3.23.100 \(\int \frac {-2-3 e^3+e^x-4 x+(-8+2 e^x) \log (\frac {1}{3} (2+3 e^3-e^x+4 x))}{12+e^3 (18-3 x)+e^x (-6+x)+22 x-4 x^2+(-2-3 e^3+e^x-4 x) \log ^2(\frac {1}{3} (2+3 e^3-e^x+4 x))} \, dx\)

Optimal. Leaf size=24 \[ \log \left (-6+x+\log ^2\left (e^3+x+\frac {1}{3} \left (2-e^x+x\right )\right )\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6741, 6684} \begin {gather*} \log \left (-x-\log ^2\left (\frac {1}{3} \left (4 x-e^x+3 e^3+2\right )\right )+6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 3*E^3 + E^x - 4*x + (-8 + 2*E^x)*Log[(2 + 3*E^3 - E^x + 4*x)/3])/(12 + E^3*(18 - 3*x) + E^x*(-6 + x)
 + 22*x - 4*x^2 + (-2 - 3*E^3 + E^x - 4*x)*Log[(2 + 3*E^3 - E^x + 4*x)/3]^2),x]

[Out]

Log[6 - x - Log[(2 + 3*E^3 - E^x + 4*x)/3]^2]

Rule 6684

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x+2 \left (1+\frac {3 e^3}{2}\right )+4 x-\left (-8+2 e^x\right ) \log \left (\frac {1}{3} \left (2+3 e^3-e^x+4 x\right )\right )}{\left (e^x-2 \left (1+\frac {3 e^3}{2}\right )-4 x\right ) \left (6-x-\log ^2\left (\frac {1}{3} \left (2+3 e^3-e^x+4 x\right )\right )\right )} \, dx\\ &=\log \left (6-x-\log ^2\left (\frac {1}{3} \left (2+3 e^3-e^x+4 x\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 30, normalized size = 1.25 \begin {gather*} \log \left (6-x-\log ^2\left (\frac {1}{3} \left (2+3 e^3-e^x+4 x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 3*E^3 + E^x - 4*x + (-8 + 2*E^x)*Log[(2 + 3*E^3 - E^x + 4*x)/3])/(12 + E^3*(18 - 3*x) + E^x*(-
6 + x) + 22*x - 4*x^2 + (-2 - 3*E^3 + E^x - 4*x)*Log[(2 + 3*E^3 - E^x + 4*x)/3]^2),x]

[Out]

Log[6 - x - Log[(2 + 3*E^3 - E^x + 4*x)/3]^2]

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fricas [A]  time = 0.55, size = 18, normalized size = 0.75 \begin {gather*} \log \left (\log \left (\frac {4}{3} \, x + e^{3} - \frac {1}{3} \, e^{x} + \frac {2}{3}\right )^{2} + x - 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-8)*log(-1/3*exp(x)+exp(3)+4/3*x+2/3)+exp(x)-3*exp(3)-4*x-2)/((exp(x)-3*exp(3)-4*x-2)*log(
-1/3*exp(x)+exp(3)+4/3*x+2/3)^2+(x-6)*exp(x)+(-3*x+18)*exp(3)-4*x^2+22*x+12),x, algorithm="fricas")

[Out]

log(log(4/3*x + e^3 - 1/3*e^x + 2/3)^2 + x - 6)

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giac [A]  time = 0.64, size = 18, normalized size = 0.75 \begin {gather*} \log \left (\log \left (\frac {4}{3} \, x + e^{3} - \frac {1}{3} \, e^{x} + \frac {2}{3}\right )^{2} + x - 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-8)*log(-1/3*exp(x)+exp(3)+4/3*x+2/3)+exp(x)-3*exp(3)-4*x-2)/((exp(x)-3*exp(3)-4*x-2)*log(
-1/3*exp(x)+exp(3)+4/3*x+2/3)^2+(x-6)*exp(x)+(-3*x+18)*exp(3)-4*x^2+22*x+12),x, algorithm="giac")

[Out]

log(log(4/3*x + e^3 - 1/3*e^x + 2/3)^2 + x - 6)

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maple [A]  time = 0.35, size = 19, normalized size = 0.79




method result size



norman \(\ln \left (\ln \left (-\frac {{\mathrm e}^{x}}{3}+{\mathrm e}^{3}+\frac {4 x}{3}+\frac {2}{3}\right )^{2}-6+x \right )\) \(19\)
risch \(\ln \left (\ln \left (-\frac {{\mathrm e}^{x}}{3}+{\mathrm e}^{3}+\frac {4 x}{3}+\frac {2}{3}\right )^{2}-6+x \right )\) \(19\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)-8)*ln(-1/3*exp(x)+exp(3)+4/3*x+2/3)+exp(x)-3*exp(3)-4*x-2)/((exp(x)-3*exp(3)-4*x-2)*ln(-1/3*exp
(x)+exp(3)+4/3*x+2/3)^2+(x-6)*exp(x)+(-3*x+18)*exp(3)-4*x^2+22*x+12),x,method=_RETURNVERBOSE)

[Out]

ln(ln(-1/3*exp(x)+exp(3)+4/3*x+2/3)^2-6+x)

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maxima [B]  time = 0.72, size = 42, normalized size = 1.75 \begin {gather*} \log \left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (4 \, x + 3 \, e^{3} - e^{x} + 2\right ) + \log \left (4 \, x + 3 \, e^{3} - e^{x} + 2\right )^{2} + x - 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-8)*log(-1/3*exp(x)+exp(3)+4/3*x+2/3)+exp(x)-3*exp(3)-4*x-2)/((exp(x)-3*exp(3)-4*x-2)*log(
-1/3*exp(x)+exp(3)+4/3*x+2/3)^2+(x-6)*exp(x)+(-3*x+18)*exp(3)-4*x^2+22*x+12),x, algorithm="maxima")

[Out]

log(log(3)^2 - 2*log(3)*log(4*x + 3*e^3 - e^x + 2) + log(4*x + 3*e^3 - e^x + 2)^2 + x - 6)

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mupad [B]  time = 5.32, size = 18, normalized size = 0.75 \begin {gather*} \ln \left ({\ln \left (\frac {4\,x}{3}+{\mathrm {e}}^3-\frac {{\mathrm {e}}^x}{3}+\frac {2}{3}\right )}^2+x-6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + 3*exp(3) - exp(x) - log((4*x)/3 + exp(3) - exp(x)/3 + 2/3)*(2*exp(x) - 8) + 2)/(22*x + exp(x)*(x -
 6) - log((4*x)/3 + exp(3) - exp(x)/3 + 2/3)^2*(4*x + 3*exp(3) - exp(x) + 2) - 4*x^2 - exp(3)*(3*x - 18) + 12)
,x)

[Out]

log(x + log((4*x)/3 + exp(3) - exp(x)/3 + 2/3)^2 - 6)

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sympy [A]  time = 0.85, size = 24, normalized size = 1.00 \begin {gather*} \log {\left (x + \log {\left (\frac {4 x}{3} - \frac {e^{x}}{3} + \frac {2}{3} + e^{3} \right )}^{2} - 6 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-8)*ln(-1/3*exp(x)+exp(3)+4/3*x+2/3)+exp(x)-3*exp(3)-4*x-2)/((exp(x)-3*exp(3)-4*x-2)*ln(-1
/3*exp(x)+exp(3)+4/3*x+2/3)**2+(x-6)*exp(x)+(-3*x+18)*exp(3)-4*x**2+22*x+12),x)

[Out]

log(x + log(4*x/3 - exp(x)/3 + 2/3 + exp(3))**2 - 6)

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