3.53.5 \(\int \frac {(-10+e^{4 x} (75+270 x-117 x^2+12 x^3)) \log (x)+(-10+2 x+e^{4 x} (75-30 x+3 x^2)) \log (\frac {e^e (2 x+e^{4 x} (-15 x+3 x^2))}{-5+x})}{-50 x+10 x^2+e^{4 x} (375 x-150 x^2+15 x^3)} \, dx\) [5205]

Optimal. Leaf size=29 \[ \frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}-\frac {2}{5-x}\right ) x\right ) \]

[Out]

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

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Rubi [F]
time = 0.99, 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 {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-10 + E^(4*x)*(75 + 270*x - 117*x^2 + 12*x^3))*Log[x] + (-10 + 2*x + E^(4*x)*(75 - 30*x + 3*x^2))*Log[(E
^E*(2*x + E^(4*x)*(-15*x + 3*x^2)))/(-5 + x)])/(-50*x + 10*x^2 + E^(4*x)*(375*x - 150*x^2 + 15*x^3)),x]

[Out]

(-4*x)/5 + (4*x*Log[x])/5 + Log[x]^2/10 - (8*Log[x]*Defer[Int][(2 + 3*E^(4*x)*(-5 + x))^(-1), x])/5 - (2*Log[x
]*Defer[Int][1/((-5 + x)*(2 - 15*E^(4*x) + 3*E^(4*x)*x)), x])/5 + Defer[Int][Log[E^E*(3*E^(4*x) + 2/(-5 + x))*
x]/x, x]/5 + (8*Defer[Int][Defer[Int][(2 + 3*E^(4*x)*(-5 + x))^(-1), x]/x, x])/5 + (2*Defer[Int][Defer[Int][1/
((2 + 3*E^(4*x)*(-5 + x))*(-5 + x)), x]/x, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{5 x} \, dx\\ &=\frac {1}{5} \int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 (-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )}+\frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx-\frac {2}{5} \int \frac {(-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx\\ &=\frac {1}{5} \int \left (\frac {(1+4 x) \log (x)}{x}+\frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx+\frac {2}{5} \int \frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx+\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=\frac {1}{5} \int \frac {(1+4 x) \log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \left (\frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x}+\frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x}\right ) \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=\frac {1}{5} \int \frac {\log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {4}{5} \int \log (x) \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=-\frac {4 x}{5}+\frac {4}{5} x \log (x)+\frac {\log ^2(x)}{10}+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.19, size = 27, normalized size = 0.93 \begin {gather*} \frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-10 + E^(4*x)*(75 + 270*x - 117*x^2 + 12*x^3))*Log[x] + (-10 + 2*x + E^(4*x)*(75 - 30*x + 3*x^2))*
Log[(E^E*(2*x + E^(4*x)*(-15*x + 3*x^2)))/(-5 + x)])/(-50*x + 10*x^2 + E^(4*x)*(375*x - 150*x^2 + 15*x^3)),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.58, size = 405, normalized size = 13.97

method	result	size
risch	\(\frac {\ln \left (x \right ) \ln \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{5}-\frac {\ln \left (x -5\right ) \ln \left (x \right )}{5}+\frac {\ln \left (x \right )^{2}}{5}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )}{10}+\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i}{x -5}\right )}{10}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{3}}{10}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )}{10}-\frac {i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{3}}{10}+\frac {\ln \left (3\right ) \ln \left (x \right )}{5}+\frac {{\mathrm e} \ln \left (x \right )}{5}\)	\(405\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*ln(((3*x^2-15*x)*exp(4*x)+2*x)*exp(exp(1))/(x-5))+((12*x^3-117*x^2+270*
x+75)*exp(4*x)-10)*ln(x))/((15*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x,method=_RETURNVERBOSE)

[Out]

