3.3.87 \(\int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} (3 x^2-3 x^3+e^{3/x} (12 x^2-4 x^3)+e^{6/x} (6 x^3-2 x^4))}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx\) [287]

3.3.87.1 Optimal result

Integrand size = 156, antiderivative size = 28 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {e^{-3 x} x^3}{5+e^{\left (2+e^{3/x} x\right )^2}} \]

output

x^3/exp(x)^3/(exp((2+x*exp(3/x))^2)+5)
 

3.3.87.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {e^{-3 x} x^3}{5+e^{\left (2+e^{3/x} x\right )^2}} \]

input

Integrate[(15*x^2 - 15*x^3 + E^(4 + 4*E^(3/x)*x + E^(6/x)*x^2)*(3*x^2 - 3* 
x^3 + E^(3/x)*(12*x^2 - 4*x^3) + E^(6/x)*(6*x^3 - 2*x^4)))/(25*E^(3*x) + 1 
0*E^(4 + 3*x + 4*E^(3/x)*x + E^(6/x)*x^2) + E^(8 + 3*x + 8*E^(3/x)*x + 2*E 
^(6/x)*x^2)),x]
 

output

x^3/(E^(3*x)*(5 + E^(2 + E^(3/x)*x)^2))
 

3.3.87.3 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.

\(\displaystyle \int \frac {-15 x^3+15 x^2+e^{e^{6/x} x^2+4 e^{3/x} x+4} \left (-3 x^3+3 x^2+e^{6/x} \left (6 x^3-2 x^4\right )+e^{3/x} \left (12 x^2-4 x^3\right )\right )}{10 e^{e^{6/x} x^2+4 e^{3/x} x+3 x+4}+e^{2 e^{6/x} x^2+8 e^{3/x} x+3 x+8}+25 e^{3 x}} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-3 x} \left (-15 x^3+15 x^2+e^{e^{6/x} x^2+4 e^{3/x} x+4} \left (-3 x^3+3 x^2+e^{6/x} \left (6 x^3-2 x^4\right )+e^{3/x} \left (12 x^2-4 x^3\right )\right )\right )}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {15 e^{-3 x} x^3}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}-\frac {3 e^{\left (e^{3/x} x+2\right )^2-3 x} x^3}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}+\frac {15 e^{-3 x} x^2}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}+\frac {12 e^{e^{6/x} x^2+4 e^{3/x} x-3 x+\frac {3}{x}+4} x^2}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}+\frac {3 e^{\left (e^{3/x} x+2\right )^2-3 x} x^2}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}-\frac {2 e^{e^{6/x} x^2+4 e^{3/x} x-3 x+\frac {6}{x}+4} x^4}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}-\frac {4 e^{e^{6/x} x^2+4 e^{3/x} x-3 x+\frac {3}{x}+4} x^3}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}+\frac {6 e^{e^{6/x} x^2+4 e^{3/x} x-3 x+\frac {6}{x}+4} x^3}{\left (e^{\left (e^{3/x} x+2\right )^2}+5\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -15 \int \frac {e^{-3 x} x^3}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx-3 \int \frac {e^{\left (e^{3/x} x+2\right )^2-3 x} x^3}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx+15 \int \frac {e^{-3 x} x^2}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx+12 \int \frac {e^{e^{6/x} x^2+4 e^{3/x} x-3 x+4+\frac {3}{x}} x^2}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx+3 \int \frac {e^{\left (e^{3/x} x+2\right )^2-3 x} x^2}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx-2 \int \frac {e^{e^{6/x} x^2+4 e^{3/x} x-3 x+4+\frac {6}{x}} x^4}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx-4 \int \frac {e^{e^{6/x} x^2+4 e^{3/x} x-3 x+4+\frac {3}{x}} x^3}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx+6 \int \frac {e^{e^{6/x} x^2+4 e^{3/x} x-3 x+4+\frac {6}{x}} x^3}{\left (5+e^{\left (e^{3/x} x+2\right )^2}\right )^2}dx\)

