\(\int \frac {e^{\frac {1}{15} (35 e^{\frac {e^6}{4}}+3 x)} (20+4 x-4 x^2)}{5-10 x+5 x^2} \, dx\) [858]

Optimal result

Integrand size = 44, antiderivative size = 31 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=\frac {4 e^{\frac {7 e^{\frac {e^6}{4}}}{3}+\frac {x}{5}} x}{1-x} \] Output:

4*x*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(1-x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=-\frac {4 e^{\frac {7 e^{\frac {e^6}{4}}}{3}+\frac {x}{5}} x}{-1+x} \] Input:

Integrate[(E^((35*E^(E^6/4) + 3*x)/15)*(20 + 4*x - 4*x^2))/(5 - 10*x + 5*x 
^2),x]
 

Output:

(-4*E^((7*E^(E^6/4))/3 + x/5)*x)/(-1 + x)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2700, 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.

\(\displaystyle \int \frac {e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )} \left (-4 x^2+4 x+20\right )}{5 x^2-10 x+5} \, dx\)

\(\Big \downarrow \) 2700

\(\displaystyle \int \left (\frac {4 e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )} (6-x)}{5 x^2-10 x+5}-\frac {4}{5} e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )}}{1-x}-4 e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )}\)

Input:

Int[(E^((35*E^(E^6/4) + 3*x)/15)*(20 + 4*x - 4*x^2))/(5 - 10*x + 5*x^2),x]
 

Output:

-4*E^((35*E^(E^6/4) + 3*x)/15) + (4*E^((35*E^(E^6/4) + 3*x)/15))/(1 - x)
 

Defintions of rubi rules used

rule 2009

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

rule 2700

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_.) + (b_.)*(x_ 
) + (c_)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d + e*x)^n), u^m/( 
a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x] && Polynomia 
lQ[u, x] && IntegerQ[m]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-1+x}\)	\(21\)
gosper	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-1+x}\)	\(23\)
norman	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-1+x}\)	\(23\)
parallelrisch	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-1+x}\)	\(23\)
orering	\(\frac {5 x \left (-1+x \right ) \left (-4 x^{2}+4 x +20\right ) {\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{\left (x^{2}-x -5\right ) \left (5 x^{2}-10 x +5\right )}\)	\(53\)
derivativedivides	\(\frac {12 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-3 x +3}-\frac {4 \,{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{5}+\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{5 \left (-3 x +3\right )}-180 \left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )-4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}-\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (1225 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}+210 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+9\right )}{15 \left (-3 x +3\right )}+4 \left (\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}}{9}+\frac {28 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{5}+\frac {11}{25}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )-420 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{225}\right )-4900 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{225}\right )+4200 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{-675 x +675}-\left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )\right )\)	\(476\)
default	\(\frac {12 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-3 x +3}-\frac {4 \,{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{5}+\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{5 \left (-3 x +3\right )}-180 \left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )-4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}-\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (1225 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}+210 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+9\right )}{15 \left (-3 x +3\right )}+4 \left (\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}}{9}+\frac {28 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{5}+\frac {11}{25}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )-420 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{225}\right )-4900 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )}{225}\right )+4200 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{-675 x +675}-\left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \operatorname {expIntegral}_{1}\left (-\frac {x}{5}+\frac {1}{5}\right )\right )\)	\(476\)

Input:

int((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x,meth 
od=_RETURNVERBOSE)
 

Output:

-4*x*exp(7/3*exp(1/4*exp(6))+1/5*x)/(-1+x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=-\frac {4 \, x e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x - 1} \] Input:

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5), 
x, algorithm="fricas")
 

Output:

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1)
 

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=- \frac {4 x e^{\frac {x}{5} + \frac {7 e^{\frac {e^{6}}{4}}}{3}}}{x - 1} \] Input:

integrate((-4*x**2+4*x+20)*exp(7/3*exp(1/4*exp(3)**2)+1/5*x)/(5*x**2-10*x+ 
5),x)
 

Output:

-4*x*exp(x/5 + 7*exp(exp(6)/4)/3)/(x - 1)
 

Maxima [F]

\[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=\int { -\frac {4 \, {\left (x^{2} - x - 5\right )} e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{5 \, {\left (x^{2} - 2 \, x + 1\right )}} \,d x } \] Input:

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5), 
x, algorithm="maxima")
 

Output:

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1) - 4*e^(7/3*e^(1/4*e^6) + 1/5)*exp 
_integral_e(2, -1/5*x + 1/5)/(x - 1) - 4*integrate(e^(1/5*x + 7/3*e^(1/4*e 
^6))/(x^2 - 2*x + 1), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=-\frac {4 \, x e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x - 1} \] Input:

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5), 
x, algorithm="giac")
 

Output:

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1)
 

Mupad [B] (verification not implemented)

Time = 2.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=-\frac {20\,x\,{\mathrm {e}}^{x/5}\,{\mathrm {e}}^{\frac {7\,{\mathrm {e}}^{\frac {{\mathrm {e}}^6}{4}}}{3}}}{5\,x-5} \] Input:

int((exp(x/5 + (7*exp(exp(6)/4))/3)*(4*x - 4*x^2 + 20))/(5*x^2 - 10*x + 5) 
,x)
 

Output:

-(20*x*exp(x/5)*exp((7*exp(exp(6)/4))/3))/(5*x - 5)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5-10 x+5 x^2} \, dx=-\frac {4 e^{\frac {7 e^{\frac {e^{6}}{4}}}{3}+\frac {x}{5}} x}{x -1} \] Input:

int((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x)
 

Output:

( - 4*e**((35*e**(e**6/4) + 3*x)/15)*x)/(x - 1)