3.23.4 \(\int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} (1250-800 x^2+e^{5+x} (625+625 x-775 x^2+400 x^3))}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx\) [2204]

3.23.4.1 Optimal result

Integrand size = 78, antiderivative size = 27 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\frac {2 x+e^{5+x} x}{\left (-1+\frac {4 x}{5}\right )^2+x}} \]

output

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

3.23.4.2 Mathematica [A] (verified)

Time = 2.97 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\frac {25 \left (2+e^{5+x}\right ) x}{25-15 x+16 x^2}} \]

input

Integrate[(E^((50*x + 25*E^(5 + x)*x)/(25 - 15*x + 16*x^2))*(1250 - 800*x^ 
2 + E^(5 + x)*(625 + 625*x - 775*x^2 + 400*x^3)))/(625 - 750*x + 1025*x^2 
- 480*x^3 + 256*x^4),x]
 

output

E^((25*(2 + E^(5 + x))*x)/(25 - 15*x + 16*x^2))
 

3.23.4.3 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.

\(\displaystyle \int \frac {e^{\frac {25 e^{x+5} x+50 x}{16 x^2-15 x+25}} \left (-800 x^2+e^{x+5} \left (400 x^3-775 x^2+625 x+625\right )+1250\right )}{256 x^4-480 x^3+1025 x^2-750 x+625} \, dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (\frac {1024 i e^{\frac {25 e^{x+5} x+50 x}{16 x^2-15 x+25}} \left (-800 x^2+e^{x+5} \left (400 x^3-775 x^2+625 x+625\right )+1250\right )}{6875 \sqrt {55} \left (-32 x+5 i \sqrt {55}+15\right )}+\frac {1024 i e^{\frac {25 e^{x+5} x+50 x}{16 x^2-15 x+25}} \left (-800 x^2+e^{x+5} \left (400 x^3-775 x^2+625 x+625\right )+1250\right )}{6875 \sqrt {55} \left (32 x+5 i \sqrt {55}-15\right )}-\frac {1024 e^{\frac {25 e^{x+5} x+50 x}{16 x^2-15 x+25}} \left (-800 x^2+e^{x+5} \left (400 x^3-775 x^2+625 x+625\right )+1250\right )}{1375 \left (-32 x+5 i \sqrt {55}+15\right )^2}-\frac {1024 e^{\frac {25 e^{x+5} x+50 x}{16 x^2-15 x+25}} \left (-800 x^2+e^{x+5} \left (400 x^3-775 x^2+625 x+625\right )+1250\right )}{1375 \left (32 x+5 i \sqrt {55}-15\right )^2}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {25 e^{\frac {25 \left (e^{x+5}+2\right ) x}{16 x^2-15 x+25}} \left (-32 x^2+e^{x+5} \left (16 x^3-31 x^2+25 x+25\right )+50\right )}{\left (16 x^2-15 x+25\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 25 \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}} \left (-32 x^2+e^{x+5} \left (16 x^3-31 x^2+25 x+25\right )+50\right )}{\left (16 x^2-15 x+25\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 25 \int \left (\frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5} \left (16 x^3-31 x^2+25 x+25\right )}{\left (16 x^2-15 x+25\right )^2}-\frac {2 e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}} \left (16 x^2-25\right )}{\left (16 x^2-15 x+25\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 25 \left (\frac {192}{55} \left (3+i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}}}{\left (-32 x+5 i \sqrt {55}+15\right )^2}dx-\frac {4096}{55} \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}}}{\left (-32 x+5 i \sqrt {55}+15\right )^2}dx+\frac {96}{55} \left (3+i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{\left (-32 x+5 i \sqrt {55}+15\right )^2}dx-\frac {2048}{55} \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{\left (-32 x+5 i \sqrt {55}+15\right )^2}dx+\frac {32 i \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{-32 x+5 i \sqrt {55}+15}dx}{5 \sqrt {55}}+\frac {1}{275} \left (275+17 i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{32 x-5 i \sqrt {55}-15}dx+\frac {192}{55} \left (3-i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}}}{\left (32 x+5 i \sqrt {55}-15\right )^2}dx-\frac {4096}{55} \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}}}{\left (32 x+5 i \sqrt {55}-15\right )^2}dx+\frac {96}{55} \left (3-i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{\left (32 x+5 i \sqrt {55}-15\right )^2}dx-\frac {2048}{55} \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{\left (32 x+5 i \sqrt {55}-15\right )^2}dx+\frac {1}{275} \left (275-17 i \sqrt {55}\right ) \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{32 x+5 i \sqrt {55}-15}dx+\frac {32 i \int \frac {e^{\frac {25 \left (2+e^{x+5}\right ) x}{16 x^2-15 x+25}+x+5}}{32 x+5 i \sqrt {55}-15}dx}{5 \sqrt {55}}\right )\)

input

Int[(E^((50*x + 25*E^(5 + x)*x)/(25 - 15*x + 16*x^2))*(1250 - 800*x^2 + E^ 
(5 + x)*(625 + 625*x - 775*x^2 + 400*x^3)))/(625 - 750*x + 1025*x^2 - 480* 
x^3 + 256*x^4),x]
 

output

$Aborted
 

3.23.4.3.1 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 2009

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

rule 2463

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u, Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt 
Q[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 
0]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

