\(\int (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))) \, dx\) [2420]

Optimal result

Integrand size = 98, antiderivative size = 23 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=-5^{e^{4+e^{(-1+4 x+\log (100))^2}}}+x^2 \] Output:

x^2-exp(ln(5)*exp(exp((4*x+2*ln(10)-1)^2)+4))
 

Mathematica [F]

\[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=\int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx \] Input:

Integrate[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[1 
00]^2))*E^(5 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2) - 8 
*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*Log 
[5]*Log[100]),x]
 

Output:

Integrate[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[1 
00]^2))*E^(5 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2) - 8 
*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*Log 
[5]*Log[100]), x]
 

Rubi [A] (verified)

Time = 5.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2009}

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.

\(\displaystyle \int \left (((8-32 x) \log (5)-8 \log (5) \log (100)) 5^{\exp \left (e^{16 x^2-8 x+(8 x-2) \log (100)+1+\log ^2(100)}+4\right )} \exp \left (16 x^2+e^{16 x^2-8 x+(8 x-2) \log (100)+1+\log ^2(100)}-8 x+(8 x-2) \log (100)+5+\log ^2(100)\right )+2 x\right ) \, dx\)

\(\Big \downarrow \) 2009

\(\displaystyle x^2-5^{e^{100^{8 x-2} e^{16 x^2-8 x+1+\log ^2(100)}+4}}\)

Input:

Int[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2) 
)*E^(5 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2) - 8*x + 1 
6*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*Log[5]*Lo 
g[100]),x]
 

Output:

-5^E^(4 + 100^(-2 + 8*x)*E^(1 - 8*x + 16*x^2 + Log[100]^2)) + x^2
 

Defintions of rubi rules used

rule 2009

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

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22

method	result	size
risch	\(-5^{{\mathrm e}^{{\mathrm e}^{\left (2 \ln \left (2\right )+2 \ln \left (5\right )+4 x -1\right )^{2}}+4}}+x^{2}\)	\(28\)
default	\(-{\mathrm e}^{\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)
norman	\(-{\mathrm e}^{\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)
parallelrisch	\(-{\mathrm e}^{\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)
parts	\(-{\mathrm e}^{\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)

Input:

int((-16*ln(5)*ln(10)+(-32*x+8)*ln(5))*exp(4*ln(10)^2+2*(8*x-2)*ln(10)+16* 
x^2-8*x+1)*exp(exp(4*ln(10)^2+2*(8*x-2)*ln(10)+16*x^2-8*x+1)+4)*exp(ln(5)* 
exp(exp(4*ln(10)^2+2*(8*x-2)*ln(10)+16*x^2-8*x+1)+4))+2*x,x,method=_RETURN 
VERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=x^{2} - 5^{e^{\left (e^{\left (16 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} \log \left (10\right ) + 4 \, \log \left (10\right )^{2} - 8 \, x + 1\right )} + 4\right )}} \] Input:

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)* 
log(10)+16*x^2-8*x+1)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+ 
4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*x, 
x, algorithm="fricas")
 

Output:

x^2 - 5^e^(e^(16*x^2 + 4*(4*x - 1)*log(10) + 4*log(10)^2 - 8*x + 1) + 4)
 

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=x^{2} - e^{e^{e^{16 x^{2} - 8 x + \left (16 x - 4\right ) \log {\left (10 \right )} + 1 + 4 \log {\left (10 \right )}^{2}} + 4} \log {\left (5 \right )}} \] Input:

integrate((-16*ln(5)*ln(10)+(-32*x+8)*ln(5))*exp(4*ln(10)**2+2*(8*x-2)*ln( 
10)+16*x**2-8*x+1)*exp(exp(4*ln(10)**2+2*(8*x-2)*ln(10)+16*x**2-8*x+1)+4)* 
exp(ln(5)*exp(exp(4*ln(10)**2+2*(8*x-2)*ln(10)+16*x**2-8*x+1)+4))+2*x,x)
 

Output:

x**2 - exp(exp(exp(16*x**2 - 8*x + (16*x - 4)*log(10) + 1 + 4*log(10)**2) 
+ 4)*log(5))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (23) = 46\).

