\(\int \frac {e^{\frac {2 (-e^5 x+x^2+(1-x) \log (4))}{\log (4)}} (-50 e^5 x+100 x^2+(25-50 x) \log (4))}{9 \log (4)} \, dx\) [2987]

Optimal result

Integrand size = 54, antiderivative size = 26 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25}{9} e^{2-2 x+\frac {2 x \left (-e^5+x\right )}{\log (4)}} x \] Output:

25/9*x*exp(1/2*x/ln(2)*(-exp(5)+x)-x+1)^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25 e^{\frac {2 \left (-e^5 x+x^2+\log (4)-x \log (4)\right )}{\log (4)}} x \left (2 e^5-4 x+\log (16)\right )}{18 \left (e^5-2 x+\log (4)\right )} \] Input:

Integrate[(E^((2*(-(E^5*x) + x^2 + (1 - x)*Log[4]))/Log[4])*(-50*E^5*x + 1 
00*x^2 + (25 - 50*x)*Log[4]))/(9*Log[4]),x]
 

Output:

(25*E^((2*(-(E^5*x) + x^2 + Log[4] - x*Log[4]))/Log[4])*x*(2*E^5 - 4*x + L 
og[16]))/(18*(E^5 - 2*x + Log[4]))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(26)=52\).

Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 27, 2725, 7292, 2726}

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.

\(\displaystyle \int \frac {e^{\frac {2 \left (x^2-e^5 x+(1-x) \log (4)\right )}{\log (4)}} \left (100 x^2-50 e^5 x+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int -25 4^{\frac {2 (1-x)}{\log (4)}} e^{-\frac {2 \left (e^5 x-x^2\right )}{\log (4)}} \left (-4 x^2+2 e^5 x-(1-2 x) \log (4)\right )dx}{9 \log (4)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {25 \int 4^{\frac {2 (1-x)}{\log (4)}} e^{-\frac {2 \left (e^5 x-x^2\right )}{\log (4)}} \left (-4 x^2+2 e^5 x-(1-2 x) \log (4)\right )dx}{9 \log (4)}\)

\(\Big \downarrow \) 2725

\(\displaystyle -\frac {25 \int e^{\frac {2 x^2}{\log (4)}-\frac {2 \left (e^5+\log (4)\right ) x}{\log (4)}+2} \left (-4 x^2+2 e^5 x-(1-2 x) \log (4)\right )dx}{9 \log (4)}\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {25 \int e^{\frac {2 x^2}{\log (4)}-\frac {2 \left (e^5+\log (4)\right ) x}{\log (4)}+2} \left (-4 x^2+2 \left (e^5+\log (4)\right ) x-\log (4)\right )dx}{9 \log (4)}\)

\(\Big \downarrow \) 2726

\(\displaystyle -\frac {25 e^{\frac {2 x^2}{\log (4)}-\frac {2 x \left (e^5+\log (4)\right )}{\log (4)}+2} \left (2 x^2-x \left (e^5+\log (4)\right )\right )}{9 \left (-2 x+e^5+\log (4)\right )}\)

Input:

Int[(E^((2*(-(E^5*x) + x^2 + (1 - x)*Log[4]))/Log[4])*(-50*E^5*x + 100*x^2 
 + (25 - 50*x)*Log[4]))/(9*Log[4]),x]
 

Output:

(-25*E^(2 + (2*x^2)/Log[4] - (2*x*(E^5 + Log[4]))/Log[4])*(2*x^2 - x*(E^5 
+ Log[4])))/(9*(E^5 - 2*x + Log[4]))
 

Defintions of rubi rules used

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 2725

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, 
 Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, 
 x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

