3.2.51 \(\int \frac {75-30 x-15 x^2+e^x (15+30 x+15 x^2)}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} (1+2 x+x^2)+e^x (-30-50 x-22 x^2-2 x^3)} \, dx\) [151]

3.2.51.1 Optimal result

Integrand size = 78, antiderivative size = 19 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=-\frac {15}{-9+e^x-x-\frac {6}{1+x}} \end {dmath*}

output

-15/(exp(x)-x-9-6/(1+x))
 

3.2.51.2 Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=\frac {15 (1+x)}{15+10 x+x^2-e^x (1+x)} \end {dmath*}

input

Integrate[(75 - 30*x - 15*x^2 + E^x*(15 + 30*x + 15*x^2))/(225 + 300*x + 1 
30*x^2 + 20*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(-30 - 50*x - 22*x^2 
 - 2*x^3)),x]
 

output

(15*(1 + x))/(15 + 10*x + x^2 - E^x*(1 + x))
 

3.2.51.3 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.

\(\displaystyle \int \frac {-15 x^2+e^x \left (15 x^2+30 x+15\right )-30 x+75}{x^4+20 x^3+130 x^2+e^{2 x} \left (x^2+2 x+1\right )+e^x \left (-2 x^3-22 x^2-50 x-30\right )+300 x+225} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {15 \left (-x^2-2 x+e^x (x+1)^2+5\right )}{\left (x^2+10 x-e^x (x+1)+15\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 15 \int \frac {-x^2-2 x+e^x (x+1)^2+5}{\left (x^2+10 x-e^x (x+1)+15\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 15 \int \left (\frac {x^3+10 x^2+23 x+20}{\left (-x^2+e^x x-10 x+e^x-15\right )^2}-\frac {x+1}{x^2-e^x x+10 x-e^x+15}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 15 \left (20 \int \frac {1}{\left (-x^2+e^x x-10 x+e^x-15\right )^2}dx+\int \frac {1}{-x^2+e^x x-10 x+e^x-15}dx+23 \int \frac {x}{\left (x^2-e^x x+10 x-e^x+15\right )^2}dx+10 \int \frac {x^2}{\left (x^2-e^x x+10 x-e^x+15\right )^2}dx-\int \frac {x}{x^2-e^x x+10 x-e^x+15}dx+\int \frac {x^3}{\left (x^2-e^x x+10 x-e^x+15\right )^2}dx\right )\)

input

Int[(75 - 30*x - 15*x^2 + E^x*(15 + 30*x + 15*x^2))/(225 + 300*x + 130*x^2 
 + 20*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(-30 - 50*x - 22*x^2 - 2*x 
^3)),x]
 

output

$Aborted
 

3.2.51.3.1 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 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] 
]
 

3.2.51.4 Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {15 x +15}{x^{2}-{\mathrm e}^{x} x +10 x -{\mathrm e}^{x}+15}\)	\(25\)
norman	\(\frac {15 x +15}{x^{2}-{\mathrm e}^{x} x +10 x -{\mathrm e}^{x}+15}\)	\(26\)
parallelrisch	\(-\frac {-15 x -15}{x^{2}-{\mathrm e}^{x} x +10 x -{\mathrm e}^{x}+15}\)	\(27\)

input

int(((15*x^2+30*x+15)*exp(x)-15*x^2-30*x+75)/((x^2+2*x+1)*exp(x)^2+(-2*x^3 
-22*x^2-50*x-30)*exp(x)+x^4+20*x^3+130*x^2+300*x+225),x,method=_RETURNVERB 
OSE)
 

output

15*(1+x)/(x^2-exp(x)*x+10*x-exp(x)+15)
 

3.2.51.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=\frac {15 \, {\left (x + 1\right )}}{x^{2} - {\left (x + 1\right )} e^{x} + 10 \, x + 15} \end {dmath*}

input

integrate(((15*x^2+30*x+15)*exp(x)-15*x^2-30*x+75)/((x^2+2*x+1)*exp(x)^2+( 
-2*x^3-22*x^2-50*x-30)*exp(x)+x^4+20*x^3+130*x^2+300*x+225),x, algorithm=\
 

output

15*(x + 1)/(x^2 - (x + 1)*e^x + 10*x + 15)
 

3.2.51.6 Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=\frac {- 15 x - 15}{- x^{2} - 10 x + \left (x + 1\right ) e^{x} - 15} \end {dmath*}

input

integrate(((15*x**2+30*x+15)*exp(x)-15*x**2-30*x+75)/((x**2+2*x+1)*exp(x)* 
*2+(-2*x**3-22*x**2-50*x-30)*exp(x)+x**4+20*x**3+130*x**2+300*x+225),x)
 

output

(-15*x - 15)/(-x**2 - 10*x + (x + 1)*exp(x) - 15)
 

3.2.51.7 Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=\frac {15 \, {\left (x + 1\right )}}{x^{2} - {\left (x + 1\right )} e^{x} + 10 \, x + 15} \end {dmath*}

input

integrate(((15*x^2+30*x+15)*exp(x)-15*x^2-30*x+75)/((x^2+2*x+1)*exp(x)^2+( 
-2*x^3-22*x^2-50*x-30)*exp(x)+x^4+20*x^3+130*x^2+300*x+225),x, algorithm=\
 

output

15*(x + 1)/(x^2 - (x + 1)*e^x + 10*x + 15)
 

3.2.51.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=\frac {15 \, {\left (x + 1\right )}}{x^{2} - x e^{x} + 10 \, x - e^{x} + 15} \end {dmath*}

input

integrate(((15*x^2+30*x+15)*exp(x)-15*x^2-30*x+75)/((x^2+2*x+1)*exp(x)^2+( 
-2*x^3-22*x^2-50*x-30)*exp(x)+x^4+20*x^3+130*x^2+300*x+225),x, algorithm=\
 

output

15*(x + 1)/(x^2 - x*e^x + 10*x - e^x + 15)
 

3.2.51.9 Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\right )} \, dx=-\frac {15\,x+15}{{\mathrm {e}}^x+x\,\left ({\mathrm {e}}^x-10\right )-x^2-15} \end {dmath*}

input

int(-(30*x - exp(x)*(30*x + 15*x^2 + 15) + 15*x^2 - 75)/(300*x + exp(2*x)* 
(2*x + x^2 + 1) + 130*x^2 + 20*x^3 + x^4 - exp(x)*(50*x + 22*x^2 + 2*x^3 + 
 30) + 225),x)
 

output

-(15*x + 15)/(exp(x) + x*(exp(x) - 10) - x^2 - 15)