\(\int \frac {(10+e^x (2-2 x)+e^{x^2} (2-4 x^2)+8 \log (2)) \log (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)})}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx\) [617]

Optimal result

Integrand size = 80, antiderivative size = 28 \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=\log ^2\left (\frac {x}{-x+\frac {1}{4} \left (5+e^x+e^{x^2}+x\right )+\log (2)}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=\log ^2\left (\frac {4 x}{e^x+e^{x^2}-3 x+5 \left (1+\frac {4 \log (2)}{5}\right )}\right ) \] Input:

Integrate[((10 + E^x*(2 - 2*x) + E^x^2*(2 - 4*x^2) + 8*Log[2])*Log[(4*x)/( 
5 + E^x + E^x^2 - 3*x + 4*Log[2])])/(5*x + E^x*x + E^x^2*x - 3*x^2 + 4*x*L 
og[2]),x]
 

Output:

Log[(4*x)/(E^x + E^x^2 - 3*x + 5*(1 + (4*Log[2])/5))]^2
 

Rubi [A] (verified)

Time = 2.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 7237}

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.

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 7237

\(\displaystyle \log ^2\left (\frac {4 x}{e^{x^2}-3 x+e^x+5+\log (16)}\right )\)

Input:

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

Output:

Log[(4*x)/(5 + E^x + E^x^2 - 3*x + Log[16])]^2
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 7237

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si 
mp[q*(y^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 5.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\ln \left (\frac {4 x}{{\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \left (2\right )-3 x +5}\right )^{2}\)	\(24\)
risch	\(\ln \left (\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}\right )^{2}-2 \ln \left (x \right ) \ln \left (\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}\right )-i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )+i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{2}+i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{2}-i \pi \ln \left (x \right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{3}+i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \left (2\right )-3 x +5\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )-i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \left (2\right )-3 x +5\right ) \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{2}-i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \left (2\right )-3 x +5\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{2}+i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \left (2\right )-3 x +5\right ) {\operatorname {csgn}\left (\frac {i x}{\ln \left (2\right )-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )}^{3}+\ln \left (x \right )^{2}\)	\(468\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=\log \left (-\frac {4 \, x}{3 \, x - e^{\left (x^{2}\right )} - e^{x} - 4 \, \log \left (2\right ) - 5}\right )^{2} \] Input:

integrate(((-4*x^2+2)*exp(x^2)+(2-2*x)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^ 
2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, a 
lgorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 2.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=\log {\left (\frac {4 x}{- 3 x + e^{x} + e^{x^{2}} + 4 \log {\left (2 \right )} + 5} \right )}^{2} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (27) = 54\).

Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.39 \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=-\log \left (x\right )^{2} + 2 \, \log \left (x\right ) \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \left (2\right ) + 5\right ) - \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \left (2\right ) + 5\right )^{2} + 2 \, {\left (\log \left (x\right ) - \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \left (2\right ) + 5\right )\right )} \log \left (-\frac {4 \, x}{3 \, x - e^{\left (x^{2}\right )} - e^{x} - 4 \, \log \left (2\right ) - 5}\right ) \] Input:

integrate(((-4*x^2+2)*exp(x^2)+(2-2*x)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^ 
2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, a 
lgorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(((-4*x^2+2)*exp(x^2)+(2-2*x)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^ 
2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, a 
lgorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx=\int \frac {\ln \left (\frac {4\,x}{{\mathrm {e}}^{x^2}-3\,x+4\,\ln \left (2\right )+{\mathrm {e}}^x+5}\right )\,\left (8\,\ln \left (2\right )-{\mathrm {e}}^{x^2}\,\left (4\,x^2-2\right )-{\mathrm {e}}^x\,\left (2\,x-2\right )+10\right )}{5\,x+x\,{\mathrm {e}}^{x^2}+4\,x\,\ln \left (2\right )+x\,{\mathrm {e}}^x-3\,x^2} \,d x \] Input:

int((log((4*x)/(exp(x^2) - 3*x + 4*log(2) + exp(x) + 5))*(8*log(2) - exp(x 
^2)*(4*x^2 - 2) - exp(x)*(2*x - 2) + 10))/(5*x + x*exp(x^2) + 4*x*log(2) + 
 x*exp(x) - 3*x^2),x)
 

Output:

int((log((4*x)/(exp(x^2) - 3*x + 4*log(2) + exp(x) + 5))*(8*log(2) - exp(x 
^2)*(4*x^2 - 2) - exp(x)*(2*x - 2) + 10))/(5*x + x*exp(x^2) + 4*x*log(2) + 
 x*exp(x) - 3*x^2), x)
 

Reduce [F]

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

int(((-4*x^2+2)*exp(x^2)+(2-2*x)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^2)+exp 
(x)+4*log(2)-3*x+5))/(exp(x^2)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x)
 

Output:

int(((-4*x^2+2)*exp(x^2)+(2-2*x)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^2)+exp 
(x)+4*log(2)-3*x+5))/(exp(x^2)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x)