3.93.72 \(\int \frac {-12+60 x-60 x^2+e^{2+x} (-5+20 x-20 x^2)+(-45+192 x-234 x^2+60 x^3+e^{2+x} (5-20 x+20 x^2)) \log (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x})}{(-45+192 x-234 x^2+60 x^3+e^{2+x} (5-20 x+20 x^2)) \log (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x})} \, dx\)

Optimal. Leaf size=29 \[ x-\log \left (\log \left (3-\frac {e^{2+x}}{3}-x+\frac {x}{5-10 x}\right )\right ) \]

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Rubi [F]  time = 4.21, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-12 + 60*x - 60*x^2 + E^(2 + x)*(-5 + 20*x - 20*x^2) + (-45 + 192*x - 234*x^2 + 60*x^3 + E^(2 + x)*(5 - 2
0*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)])/((-45 + 192*x - 234*x^2 + 60*x
^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]),x]

[Out]

x - Defer[Int][Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]^(-1), x] + 60*Defer[Int][1/((45
- 5*E^(2 + x) - 102*x + 10*E^(2 + x)*x + 30*x^2)*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*
x)]), x] - 132*Defer[Int][x/((45 - 5*E^(2 + x) - 102*x + 10*E^(2 + x)*x + 30*x^2)*Log[(-45 + E^(2 + x)*(5 - 10
*x) + 102*x - 30*x^2)/(-15 + 30*x)]), x] + 30*Defer[Int][x^2/((45 - 5*E^(2 + x) - 102*x + 10*E^(2 + x)*x + 30*
x^2)*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]), x] + 3*Defer[Int][1/((-1 + 2*x)*(45 - 5
*E^(2 + x) - 102*x + 10*E^(2 + x)*x + 30*x^2)*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)])
, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12-60 x+60 x^2-e^{2+x} \left (-5+20 x-20 x^2\right )-\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{(1-2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ &=\int \left (\frac {3 \left (-19+84 x-98 x^2+20 x^3\right )}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {-1+\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx\\ &=3 \int \frac {-19+84 x-98 x^2+20 x^3}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+\int \frac {-1+\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ &=3 \int \left (\frac {20}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}-\frac {44 x}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {10 x^2}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {1}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx+\int \left (1-\frac {1}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx\\ &=x+3 \int \frac {1}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+30 \int \frac {x^2}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+60 \int \frac {1}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx-132 \int \frac {x}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx-\int \frac {1}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 35, normalized size = 1.21 \begin {gather*} x-\log \left (\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 + 60*x - 60*x^2 + E^(2 + x)*(-5 + 20*x - 20*x^2) + (-45 + 192*x - 234*x^2 + 60*x^3 + E^(2 + x)*
(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)])/((-45 + 192*x - 234*x^2
+ 60*x^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]),x]

[Out]

x - Log[Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]]

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fricas [A]  time = 0.48, size = 36, normalized size = 1.24 \begin {gather*} x - \log \left (\log \left (-\frac {30 \, x^{2} + 5 \, {\left (2 \, x - 1\right )} e^{\left (x + 2\right )} - 102 \, x + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-1
5))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5
)*exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="fricas")

[Out]

x - log(log(-1/15*(30*x^2 + 5*(2*x - 1)*e^(x + 2) - 102*x + 45)/(2*x - 1)))

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giac [A]  time = 0.69, size = 38, normalized size = 1.31 \begin {gather*} x - \log \left (\log \left (-\frac {30 \, x^{2} + 10 \, x e^{\left (x + 2\right )} - 102 \, x - 5 \, e^{\left (x + 2\right )} + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-1
5))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5
)*exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="giac")

[Out]

x - log(log(-1/15*(30*x^2 + 10*x*e^(x + 2) - 102*x - 5*e^(x + 2) + 45)/(2*x - 1)))

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maple [A]  time = 0.22, size = 35, normalized size = 1.21

method	result	size
norman	\(x -\ln \left (\ln \left (\frac {\left (-10 x +5\right ) {\mathrm e}^{2+x}-30 x^{2}+102 x -45}{30 x -15}\right )\right )\)	\(35\)
risch	\(x -\ln \left (\ln \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )+\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x -\frac {1}{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x -\frac {1}{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{3}+2 \pi +2 i \ln \left (x -\frac {1}{2}\right )\right )}{2}\right )\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*ln(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15))+(-2
0*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/ln(((-10*x+5)*exp(2+
x)-30*x^2+102*x-45)/(30*x-15)),x,method=_RETURNVERBOSE)

[Out]

x-ln(ln(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15)))

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maxima [A]  time = 1.47, size = 47, normalized size = 1.62 \begin {gather*} x - \log \left (-\log \relax (5) - \log \relax (3) + \log \left (-30 \, x^{2} - 5 \, {\left (2 \, x e^{2} - e^{2}\right )} e^{x} + 102 \, x - 45\right ) - \log \left (2 \, x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-1
5))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5
)*exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="maxima")

[Out]

x - log(-log(5) - log(3) + log(-30*x^2 - 5*(2*x*e^2 - e^2)*e^x + 102*x - 45) - log(2*x - 1))

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mupad [B]  time = 6.91, size = 35, normalized size = 1.21 \begin {gather*} x-\ln \left (\ln \left (-\frac {30\,x^2-102\,x+{\mathrm {e}}^2\,{\mathrm {e}}^x\,\left (10\,x-5\right )+45}{30\,x-15}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + 2)*(20*x^2 - 20*x + 5) - 60*x - log(-(exp(x + 2)*(10*x - 5) - 102*x + 30*x^2 + 45)/(30*x - 15))*
(192*x + exp(x + 2)*(20*x^2 - 20*x + 5) - 234*x^2 + 60*x^3 - 45) + 60*x^2 + 12)/(log(-(exp(x + 2)*(10*x - 5) -
102*x + 30*x^2 + 45)/(30*x - 15))*(192*x + exp(x + 2)*(20*x^2 - 20*x + 5) - 234*x^2 + 60*x^3 - 45)),x)

[Out]

x - log(log(-(30*x^2 - 102*x + exp(2)*exp(x)*(10*x - 5) + 45)/(30*x - 15)))

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sympy [A]  time = 0.77, size = 29, normalized size = 1.00 \begin {gather*} x - \log {\left (\log {\left (\frac {- 30 x^{2} + 102 x + \left (5 - 10 x\right ) e^{x + 2} - 45}{30 x - 15} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x**2-20*x+5)*exp(2+x)+60*x**3-234*x**2+192*x-45)*ln(((-10*x+5)*exp(2+x)-30*x**2+102*x-45)/(30*
x-15))+(-20*x**2+20*x-5)*exp(2+x)-60*x**2+60*x-12)/((20*x**2-20*x+5)*exp(2+x)+60*x**3-234*x**2+192*x-45)/ln(((
-10*x+5)*exp(2+x)-30*x**2+102*x-45)/(30*x-15)),x)

[Out]

x - log(log((-30*x**2 + 102*x + (5 - 10*x)*exp(x + 2) - 45)/(30*x - 15)))

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