rule 7292

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {25 x \,{\mathrm e}^{-\frac {2 x \ln \left (2\right )+x \,{\mathrm e}^{5}-x^{2}-2 \ln \left (2\right )}{\ln \left (2\right )}}}{9}\)	\(30\)
norman	\(\frac {25 x \,{\mathrm e}^{\frac {2 \left (1-x \right ) \ln \left (2\right )-x \,{\mathrm e}^{5}+x^{2}}{\ln \left (2\right )}}}{9}\)	\(31\)
parallelrisch	\(\frac {25 x \,{\mathrm e}^{\frac {2 \left (1-x \right ) \ln \left (2\right )-x \,{\mathrm e}^{5}+x^{2}}{\ln \left (2\right )}}}{9}\)	\(31\)
gosper	\(\frac {25 \,{\mathrm e}^{-\frac {2 x \ln \left (2\right )+\ln \left (\frac {1}{4}\right )+x \,{\mathrm e}^{5}-x^{2}}{\ln \left (2\right )}} x}{9}\)	\(32\)
default	\(\frac {50 \,{\mathrm e}^{2} \left (\frac {\ln \left (2\right ) x \,{\mathrm e}^{\frac {x^{2}}{\ln \left (2\right )}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) x}}{2}-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \ln \left (2\right ) \left (\frac {\ln \left (2\right ) {\mathrm e}^{\frac {x^{2}}{\ln \left (2\right )}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) x}}{2}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \ln \left (2\right )^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right )^{2} \ln \left (2\right )}{4}} \operatorname {erf}\left (\frac {i x}{\sqrt {\ln \left (2\right )}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \sqrt {\ln \left (2\right )}}{2}\right )}{4}\right )}{2}+\frac {i \ln \left (2\right )^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right )^{2} \ln \left (2\right )}{4}} \operatorname {erf}\left (\frac {i x}{\sqrt {\ln \left (2\right )}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \sqrt {\ln \left (2\right )}}{2}\right )}{4}\right )-\frac {25 i {\mathrm e}^{2} \ln \left (2\right )^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right )^{2} \ln \left (2\right )}{4}} \operatorname {erf}\left (\frac {i x}{\sqrt {\ln \left (2\right )}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \sqrt {\ln \left (2\right )}}{2}\right )}{2}-25 \,{\mathrm e}^{2} {\mathrm e}^{5} \left (\frac {\ln \left (2\right ) {\mathrm e}^{\frac {x^{2}}{\ln \left (2\right )}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) x}}{2}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \ln \left (2\right )^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right )^{2} \ln \left (2\right )}{4}} \operatorname {erf}\left (\frac {i x}{\sqrt {\ln \left (2\right )}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \sqrt {\ln \left (2\right )}}{2}\right )}{4}\right )-50 \,{\mathrm e}^{2} \ln \left (2\right ) \left (\frac {\ln \left (2\right ) {\mathrm e}^{\frac {x^{2}}{\ln \left (2\right )}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) x}}{2}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \ln \left (2\right )^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right )^{2} \ln \left (2\right )}{4}} \operatorname {erf}\left (\frac {i x}{\sqrt {\ln \left (2\right )}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \left (2\right )}\right ) \sqrt {\ln \left (2\right )}}{2}\right )}{4}\right )}{9 \ln \left (2\right )}\)	\(457\)

Input:

int(1/18*(2*(-50*x+25)*ln(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)*ln(2)-x 
*exp(5)+x^2)/ln(2))^2/ln(2),x,method=_RETURNVERBOSE)
 

Output:

25/9*x*exp(-(2*x*ln(2)+x*exp(5)-x^2-2*ln(2))/ln(2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25}{9} \, x e^{\left (\frac {x^{2} - x e^{5} - 2 \, {\left (x - 1\right )} \log \left (2\right )}{\log \left (2\right )}\right )} \] Input:

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)* 
log(2)-x*exp(5)+x^2)/log(2))^2/log(2),x, algorithm="fricas")
 

Output:

25/9*x*e^((x^2 - x*e^5 - 2*(x - 1)*log(2))/log(2))
 

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25 x e^{\frac {2 \left (\frac {x^{2}}{2} - \frac {x e^{5}}{2} + \frac {\left (2 - 2 x\right ) \log {\left (2 \right )}}{2}\right )}{\log {\left (2 \right )}}}}{9} \] Input:

integrate(1/18*(2*(-50*x+25)*ln(2)-50*x*exp(5)+100*x**2)*exp(1/2*(2*(1-x)* 
ln(2)-x*exp(5)+x**2)/ln(2))**2/ln(2),x)
 

Output:

25*x*exp(2*(x**2/2 - x*exp(5)/2 + (2 - 2*x)*log(2)/2)/log(2))/9
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.19 (sec) , antiderivative size = 639, normalized size of antiderivative = 24.58 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\text {Too large to display} \] Input:

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)* 
log(2)-x*exp(5)+x^2)/log(2))^2/log(2),x, algorithm="maxima")
 

Output:

