3.4.75 \(\int \frac {(-2 x^3+10 x^4) \log ^2(\frac {4}{x})-8 x^3 \log ^2(\frac {4}{x}) \log (x)+e^{\frac {1}{2} (-10+e^{\frac {3}{\log (\frac {4}{x})}})} ((-2 x+6 x^2) \log ^2(\frac {4}{x})-4 x \log ^2(\frac {4}{x}) \log (x)+e^{\frac {3}{\log (\frac {4}{x})}} (3 x^2-3 x \log (x)))}{2 \log ^2(\frac {4}{x})} \, dx\) [375]

3.4.75.1 Optimal result

Integrand size = 125, antiderivative size = 34 \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=x^2 \left (e^{-5+\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}}+x^2\right ) (x-\log (x)) \]

output

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

3.4.75.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=x^5+e^{-5+\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} x^2 (x-\log (x))-x^4 \log (x) \]

input

Integrate[((-2*x^3 + 10*x^4)*Log[4/x]^2 - 8*x^3*Log[4/x]^2*Log[x] + E^((-1 
0 + E^(3/Log[4/x]))/2)*((-2*x + 6*x^2)*Log[4/x]^2 - 4*x*Log[4/x]^2*Log[x] 
+ E^(3/Log[4/x])*(3*x^2 - 3*x*Log[x])))/(2*Log[4/x]^2),x]
 

output

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

3.4.75.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (e^{\frac {3}{\log \left (\frac {4}{x}\right )}}-10\right )} \left (\left (6 x^2-2 x\right ) \log ^2\left (\frac {4}{x}\right )+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )-4 x \log (x) \log ^2\left (\frac {4}{x}\right )\right )+\left (10 x^4-2 x^3\right ) \log ^2\left (\frac {4}{x}\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {8 \log ^2\left (\frac {4}{x}\right ) \log (x) x^3+2 \left (x^3-5 x^4\right ) \log ^2\left (\frac {4}{x}\right )+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (2 \left (x-3 x^2\right ) \log ^2\left (\frac {4}{x}\right )+4 x \log (x) \log ^2\left (\frac {4}{x}\right )-3 e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (x^2-x \log (x)\right )\right )}{\log ^2\left (\frac {4}{x}\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {8 \log ^2\left (\frac {4}{x}\right ) \log (x) x^3+2 \left (x^3-5 x^4\right ) \log ^2\left (\frac {4}{x}\right )+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (2 \left (x-3 x^2\right ) \log ^2\left (\frac {4}{x}\right )+4 x \log (x) \log ^2\left (\frac {4}{x}\right )-3 e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (x^2-x \log (x)\right )\right )}{\log ^2\left (\frac {4}{x}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (-\frac {3 e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}-5+\frac {3}{\log \left (\frac {4}{x}\right )}} x (x-\log (x))}{\log ^2\left (\frac {4}{x}\right )}-\frac {2 x \left (5 e^5 x^3-4 e^5 \log (x) x^2-e^5 x^2+3 e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} x-e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}}-2 e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} \log (x)\right )}{e^5}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (3 \int \frac {e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}-5+\frac {3}{\log \left (\frac {4}{x}\right )}} x^2}{\log ^2\left (\frac {4}{x}\right )}dx+\frac {6 \int e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} x^2dx}{e^5}-3 \int \frac {e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}-5+\frac {3}{\log \left (\frac {4}{x}\right )}} x \log (x)}{\log ^2\left (\frac {4}{x}\right )}dx-\frac {2 \int e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} xdx}{e^5}-\frac {4 \int e^{\frac {1}{2} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}} x \log (x)dx}{e^5}+2 x^5-2 x^4 \log (x)\right )\)

input

Int[((-2*x^3 + 10*x^4)*Log[4/x]^2 - 8*x^3*Log[4/x]^2*Log[x] + E^((-10 + E^ 
(3/Log[4/x]))/2)*((-2*x + 6*x^2)*Log[4/x]^2 - 4*x*Log[4/x]^2*Log[x] + E^(3 
/Log[4/x])*(3*x^2 - 3*x*Log[x])))/(2*Log[4/x]^2),x]
 

output

$Aborted
 

3.4.75.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

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

3.4.75.4 Maple [A] (verified)

Time = 12.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21

method	result	size
risch	\(x^{5}-x^{4} \ln \left (x \right )+x^{2} \left (x -\ln \left (x \right )\right ) {\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{2 \ln \left (2\right )-\ln \left (x \right )}}}{2}-5}\)	\(41\)
parallelrisch	\(-x^{4} \ln \left (x \right )+x^{5}-x^{2} \ln \left (x \right ) {\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{\ln \left (\frac {4}{x}\right )}}}{2}-5}+{\mathrm e}^{\frac {{\mathrm e}^{\frac {3}{\ln \left (\frac {4}{x}\right )}}}{2}-5} x^{3}\)	\(55\)