Time = 0.62 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=x^{2} - 5^{e^{\left (\frac {1}{625} \cdot 2^{8 \, \log \left (5\right ) - 4} e^{\left (16 \, x^{2} + 16 \, x \log \left (5\right ) + 4 \, \log \left (5\right )^{2} + 16 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 8 \, x + 1\right )} + 4\right )}} \] Input:

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)* 
log(10)+16*x^2-8*x+1)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+ 
4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*x, 
x, algorithm="maxima")
 

Output:

x^2 - 5^e^(1/625*2^(8*log(5) - 4)*e^(16*x^2 + 16*x*log(5) + 4*log(5)^2 + 1 
6*x*log(2) + 4*log(2)^2 - 8*x + 1) + 4)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=x^{2} - 5^{e^{\left (e^{\left (16 \, x^{2} + 16 \, x \log \left (10\right ) + 4 \, \log \left (10\right )^{2} - 8 \, x - 4 \, \log \left (10\right ) + 1\right )} + 4\right )}} \] Input:

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)* 
log(10)+16*x^2-8*x+1)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+ 
4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*x, 
x, algorithm="giac")
 

Output:

x^2 - 5^e^(e^(16*x^2 + 16*x*log(10) + 4*log(10)^2 - 8*x - 4*log(10) + 1) + 
 4)
 

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=x^2-5^{{\mathrm {e}}^{\frac {{10}^{16\,x}\,{\mathrm {e}}^{-8\,x}\,\mathrm {e}\,{\mathrm {e}}^{4\,{\ln \left (10\right )}^2}\,{\mathrm {e}}^{16\,x^2}}{10000}+4}} \] Input:

int(2*x - exp(2*log(10)*(8*x - 2) - 8*x + 4*log(10)^2 + 16*x^2 + 1)*exp(ex 
p(2*log(10)*(8*x - 2) - 8*x + 4*log(10)^2 + 16*x^2 + 1) + 4)*exp(exp(exp(2 
*log(10)*(8*x - 2) - 8*x + 4*log(10)^2 + 16*x^2 + 1) + 4)*log(5))*(log(5)* 
(32*x - 8) + 16*log(5)*log(10)),x)
 

Output:

x^2 - 5^exp((10^(16*x)*exp(-8*x)*exp(1)*exp(4*log(10)^2)*exp(16*x^2))/1000 
0 + 4)
 

Reduce [F]

\[ \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx=-\frac {2 e^{4 \mathrm {log}\left (10\right )^{2}} \left (\int \frac {e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +10000 e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +80000 e^{8 x} x}{10000 e^{8 x}}} \mathrm {log}\left (5\right ) e^{4}+160000 e^{8 x} x^{2}}{10000 e^{8 x}}} 10^{16 x} x}{e^{8 x}}d x \right ) \mathrm {log}\left (5\right ) e^{5}}{625}-\frac {e^{4 \mathrm {log}\left (10\right )^{2}} \left (\int \frac {e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +10000 e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +80000 e^{8 x} x}{10000 e^{8 x}}} \mathrm {log}\left (5\right ) e^{4}+160000 e^{8 x} x^{2}}{10000 e^{8 x}}} 10^{16 x}}{e^{8 x}}d x \right ) \mathrm {log}\left (10\right ) \mathrm {log}\left (5\right ) e^{5}}{625}+\frac {e^{4 \mathrm {log}\left (10\right )^{2}} \left (\int \frac {e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +10000 e^{\frac {e^{4 \mathrm {log}\left (10\right )^{2}+16 x^{2}} 10^{16 x} e +80000 e^{8 x} x}{10000 e^{8 x}}} \mathrm {log}\left (5\right ) e^{4}+160000 e^{8 x} x^{2}}{10000 e^{8 x}}} 10^{16 x}}{e^{8 x}}d x \right ) \mathrm {log}\left (5\right ) e^{5}}{1250}+x^{2} \] Input:

int((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)*log(10 
)+16*x^2-8*x+1)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4)*exp 
(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*x,x)
 

Output:

( - 4*e**(4*log(10)**2)*int((e**((e**(4*log(10)**2 + 16*x**2)*10**(16*x)*e 
 + 10000*e**((e**(4*log(10)**2 + 16*x**2)*10**(16*x)*e + 80000*e**(8*x)*x) 
/(10000*e**(8*x)))*log(5)*e**4 + 160000*e**(8*x)*x**2)/(10000*e**(8*x)))*1 
0**(16*x)*x)/e**(8*x),x)*log(5)*e**5 - 2*e**(4*log(10)**2)*int((e**((e**(4 
*log(10)**2 + 16*x**2)*10**(16*x)*e + 10000*e**((e**(4*log(10)**2 + 16*x** 
2)*10**(16*x)*e + 80000*e**(8*x)*x)/(10000*e**(8*x)))*log(5)*e**4 + 160000 
*e**(8*x)*x**2)/(10000*e**(8*x)))*10**(16*x))/e**(8*x),x)*log(10)*log(5)*e 
**5 + e**(4*log(10)**2)*int((e**((e**(4*log(10)**2 + 16*x**2)*10**(16*x)*e 
 + 10000*e**((e**(4*log(10)**2 + 16*x**2)*10**(16*x)*e + 80000*e**(8*x)*x) 
/(10000*e**(8*x)))*log(5)*e**4 + 160000*e**(8*x)*x**2)/(10000*e**(8*x)))*1 
0**(16*x))/e**(8*x),x)*log(5)*e**5 + 1250*x**2)/1250