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

3.23.4.4 Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
risch	\({\mathrm e}^{\frac {25 x \left ({\mathrm e}^{5+x}+2\right )}{16 x^{2}-15 x +25}}\)	\(23\)
parallelrisch	\({\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}\)	\(26\)
norman	\(\frac {-15 x \,{\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}+16 x^{2} {\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}+25 \,{\mathrm e}^{\frac {25 x \,{\mathrm e}^{5+x}+50 x}{16 x^{2}-15 x +25}}}{16 x^{2}-15 x +25}\)	\(100\)

input

int(((400*x^3-775*x^2+625*x+625)*exp(5+x)-800*x^2+1250)*exp((25*x*exp(5+x) 
+50*x)/(16*x^2-15*x+25))/(256*x^4-480*x^3+1025*x^2-750*x+625),x,method=_RE 
TURNVERBOSE)
 

output

exp(25*x*(exp(5+x)+2)/(16*x^2-15*x+25))
 

3.23.4.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\left (\frac {25 \, {\left (x e^{\left (x + 5\right )} + 2 \, x\right )}}{16 \, x^{2} - 15 \, x + 25}\right )} \]

input

integrate(((400*x^3-775*x^2+625*x+625)*exp(5+x)-800*x^2+1250)*exp((25*x*ex 
p(5+x)+50*x)/(16*x^2-15*x+25))/(256*x^4-480*x^3+1025*x^2-750*x+625),x, alg 
orithm=\
 

output

e^(25*(x*e^(x + 5) + 2*x)/(16*x^2 - 15*x + 25))
 

3.23.4.6 Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\frac {25 x e^{x + 5} + 50 x}{16 x^{2} - 15 x + 25}} \]

input

integrate(((400*x**3-775*x**2+625*x+625)*exp(5+x)-800*x**2+1250)*exp((25*x 
*exp(5+x)+50*x)/(16*x**2-15*x+25))/(256*x**4-480*x**3+1025*x**2-750*x+625) 
,x)
 

output

exp((25*x*exp(x + 5) + 50*x)/(16*x**2 - 15*x + 25))
 

3.23.4.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\left (\frac {25 \, x e^{\left (x + 5\right )}}{16 \, x^{2} - 15 \, x + 25} + \frac {50 \, x}{16 \, x^{2} - 15 \, x + 25}\right )} \]

input

integrate(((400*x^3-775*x^2+625*x+625)*exp(5+x)-800*x^2+1250)*exp((25*x*ex 
p(5+x)+50*x)/(16*x^2-15*x+25))/(256*x^4-480*x^3+1025*x^2-750*x+625),x, alg 
orithm=\
 

output

e^(25*x*e^(x + 5)/(16*x^2 - 15*x + 25) + 50*x/(16*x^2 - 15*x + 25))
 

3.23.4.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx=e^{\left (\frac {25 \, x e^{\left (x + 5\right )}}{16 \, x^{2} - 15 \, x + 25} + \frac {50 \, x}{16 \, x^{2} - 15 \, x + 25}\right )} \]

input

integrate(((400*x^3-775*x^2+625*x+625)*exp(5+x)-800*x^2+1250)*exp((25*x*ex 
p(5+x)+50*x)/(16*x^2-15*x+25))/(256*x^4-480*x^3+1025*x^2-750*x+625),x, alg 
orithm=\
 

output

e^(25*x*e^(x + 5)/(16*x^2 - 15*x + 25) + 50*x/(16*x^2 - 15*x + 25))
 

3.23.4.9 Mupad [B] (verification not implemented)

Time = 13.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {50 x+25 e^{5+x} x}{25-15 x+16 x^2}} \left (1250-800 x^2+e^{5+x} \left (625+625 x-775 x^2+400 x^3\right )\right )}{625-750 x+1025 x^2-480 x^3+256 x^4} \, dx={\mathrm {e}}^{\frac {50\,x+25\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{16\,x^2-15\,x+25}} \]

input

int((exp((50*x + 25*x*exp(x + 5))/(16*x^2 - 15*x + 25))*(exp(x + 5)*(625*x 
 - 775*x^2 + 400*x^3 + 625) - 800*x^2 + 1250))/(1025*x^2 - 750*x - 480*x^3 
 + 256*x^4 + 625),x)
 

output

exp((50*x + 25*x*exp(5)*exp(x))/(16*x^2 - 15*x + 25))