input

Int[(15*x^2 - 15*x^3 + E^(4 + 4*E^(3/x)*x + E^(6/x)*x^2)*(3*x^2 - 3*x^3 + 
E^(3/x)*(12*x^2 - 4*x^3) + E^(6/x)*(6*x^3 - 2*x^4)))/(25*E^(3*x) + 10*E^(4 
 + 3*x + 4*E^(3/x)*x + E^(6/x)*x^2) + E^(8 + 3*x + 8*E^(3/x)*x + 2*E^(6/x) 
*x^2)),x]
 

output

$Aborted
 

3.3.87.3.1 Defintions of rubi rules used

rule 2009

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

rule 7292

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

rule 7293

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

3.3.87.4 Maple [A] (verified)

Time = 3.76 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25

method	result	size
risch	\(\frac {x^{3} {\mathrm e}^{-3 x}}{{\mathrm e}^{x^{2} {\mathrm e}^{\frac {6}{x}}+4 x \,{\mathrm e}^{\frac {3}{x}}+4}+5}\)	\(35\)
parallelrisch	\(\frac {x^{3} {\mathrm e}^{-3 x}}{{\mathrm e}^{x^{2} {\mathrm e}^{\frac {6}{x}}+4 x \,{\mathrm e}^{\frac {3}{x}}+4}+5}\)	\(37\)

input

int((((-2*x^4+6*x^3)*exp(3/x)^2+(-4*x^3+12*x^2)*exp(3/x)-3*x^3+3*x^2)*exp( 
x^2*exp(3/x)^2+4*x*exp(3/x)+4)-15*x^3+15*x^2)/(exp(x)^3*exp(x^2*exp(3/x)^2 
+4*x*exp(3/x)+4)^2+10*exp(x)^3*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)+25*exp(x 
)^3),x,method=_RETURNVERBOSE)
 

output

x^3*exp(-3*x)/(exp(x^2*exp(6/x)+4*x*exp(3/x)+4)+5)
 

3.3.87.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {x^{3}}{e^{\left (x^{2} e^{\frac {6}{x}} + 4 \, x e^{\frac {3}{x}} + 3 \, x + 4\right )} + 5 \, e^{\left (3 \, x\right )}} \]

input

integrate((((-2*x^4+6*x^3)*exp(3/x)^2+(-4*x^3+12*x^2)*exp(3/x)-3*x^3+3*x^2 
)*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)-15*x^3+15*x^2)/(exp(x)^3*exp(x^2*exp( 
3/x)^2+4*x*exp(3/x)+4)^2+10*exp(x)^3*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)+25 
*exp(x)^3),x, algorithm=\
 

output

x^3/(e^(x^2*e^(6/x) + 4*x*e^(3/x) + 3*x + 4) + 5*e^(3*x))
 

3.3.87.6 Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {x^{3}}{e^{3 x} e^{x^{2} e^{\frac {6}{x}} + 4 x e^{\frac {3}{x}} + 4} + 5 e^{3 x}} \]

input

integrate((((-2*x**4+6*x**3)*exp(3/x)**2+(-4*x**3+12*x**2)*exp(3/x)-3*x**3 
+3*x**2)*exp(x**2*exp(3/x)**2+4*x*exp(3/x)+4)-15*x**3+15*x**2)/(exp(x)**3* 
exp(x**2*exp(3/x)**2+4*x*exp(3/x)+4)**2+10*exp(x)**3*exp(x**2*exp(3/x)**2+ 
4*x*exp(3/x)+4)+25*exp(x)**3),x)
 

output

x**3/(exp(3*x)*exp(x**2*exp(6/x) + 4*x*exp(3/x) + 4) + 5*exp(3*x))
 