input

int(1/2*(((-3*x*ln(x)+3*x^2)*exp(3/ln(4/x))-4*x*ln(4/x)^2*ln(x)+(6*x^2-2*x 
)*ln(4/x)^2)*exp(1/2*exp(3/ln(4/x))-5)-8*x^3*ln(4/x)^2*ln(x)+(10*x^4-2*x^3 
)*ln(4/x)^2)/ln(4/x)^2,x,method=_RETURNVERBOSE)
 

output

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

3.4.75.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=x^{5} - 2 \, x^{4} \log \left (2\right ) + x^{4} \log \left (\frac {4}{x}\right ) + {\left (x^{3} - 2 \, x^{2} \log \left (2\right ) + x^{2} \log \left (\frac {4}{x}\right )\right )} e^{\left (\frac {1}{2} \, e^{\frac {3}{\log \left (\frac {4}{x}\right )}} - 5\right )} \]

input

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+ 
(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x) 
+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm=\
 

output

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

3.4.75.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=\text {Timed out} \]

input

integrate(1/2*(((-3*x*ln(x)+3*x**2)*exp(3/ln(4/x))-4*x*ln(4/x)**2*ln(x)+(6 
*x**2-2*x)*ln(4/x)**2)*exp(1/2*exp(3/ln(4/x))-5)-8*x**3*ln(4/x)**2*ln(x)+( 
10*x**4-2*x**3)*ln(4/x)**2)/ln(4/x)**2,x)
 

output

Timed out
 

3.4.75.7 Maxima [F]

\[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=\int { -\frac {8 \, x^{3} \log \left (x\right ) \log \left (\frac {4}{x}\right )^{2} - 2 \, {\left (5 \, x^{4} - x^{3}\right )} \log \left (\frac {4}{x}\right )^{2} + {\left (4 \, x \log \left (x\right ) \log \left (\frac {4}{x}\right )^{2} - 2 \, {\left (3 \, x^{2} - x\right )} \log \left (\frac {4}{x}\right )^{2} - 3 \, {\left (x^{2} - x \log \left (x\right )\right )} e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} e^{\left (\frac {1}{2} \, e^{\frac {3}{\log \left (\frac {4}{x}\right )}} - 5\right )}}{2 \, \log \left (\frac {4}{x}\right )^{2}} \,d x } \]

input

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+ 
(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x) 
+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm=\
 

output

x^5 - x^4*log(x) - 1/2*integrate(-(3*x^2 + 2*(12*x^2*log(2)^2 - 2*x*log(x) 
^3 - 4*x*log(2)^2 + (3*x^2 + x*(8*log(2) - 1))*log(x)^2 - 4*(3*x^2*log(2) 
+ (2*log(2)^2 - log(2))*x)*log(x))*e^(-3/(2*log(2) - log(x))) - 3*x*log(x) 
)*e^(3/(2*log(2) - log(x)) + 1/2*e^(3/(2*log(2) - log(x))))/(4*e^5*log(2)^ 
2 - 4*e^5*log(2)*log(x) + e^5*log(x)^2), x)
 

3.4.75.8 Giac [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76 \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=x^{5} - x^{4} \log \left (x\right ) + x^{3} e^{\left (\frac {1}{2} \, e^{\left (\frac {3}{2 \, \log \left (2\right ) - \log \left (x\right )}\right )} - 5\right )} - x^{2} e^{\left (\frac {1}{2} \, e^{\left (\frac {3}{2 \, \log \left (2\right ) - \log \left (x\right )}\right )} - 5\right )} \log \left (x\right ) \]

input

integrate(1/2*(((-3*x*log(x)+3*x^2)*exp(3/log(4/x))-4*x*log(4/x)^2*log(x)+ 
(6*x^2-2*x)*log(4/x)^2)*exp(1/2*exp(3/log(4/x))-5)-8*x^3*log(4/x)^2*log(x) 
+(10*x^4-2*x^3)*log(4/x)^2)/log(4/x)^2,x, algorithm=\
 

output

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

3.4.75.9 Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-2 x^3+10 x^4\right ) \log ^2\left (\frac {4}{x}\right )-8 x^3 \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {1}{2} \left (-10+e^{\frac {3}{\log \left (\frac {4}{x}\right )}}\right )} \left (\left (-2 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )-4 x \log ^2\left (\frac {4}{x}\right ) \log (x)+e^{\frac {3}{\log \left (\frac {4}{x}\right )}} \left (3 x^2-3 x \log (x)\right )\right )}{2 \log ^2\left (\frac {4}{x}\right )} \, dx=x^5-x^4\,\ln \left (x\right )-{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {3}{\ln \left (\frac {4}{x}\right )}}}{2}-5}\,\left (x^2\,\ln \left (x\right )-x^3\right ) \]

input

int(-((log(4/x)^2*(2*x^3 - 10*x^4))/2 + (exp(exp(3/log(4/x))/2 - 5)*(exp(3 
/log(4/x))*(3*x*log(x) - 3*x^2) + log(4/x)^2*(2*x - 6*x^2) + 4*x*log(4/x)^ 
2*log(x)))/2 + 4*x^3*log(4/x)^2*log(x))/log(4/x)^2,x)
 

output

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