-25/36*(2*(sqrt(pi)*(2*x/log(2) - (e^5 + 2*log(2))/log(2))*(erf(1/2*sqrt(- 
(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2))) - 1)*(e^5 + 2*log(2))*sq 
rt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2)) + 2*2^(1 
/4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2)*sqrt(log(2)))*e^(-1/4*(e^5 + 
2*log(2))^2/log(2) + 2)*log(2)^(3/2) - 2*sqrt(pi)*erf(x*sqrt(-1/log(2)) + 
1/2*(e^5 + 2*log(2))/(sqrt(-1/log(2))*log(2)))*e^(-1/4*(e^5 + 2*log(2))^2/ 
log(2) + 2)*log(2)/sqrt(-1/log(2)) + (sqrt(pi)*(2*x/log(2) - (e^5 + 2*log( 
2))/log(2))*(erf(1/2*sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2) 
)) - 1)*(e^5 + 2*log(2))*sqrt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*log(2)) 
/log(2))^2*log(2)) + 2*2^(1/4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2)*sq 
rt(log(2)))*e^(-1/4*(e^5 + 2*log(2))^2/log(2) + 7)*sqrt(log(2)) + (4*(2*x/ 
log(2) - (e^5 + 2*log(2))/log(2))^3*gamma(3/2, -1/4*(2*x/log(2) - (e^5 + 2 
*log(2))/log(2))^2*log(2))*log(2)^(5/2)/(-(2*x/log(2) - (e^5 + 2*log(2))/l 
og(2))^2*log(2))^(3/2) - sqrt(pi)*(2*x/log(2) - (e^5 + 2*log(2))/log(2))*( 
erf(1/2*sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2))) - 1)*(e^5 
+ 2*log(2))^2*sqrt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2* 
log(2)) - 4*2^(1/4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2)*(e^5 + 2*log( 
2))*sqrt(log(2)))*e^(-1/4*(e^5 + 2*log(2))^2/log(2) + 2)*sqrt(log(2)))/log 
(2)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 213, normalized size of antiderivative = 8.19 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=-\frac {25 \, {\left (-i \, \sqrt {\pi } {\left (2 \, e^{5} \log \left (2\right ) + e^{10}\right )} \operatorname {erf}\left (-\frac {i \, {\left (2 \, x - e^{5} - 2 \, \log \left (2\right )\right )}}{2 \, \sqrt {\log \left (2\right )}}\right ) e^{\left (-\frac {4 \, e^{5} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + e^{10} - 8 \, \log \left (2\right )}{4 \, \log \left (2\right )}\right )} \sqrt {\log \left (2\right )} + i \, \sqrt {\pi } {\left (e^{5} + 2 \, \log \left (2\right )\right )} \operatorname {erf}\left (-\frac {i \, {\left (2 \, x - e^{5} - 2 \, \log \left (2\right )\right )}}{2 \, \sqrt {\log \left (2\right )}}\right ) e^{\left (-\frac {4 \, e^{5} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + e^{10} - 28 \, \log \left (2\right )}{4 \, \log \left (2\right )}\right )} \sqrt {\log \left (2\right )} - 2 \, {\left ({\left (2 \, x - e^{5} - 2 \, \log \left (2\right )\right )} \log \left (2\right ) + 2 \, e^{5} \log \left (2\right ) + 2 \, \log \left (2\right )^{2}\right )} e^{\left (\frac {x^{2} - x e^{5} - 2 \, x \log \left (2\right ) + 2 \, \log \left (2\right )}{\log \left (2\right )}\right )} + 2 \, e^{\left (\frac {x^{2} - x e^{5} - 2 \, x \log \left (2\right ) + 7 \, \log \left (2\right )}{\log \left (2\right )}\right )} \log \left (2\right )\right )}}{36 \, \log \left (2\right )} \] Input:

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)* 
log(2)-x*exp(5)+x^2)/log(2))^2/log(2),x, algorithm="giac")
 

Output:

-25/36*(-I*sqrt(pi)*(2*e^5*log(2) + e^10)*erf(-1/2*I*(2*x - e^5 - 2*log(2) 
)/sqrt(log(2)))*e^(-1/4*(4*e^5*log(2) + 4*log(2)^2 + e^10 - 8*log(2))/log( 
2))*sqrt(log(2)) + I*sqrt(pi)*(e^5 + 2*log(2))*erf(-1/2*I*(2*x - e^5 - 2*l 
og(2))/sqrt(log(2)))*e^(-1/4*(4*e^5*log(2) + 4*log(2)^2 + e^10 - 28*log(2) 
)/log(2))*sqrt(log(2)) - 2*((2*x - e^5 - 2*log(2))*log(2) + 2*e^5*log(2) + 
 2*log(2)^2)*e^((x^2 - x*e^5 - 2*x*log(2) + 2*log(2))/log(2)) + 2*e^((x^2 
- x*e^5 - 2*x*log(2) + 7*log(2))/log(2))*log(2))/log(2)
 

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25\,x\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {x^2}{\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{\ln \left (2\right )}}}{9} \] Input:

int(-(exp(-(2*((x*exp(5))/2 + log(2)*(x - 1) - x^2/2))/log(2))*(2*log(2)*( 
50*x - 25) + 50*x*exp(5) - 100*x^2))/(18*log(2)),x)
 

Output:

(25*x*exp(-2*x)*exp(2)*exp(x^2/log(2))*exp(-(x*exp(5))/log(2)))/9
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right )}{9 \log (4)} \, dx=\frac {25 e^{\frac {x^{2}}{\mathrm {log}\left (2\right )}} e^{2} x}{9 e^{\frac {2 \,\mathrm {log}\left (2\right ) x +e^{5} x}{\mathrm {log}\left (2\right )}}} \] Input:

int(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)*log(2) 
-x*exp(5)+x^2)/log(2))^2/log(2),x)
 

Output:

(25*e**(x**2/log(2))*e**2*x)/(9*e**((2*log(2)*x + e**5*x)/log(2)))