3))/x)+8)+((x^3+2*x^2+x)*exp(3)-x^4-2*x^3-x^2)*ln((x-exp(3))/x)+(2*x^3+4
*x^2+2*x)*exp(3)-2*x^4-4*x^3-2*x^2),x)
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maxima [B] time = 0.68, size = 57, normalized size = 1.97 \begin {gather*} \frac {x^{2} - x {\left (2 \, \log \relax (2) - 1\right )} - x \log \left (\log \left (x - e^{3}\right ) - \log \relax (x) + 2\right ) - 5}{x - 2 \, \log \relax (2) - \log \left (\log \left (x - e^{3}\right ) - \log \relax (x) + 2\right ) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((x*exp(3)-x^2)*log((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp
(3)+2*x^3+2*x^2)*log((x-exp(3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+6*x
)*exp(3)-x^4-2*x^3-6*x^2)*log((x-exp(3))/x)+(2*x^3+4*x^2+12*x+5)*exp(3)-2*x^4-4*x^3-12*x^2)/(((x*exp(3)-x^2)*l
og((x-exp(3))/x)+2*x*exp(3)-2*x^2)*log(4*log((x-exp(3))/x)+8)^2+(((-2*x^2-2*x)*exp(3)+2*x^3+2*x^2)*log((x-exp(
3))/x)+(-4*x^2-4*x)*exp(3)+4*x^3+4*x^2)*log(4*log((x-exp(3))/x)+8)+((x^3+2*x^2+x)*exp(3)-x^4-2*x^3-x^2)*log((x
-exp(3))/x)+(2*x^3+4*x^2+2*x)*exp(3)-2*x^4-4*x^3-2*x^2),x, algorithm="maxima")
[Out]
(x^2 - x*(2*log(2) - 1) - x*log(log(x - e^3) - log(x) + 2) - 5)/(x - 2*log(2) - log(log(x - e^3) - log(x) + 2)
+ 1)
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mupad [B] time = 1.46, size = 27, normalized size = 0.93 \begin {gather*} x-\frac {5}{x-\ln \left (4\,\ln \left (\frac {x-{\mathrm {e}}^3}{x}\right )+8\right )+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((log((x - exp(3))/x)*(6*x^2 - exp(3)*(6*x + 2*x^2 + x^3) + 2*x^3 + x^4) - log(4*log((x - exp(3))/x) + 8)*(
log((x - exp(3))/x)*(2*x^2 - exp(3)*(2*x + 2*x^2) + 2*x^3) - exp(3)*(4*x + 4*x^2) + 4*x^2 + 4*x^3) - exp(3)*(1
2*x + 4*x^2 + 2*x^3 + 5) - log(4*log((x - exp(3))/x) + 8)^2*(2*x*exp(3) + log((x - exp(3))/x)*(x*exp(3) - x^2)
- 2*x^2) + 12*x^2 + 4*x^3 + 2*x^4)/(log((x - exp(3))/x)*(x^2 - exp(3)*(x + 2*x^2 + x^3) + 2*x^3 + x^4) - exp(
3)*(2*x + 4*x^2 + 2*x^3) - log(4*log((x - exp(3))/x) + 8)^2*(2*x*exp(3) + log((x - exp(3))/x)*(x*exp(3) - x^2)
- 2*x^2) - log(4*log((x - exp(3))/x) + 8)*(log((x - exp(3))/x)*(2*x^2 - exp(3)*(2*x + 2*x^2) + 2*x^3) - exp(3
)*(4*x + 4*x^2) + 4*x^2 + 4*x^3) + 2*x^2 + 4*x^3 + 2*x^4),x)
[Out]
x - 5/(x - log(4*log((x - exp(3))/x) + 8) + 1)
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sympy [A] time = 0.53, size = 19, normalized size = 0.66 \begin {gather*} x + \frac {5}{- x + \log {\left (4 \log {\left (\frac {x - e^{3}}{x} \right )} + 8 \right )} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((x*exp(3)-x**2)*ln((x-exp(3))/x)+2*x*exp(3)-2*x**2)*ln(4*ln((x-exp(3))/x)+8)**2+(((-2*x**2-2*x)*ex
p(3)+2*x**3+2*x**2)*ln((x-exp(3))/x)+(-4*x**2-4*x)*exp(3)+4*x**3+4*x**2)*ln(4*ln((x-exp(3))/x)+8)+((x**3+2*x**
2+6*x)*exp(3)-x**4-2*x**3-6*x**2)*ln((x-exp(3))/x)+(2*x**3+4*x**2+12*x+5)*exp(3)-2*x**4-4*x**3-12*x**2)/(((x*e
xp(3)-x**2)*ln((x-exp(3))/x)+2*x*exp(3)-2*x**2)*ln(4*ln((x-exp(3))/x)+8)**2+(((-2*x**2-2*x)*exp(3)+2*x**3+2*x*
*2)*ln((x-exp(3))/x)+(-4*x**2-4*x)*exp(3)+4*x**3+4*x**2)*ln(4*ln((x-exp(3))/x)+8)+((x**3+2*x**2+x)*exp(3)-x**4
-2*x**3-x**2)*ln((x-exp(3))/x)+(2*x**3+4*x**2+2*x)*exp(3)-2*x**4-4*x**3-2*x**2),x)
[Out]
x + 5/(-x + log(4*log((x - exp(3))/x) + 8) - 1)
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