3.3.87.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {x^{3}}{e^{\left (x^{2} e^{\frac {6}{x}} + 4 \, x e^{\frac {3}{x}} + 3 \, x + 4\right )} + 5 \, e^{\left (3 \, x\right )}} \]

input

integrate((((-2*x^4+6*x^3)*exp(3/x)^2+(-4*x^3+12*x^2)*exp(3/x)-3*x^3+3*x^2 
)*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)-15*x^3+15*x^2)/(exp(x)^3*exp(x^2*exp( 
3/x)^2+4*x*exp(3/x)+4)^2+10*exp(x)^3*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)+25 
*exp(x)^3),x, algorithm=\
 

output

x^3/(e^(x^2*e^(6/x) + 4*x*e^(3/x) + 3*x + 4) + 5*e^(3*x))
 

3.3.87.8 Giac [F]

\[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\int { -\frac {15 \, x^{3} - 15 \, x^{2} + {\left (3 \, x^{3} - 3 \, x^{2} + 2 \, {\left (x^{4} - 3 \, x^{3}\right )} e^{\frac {6}{x}} + 4 \, {\left (x^{3} - 3 \, x^{2}\right )} e^{\frac {3}{x}}\right )} e^{\left (x^{2} e^{\frac {6}{x}} + 4 \, x e^{\frac {3}{x}} + 4\right )}}{e^{\left (2 \, x^{2} e^{\frac {6}{x}} + 8 \, x e^{\frac {3}{x}} + 3 \, x + 8\right )} + 10 \, e^{\left (x^{2} e^{\frac {6}{x}} + 4 \, x e^{\frac {3}{x}} + 3 \, x + 4\right )} + 25 \, e^{\left (3 \, x\right )}} \,d x } \]

input

integrate((((-2*x^4+6*x^3)*exp(3/x)^2+(-4*x^3+12*x^2)*exp(3/x)-3*x^3+3*x^2 
)*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)-15*x^3+15*x^2)/(exp(x)^3*exp(x^2*exp( 
3/x)^2+4*x*exp(3/x)+4)^2+10*exp(x)^3*exp(x^2*exp(3/x)^2+4*x*exp(3/x)+4)+25 
*exp(x)^3),x, algorithm=\
 

output

integrate(-(15*x^3 - 15*x^2 + (3*x^3 - 3*x^2 + 2*(x^4 - 3*x^3)*e^(6/x) + 4 
*(x^3 - 3*x^2)*e^(3/x))*e^(x^2*e^(6/x) + 4*x*e^(3/x) + 4))/(e^(2*x^2*e^(6/ 
x) + 8*x*e^(3/x) + 3*x + 8) + 10*e^(x^2*e^(6/x) + 4*x*e^(3/x) + 3*x + 4) + 
 25*e^(3*x)), x)
 

3.3.87.9 Mupad [B] (verification not implemented)

Time = 9.75 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {15 x^2-15 x^3+e^{4+4 e^{3/x} x+e^{6/x} x^2} \left (3 x^2-3 x^3+e^{3/x} \left (12 x^2-4 x^3\right )+e^{6/x} \left (6 x^3-2 x^4\right )\right )}{25 e^{3 x}+10 e^{4+3 x+4 e^{3/x} x+e^{6/x} x^2}+e^{8+3 x+8 e^{3/x} x+2 e^{6/x} x^2}} \, dx=\frac {x^3}{5\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{6/x}}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{3/x}}} \]

input

int((exp(4*x*exp(3/x) + x^2*exp(6/x) + 4)*(exp(6/x)*(6*x^3 - 2*x^4) + exp( 
3/x)*(12*x^2 - 4*x^3) + 3*x^2 - 3*x^3) + 15*x^2 - 15*x^3)/(25*exp(3*x) + 1 
0*exp(4*x*exp(3/x) + x^2*exp(6/x) + 4)*exp(3*x) + exp(8*x*exp(3/x) + 2*x^2 
*exp(6/x) + 8)*exp(3*x)),x)
 

output

x^3/(5*exp(3*x) + exp(x^2*exp(6/x))*exp(3*x)*exp(4)*exp(4*x*exp(3/x)))