Integrand size = 89, antiderivative size = 24 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{1+\left (1+e^x+e^{\frac {2 x}{5+x}}-3 x\right ) x} \] Output:
exp(1+(exp(x)-3*x+1+exp(2*x/(5+x)))*x)
Time = 5.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{1+x+e^x x+e^{2-\frac {10}{5+x}} x-3 x^2} \] Input:
Integrate[(E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)*(25 - 140*x - 5 9*x^2 - 6*x^3 + E^((2*x)/(5 + x))*(25 + 20*x + x^2) + E^x*(25 + 35*x + 11* x^2 + x^3)))/(25 + 10*x + x^2),x]
Output:
E^(1 + x + E^x*x + E^(2 - 10/(5 + x))*x - 3*x^2)
Time = 2.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2007, 7257}
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.
\(\displaystyle \int \frac {e^{-3 x^2+e^x x+e^{\frac {2 x}{x+5}} x+x+1} \left (-6 x^3-59 x^2+e^{\frac {2 x}{x+5}} \left (x^2+20 x+25\right )+e^x \left (x^3+11 x^2+35 x+25\right )-140 x+25\right )}{x^2+10 x+25} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-3 x^2+e^x x+e^{\frac {2 x}{x+5}} x+x+1} \left (-6 x^3-59 x^2+e^{\frac {2 x}{x+5}} \left (x^2+20 x+25\right )+e^x \left (x^3+11 x^2+35 x+25\right )-140 x+25\right )}{(x+5)^2}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{-3 x^2+e^x x+e^{\frac {2 x}{x+5}} x+x+1}\) |
Input:
Int[(E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)*(25 - 140*x - 59*x^2 - 6*x^3 + E^((2*x)/(5 + x))*(25 + 20*x + x^2) + E^x*(25 + 35*x + 11*x^2 + x^3)))/(25 + 10*x + x^2),x]
Output:
E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 1.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {2 x}{5+x}}-3 x^{2}+x +1}\) | \(25\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {2 x}{5+x}}-3 x^{2}+x +1}\) | \(25\) |
Input:
int(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2 -140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x,method =_RETURNVERBOSE)
Output:
exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (\frac {2 \, x}{x + 5}\right )} + x + 1\right )} \] Input:
integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3- 59*x^2-140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="fricas")
Output:
e^(-3*x^2 + x*e^x + x*e^(2*x/(x + 5)) + x + 1)
Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{- 3 x^{2} + x e^{x} + x e^{\frac {2 x}{x + 5}} + x + 1} \] Input:
integrate(((x**3+11*x**2+35*x+25)*exp(x)+(x**2+20*x+25)*exp(2*x/(5+x))-6*x **3-59*x**2-140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x**2+x+1)/(x**2+10*x +25),x)
Output:
exp(-3*x**2 + x*exp(x) + x*exp(2*x/(x + 5)) + x + 1)
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (-\frac {10}{x + 5} + 2\right )} + x + 1\right )} \] Input:
integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3- 59*x^2-140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="maxima")
Output:
e^(-3*x^2 + x*e^x + x*e^(-10/(x + 5) + 2) + x + 1)
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx=e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (\frac {2 \, x}{x + 5}\right )} + x + 1\right )} \] Input:
integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3- 59*x^2-140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="giac")
Output:
e^(-3*x^2 + x*e^x + x*e^(2*x/(x + 5)) + x + 1)
Time = 1.77 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\mathrm {e}\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{\frac {2\,x}{x+5}}}\,{\mathrm {e}}^x \] Input:
int(-(exp(x + x*exp((2*x)/(x + 5)) + x*exp(x) - 3*x^2 + 1)*(140*x - exp(x) *(35*x + 11*x^2 + x^3 + 25) + 59*x^2 + 6*x^3 - exp((2*x)/(x + 5))*(20*x + x^2 + 25) - 25))/(10*x + x^2 + 25),x)
Output:
exp(x*exp(x))*exp(1)*exp(-3*x^2)*exp(x*exp((2*x)/(x + 5)))*exp(x)
\[ \int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{25+10 x+x^2} \, dx =\text {Too large to display} \] Input:
int(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2 -140*x+25)*exp(exp(x)*x+x*exp(2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x)
Output:
e*(25*int(e**((e**((x**2 + 5*x + 10)/(x + 5))*x + e**(10/(x + 5))*x + e**2 *x)/e**(10/(x + 5)))/(e**((3*x**3 + 15*x**2 + 10)/(x + 5))*x**2 + 10*e**(( 3*x**3 + 15*x**2 + 10)/(x + 5))*x + 25*e**((3*x**3 + 15*x**2 + 10)/(x + 5) )),x)*e**2 + 25*int(e**((e**((x**2 + 5*x + 10)/(x + 5))*x + e**(10/(x + 5) )*x + e**2*x)/e**(10/(x + 5)))/(e**(3*x**2)*x**2 + 10*e**(3*x**2)*x + 25*e **(3*x**2)),x) + 25*int(e**((e**((x**2 + 5*x + 10)/(x + 5))*x + 2*e**(10/( x + 5))*x + e**2*x)/e**(10/(x + 5)))/(e**(3*x**2)*x**2 + 10*e**(3*x**2)*x + 25*e**(3*x**2)),x) - 6*int((e**((e**((x**2 + 5*x + 10)/(x + 5))*x + e**( 10/(x + 5))*x + e**2*x)/e**(10/(x + 5)))*x**3)/(e**(3*x**2)*x**2 + 10*e**( 3*x**2)*x + 25*e**(3*x**2)),x) + int((e**((e**((x**2 + 5*x + 10)/(x + 5))* x + e**(10/(x + 5))*x + e**2*x)/e**(10/(x + 5)))*x**2)/(e**((3*x**3 + 15*x **2 + 10)/(x + 5))*x**2 + 10*e**((3*x**3 + 15*x**2 + 10)/(x + 5))*x + 25*e **((3*x**3 + 15*x**2 + 10)/(x + 5))),x)*e**2 - 59*int((e**((e**((x**2 + 5* x + 10)/(x + 5))*x + e**(10/(x + 5))*x + e**2*x)/e**(10/(x + 5)))*x**2)/(e **(3*x**2)*x**2 + 10*e**(3*x**2)*x + 25*e**(3*x**2)),x) + 20*int((e**((e** ((x**2 + 5*x + 10)/(x + 5))*x + e**(10/(x + 5))*x + e**2*x)/e**(10/(x + 5) ))*x)/(e**((3*x**3 + 15*x**2 + 10)/(x + 5))*x**2 + 10*e**((3*x**3 + 15*x** 2 + 10)/(x + 5))*x + 25*e**((3*x**3 + 15*x**2 + 10)/(x + 5))),x)*e**2 - 14 0*int((e**((e**((x**2 + 5*x + 10)/(x + 5))*x + e**(10/(x + 5))*x + e**2*x) /e**(10/(x + 5)))*x)/(e**(3*x**2)*x**2 + 10*e**(3*x**2)*x + 25*e**(3*x*...