1/5*ln(x)*ln(x*exp(4*x)-5*exp(4*x)+2/3)-1/5*ln(x-5)*ln(x)+1/5*ln(x)^2+1/10*I*Pi*ln(x)*csgn(I*(x*exp(4*x)-5*exp
(4*x)+2/3))*csgn(I/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))^2+1/10*I*Pi*ln(x)*csgn(I/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/
3))*csgn(I*x/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))^2+1/10*I*Pi*ln(x)*csgn(I/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))^2*
csgn(I/(x-5))+1/10*I*Pi*ln(x)*csgn(I*x)*csgn(I*x/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))^2-1/10*I*Pi*ln(x)*csgn(I*(
x*exp(4*x)-5*exp(4*x)+2/3))*csgn(I/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))*csgn(I/(x-5))-1/10*I*Pi*ln(x)*csgn(I/(x-
5)*(x*exp(4*x)-5*exp(4*x)+2/3))^3-1/10*I*Pi*ln(x)*csgn(I*x)*csgn(I/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))*csgn(I*x
/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))-1/10*I*Pi*ln(x)*csgn(I*x/(x-5)*(x*exp(4*x)-5*exp(4*x)+2/3))^3+1/5*ln(3)*ln
(x)+1/5*exp(1)*ln(x)

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Maxima [A]
time = 0.33, size = 37, normalized size = 1.28 \begin {gather*} \frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2*x)*exp(exp(1))/(-5+x))+((12*x^3-117*
x^2+270*x+75)*exp(4*x)-10)*log(x))/((15*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 31, normalized size = 1.07 \begin {gather*} \frac {1}{5} \, \log \left (x\right ) \log \left (\frac {{\left (3 \, {\left (x^{2} - 5 \, x\right )} e^{\left (4 \, x\right )} + 2 \, x\right )} e^{e}}{x - 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2*x)*exp(exp(1))/(-5+x))+((12*x^3-117*
x^2+270*x+75)*exp(4*x)-10)*log(x))/((15*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.53, size = 31, normalized size = 1.07 \begin {gather*} \frac {\log {\left (x \right )} \log {\left (\frac {\left (2 x + \left (3 x^{2} - 15 x\right ) e^{4 x}\right ) e^{e}}{x - 5} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2-30*x+75)*exp(4*x)+2*x-10)*ln(((3*x**2-15*x)*exp(4*x)+2*x)*exp(exp(1))/(-5+x))+((12*x**3-11
7*x**2+270*x+75)*exp(4*x)-10)*ln(x))/((15*x**3-150*x**2+375*x)*exp(4*x)+10*x**2-50*x),x)

[Out]

log(x)*log((2*x + (3*x**2 - 15*x)*exp(4*x))*exp(E)/(x - 5))/5

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Giac [A]
time = 0.45, size = 41, normalized size = 1.41 \begin {gather*} \frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, x e^{\left (4 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*exp(4*x)+2*x-10)*log(((3*x^2-15*x)*exp(4*x)+2*x)*exp(exp(1))/(-5+x))+((12*x^3-117*
x^2+270*x+75)*exp(4*x)-10)*log(x))/((15*x^3-150*x^2+375*x)*exp(4*x)+10*x^2-50*x),x, algorithm="giac")

[Out]

1/5*e*log(x) + 1/5*log(3*x*e^(4*x) - 15*e^(4*x) + 2)*log(x) - 1/5*log(x - 5)*log(x) + 1/5*log(x)^2

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Mupad [B]
time = 3.54, size = 33, normalized size = 1.14 \begin {gather*} \frac {\ln \left (x\right )\,\left (\ln \left (\frac {2\,x-{\mathrm {e}}^{4\,x}\,\left (15\,x-3\,x^2\right )}{x-5}\right )+\mathrm {e}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((exp(exp(1))*(2*x - exp(4*x)*(15*x - 3*x^2)))/(x - 5))*(2*x + exp(4*x)*(3*x^2 - 30*x + 75) - 10) + lo
g(x)*(exp(4*x)*(270*x - 117*x^2 + 12*x^3 + 75) - 10))/(exp(4*x)*(375*x - 150*x^2 + 15*x^3) - 50*x + 10*x^2),x)

[Out]

(log(x)*(log((2*x - exp(4*x)*(15*x - 3*x^2))/(x - 5)) + exp(1)))/5

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