3.13.94 \(\int \frac {-560+840 x+e^x (112-56 x-84 x^2)}{400 x^2-600 x^3+225 x^4+e^x (-160 x^2+240 x^3-90 x^4)+e^{2 x} (16 x^2-24 x^3+9 x^4)} \, dx\) [1294]

3.13.94.1 Optimal result

Integrand size = 80, antiderivative size = 25 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=\frac {28}{3 \left (5-e^x\right ) \left (\frac {4}{3}-x\right ) x} \end {dmath*}

output

84/(45-9*exp(x))/(4/3-x)/x
 

3.13.94.2 Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=\frac {28}{\left (-5+e^x\right ) x (-4+3 x)} \end {dmath*}

input

Integrate[(-560 + 840*x + E^x*(112 - 56*x - 84*x^2))/(400*x^2 - 600*x^3 + 
225*x^4 + E^x*(-160*x^2 + 240*x^3 - 90*x^4) + E^(2*x)*(16*x^2 - 24*x^3 + 9 
*x^4)),x]
 

output

28/((-5 + E^x)*x*(-4 + 3*x))
 

3.13.94.3 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.

\(\displaystyle \int \frac {e^x \left (-84 x^2-56 x+112\right )+840 x-560}{225 x^4-600 x^3+400 x^2+e^x \left (-90 x^4+240 x^3-160 x^2\right )+e^{2 x} \left (9 x^4-24 x^3+16 x^2\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {28 \left (-e^x \left (3 x^2+2 x-4\right )+30 x-20\right )}{\left (5-e^x\right )^2 (4-3 x)^2 x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 28 \int -\frac {-30 x-e^x \left (-3 x^2-2 x+4\right )+20}{\left (5-e^x\right )^2 (4-3 x)^2 x^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -28 \int \frac {-30 x-e^x \left (-3 x^2-2 x+4\right )+20}{\left (5-e^x\right )^2 (4-3 x)^2 x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -28 \int \left (\frac {3 x^2+2 x-4}{\left (-5+e^x\right ) x^2 (3 x-4)^2}+\frac {5}{\left (-5+e^x\right )^2 x (3 x-4)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -28 \left (-\frac {1}{4} \int \frac {1}{\left (-5+e^x\right ) x^2}dx-\frac {5}{4} \int \frac {1}{\left (-5+e^x\right )^2 x}dx-\frac {1}{4} \int \frac {1}{\left (-5+e^x\right ) x}dx+\frac {9}{4} \int \frac {1}{\left (-5+e^x\right ) (3 x-4)^2}dx+\frac {15}{4} \int \frac {1}{\left (-5+e^x\right )^2 (3 x-4)}dx+\frac {3}{4} \int \frac {1}{\left (-5+e^x\right ) (3 x-4)}dx\right )\)

input

Int[(-560 + 840*x + E^x*(112 - 56*x - 84*x^2))/(400*x^2 - 600*x^3 + 225*x^ 
4 + E^x*(-160*x^2 + 240*x^3 - 90*x^4) + E^(2*x)*(16*x^2 - 24*x^3 + 9*x^4)) 
,x]
 

output

$Aborted
 

3.13.94.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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] 
]
 

3.13.94.4 Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
norman	\(\frac {28}{x \left ({\mathrm e}^{x}-5\right ) \left (-4+3 x \right )}\)	\(19\)
risch	\(\frac {28}{x \left ({\mathrm e}^{x}-5\right ) \left (-4+3 x \right )}\)	\(19\)
parallelrisch	\(\frac {28}{x \left ({\mathrm e}^{x}-5\right ) \left (-4+3 x \right )}\)	\(19\)

input

int(((-84*x^2-56*x+112)*exp(x)+840*x-560)/((9*x^4-24*x^3+16*x^2)*exp(x)^2+ 
(-90*x^4+240*x^3-160*x^2)*exp(x)+225*x^4-600*x^3+400*x^2),x,method=_RETURN 
VERBOSE)
 

output

28/x/(exp(x)-5)/(-4+3*x)
 

3.13.94.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=-\frac {28}{15 \, x^{2} - {\left (3 \, x^{2} - 4 \, x\right )} e^{x} - 20 \, x} \end {dmath*}

input

integrate(((-84*x^2-56*x+112)*exp(x)+840*x-560)/((9*x^4-24*x^3+16*x^2)*exp 
(x)^2+(-90*x^4+240*x^3-160*x^2)*exp(x)+225*x^4-600*x^3+400*x^2),x, algorit 
hm=\
 

output

-28/(15*x^2 - (3*x^2 - 4*x)*e^x - 20*x)
 

3.13.94.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=\frac {252}{- 135 x^{2} + 180 x + \left (27 x^{2} - 36 x\right ) e^{x}} \end {dmath*}

input

integrate(((-84*x**2-56*x+112)*exp(x)+840*x-560)/((9*x**4-24*x**3+16*x**2) 
*exp(x)**2+(-90*x**4+240*x**3-160*x**2)*exp(x)+225*x**4-600*x**3+400*x**2) 
,x)
 

output

252/(-135*x**2 + 180*x + (27*x**2 - 36*x)*exp(x))
 

3.13.94.7 Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=-\frac {28}{15 \, x^{2} - {\left (3 \, x^{2} - 4 \, x\right )} e^{x} - 20 \, x} \end {dmath*}

input

integrate(((-84*x^2-56*x+112)*exp(x)+840*x-560)/((9*x^4-24*x^3+16*x^2)*exp 
(x)^2+(-90*x^4+240*x^3-160*x^2)*exp(x)+225*x^4-600*x^3+400*x^2),x, algorit 
hm=\
 

output

-28/(15*x^2 - (3*x^2 - 4*x)*e^x - 20*x)
 

3.13.94.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=\frac {28}{3 \, x^{2} e^{x} - 15 \, x^{2} - 4 \, x e^{x} + 20 \, x} \end {dmath*}

input

integrate(((-84*x^2-56*x+112)*exp(x)+840*x-560)/((9*x^4-24*x^3+16*x^2)*exp 
(x)^2+(-90*x^4+240*x^3-160*x^2)*exp(x)+225*x^4-600*x^3+400*x^2),x, algorit 
hm=\
 

output

28/(3*x^2*e^x - 15*x^2 - 4*x*e^x + 20*x)
 

3.13.94.9 Mupad [B] (verification not implemented)

Time = 16.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {-560+840 x+e^x \left (112-56 x-84 x^2\right )}{400 x^2-600 x^3+225 x^4+e^x \left (-160 x^2+240 x^3-90 x^4\right )+e^{2 x} \left (16 x^2-24 x^3+9 x^4\right )} \, dx=-\frac {28\,\left (4\,x-3\,x^2\right )}{x^2\,{\left (3\,x-4\right )}^2\,\left ({\mathrm {e}}^x-5\right )} \end {dmath*}

input

int(-(exp(x)*(56*x + 84*x^2 - 112) - 840*x + 560)/(exp(2*x)*(16*x^2 - 24*x 
^3 + 9*x^4) - exp(x)*(160*x^2 - 240*x^3 + 90*x^4) + 400*x^2 - 600*x^3 + 22 
5*x^4),x)
 

output

-(28*(4*x - 3*x^2))/(x^2*(3*x - 4)^2*(exp(x) - 5))