Integrand size = 171, antiderivative size = 30 \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\frac {x}{5}+\frac {1}{1+x}-\frac {3^x x}{-e^{1+x}+5 x} \]
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\frac {1}{5} \left (x+\frac {5\ 3^x x}{e^{1+x}-5 x}+\frac {5}{1+x}\right ) \]
Integrate[(-100*x^2 + 50*x^3 + 25*x^4 + E^(2 + 2*x)*(-4 + 2*x + x^2) + E^( 1 + x)*(40*x - 20*x^2 - 10*x^3) + 3^x*((-25*x^2 - 50*x^3 - 25*x^4)*Log[3] + E^(1 + x)*(5 + 5*x - 5*x^2 - 5*x^3 + (5*x + 10*x^2 + 5*x^3)*Log[3])))/(1 25*x^2 + 250*x^3 + 125*x^4 + E^(2 + 2*x)*(5 + 10*x + 5*x^2) + E^(1 + x)*(- 50*x - 100*x^2 - 50*x^3)),x]
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.
| \(\displaystyle \int \frac {25 x^4+50 x^3-100 x^2+e^{2 x+2} \left (x^2+2 x-4\right )+e^{x+1} \left (-10 x^3-20 x^2+40 x\right )+3^x \left (e^{x+1} \left (-5 x^3-5 x^2+\left (5 x^3+10 x^2+5 x\right ) \log (3)+5 x+5\right )+\left (-25 x^4-50 x^3-25 x^2\right ) \log (3)\right )}{125 x^4+250 x^3+125 x^2+e^{2 x+2} \left (5 x^2+10 x+5\right )+e^{x+1} \left (-50 x^3-100 x^2-50 x\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {25 x^4+50 x^3-100 x^2+e^{2 x+2} \left (x^2+2 x-4\right )+e^{x+1} \left (-10 x^3-20 x^2+40 x\right )+3^x \left (e^{x+1} \left (-5 x^3-5 x^2+\left (5 x^3+10 x^2+5 x\right ) \log (3)+5 x+5\right )+\left (-25 x^4-50 x^3-25 x^2\right ) \log (3)\right )}{5 \left (e^{x+1}-5 x\right )^2 (x+1)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int -\frac {-25 x^4-50 x^3+100 x^2+e^{2 x+2} \left (-x^2-2 x+4\right )-10 e^{x+1} \left (-x^3-2 x^2+4 x\right )+5\ 3^x \left (5 \left (x^4+2 x^3+x^2\right ) \log (3)-e^{x+1} \left (-x^3-x^2+x+\left (x^3+2 x^2+x\right ) \log (3)+1\right )\right )}{\left (e^{x+1}-5 x\right )^2 (x+1)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{5} \int \frac {-25 x^4-50 x^3+100 x^2+e^{2 x+2} \left (-x^2-2 x+4\right )-10 e^{x+1} \left (-x^3-2 x^2+4 x\right )+5\ 3^x \left (5 \left (x^4+2 x^3+x^2\right ) \log (3)-e^{x+1} \left (-x^3-x^2+x+\left (x^3+2 x^2+x\right ) \log (3)+1\right )\right )}{\left (e^{x+1}-5 x\right )^2 (x+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{5} \int \left (\frac {25\ 3^x (x-1) x}{\left (e^{x+1}-5 x\right )^2}+\frac {-x^2-2 x+4}{(x+1)^2}+\frac {5\ 3^x (x (1-\log (3))-1)}{e^{x+1}-5 x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (-25 \int \frac {3^x x^2}{\left (e^{x+1}-5 x\right )^2}dx+5 \int \frac {3^x}{e^{x+1}-5 x}dx+25 \int \frac {3^x x}{\left (e^{x+1}-5 x\right )^2}dx-5 (1-\log (3)) \int \frac {3^x x}{e^{x+1}-5 x}dx+x+\frac {5}{x+1}\right )\) |
Int[(-100*x^2 + 50*x^3 + 25*x^4 + E^(2 + 2*x)*(-4 + 2*x + x^2) + E^(1 + x) *(40*x - 20*x^2 - 10*x^3) + 3^x*((-25*x^2 - 50*x^3 - 25*x^4)*Log[3] + E^(1 + x)*(5 + 5*x - 5*x^2 - 5*x^3 + (5*x + 10*x^2 + 5*x^3)*Log[3])))/(125*x^2 + 250*x^3 + 125*x^4 + E^(2 + 2*x)*(5 + 10*x + 5*x^2) + E^(1 + x)*(-50*x - 100*x^2 - 50*x^3)),x]
3.24.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(65\) vs. \(2(29)=58\).
Time = 1.95 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.20
| method | result | size |
| norman | \(\frac {x^{3}+4 x -\frac {4 \,{\mathrm e}^{1+x}}{5}-x \,{\mathrm e}^{x \ln \left (3\right )}-x^{2} {\mathrm e}^{x \ln \left (3\right )}-\frac {x^{2} {\mathrm e}^{1+x}}{5}}{5 x^{2}-x \,{\mathrm e}^{1+x}+5 x -{\mathrm e}^{1+x}}\) | \(66\) |
| parallelrisch | \(\frac {5 x^{3}-x^{2} {\mathrm e}^{1+x}-5 x^{2} {\mathrm e}^{x \ln \left (3\right )}-5 x \,{\mathrm e}^{x \ln \left (3\right )}+20 x -4 \,{\mathrm e}^{1+x}}{25 x^{2}-5 x \,{\mathrm e}^{1+x}+25 x -5 \,{\mathrm e}^{1+x}}\) | \(69\) |
| parts | \(\frac {x^{3}+4 x -\frac {4 \,{\mathrm e}^{1+x}}{5}-\frac {x^{2} {\mathrm e}^{1+x}}{5}}{5 x^{2}-x \,{\mathrm e}^{1+x}+5 x -{\mathrm e}^{1+x}}+\frac {-x \,{\mathrm e}^{x \ln \left (3\right )}-x^{2} {\mathrm e}^{x \ln \left (3\right )}}{5 x^{2}-x \,{\mathrm e}^{1+x}+5 x -{\mathrm e}^{1+x}}\) | \(93\) |
int(((((5*x^3+10*x^2+5*x)*ln(3)-5*x^3-5*x^2+5*x+5)*exp(1+x)+(-25*x^4-50*x^ 3-25*x^2)*ln(3))*exp(x*ln(3))+(x^2+2*x-4)*exp(1+x)^2+(-10*x^3-20*x^2+40*x) *exp(1+x)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(1+x)^2+(-50*x^3-100*x ^2-50*x)*exp(1+x)+125*x^4+250*x^3+125*x^2),x,method=_RETURNVERBOSE)
(x^3+4*x-4/5*exp(1+x)-x*exp(x*ln(3))-x^2*exp(x*ln(3))-1/5*x^2*exp(1+x))/(5 *x^2-x*exp(1+x)+5*x-exp(1+x))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} + x + 5\right )} e^{\left (x + 1\right )} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x + 1\right )} e^{\left (x + 1\right )} + 5 \, x\right )}} \]
integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(1+x)+(-25*x^ 4-50*x^3-25*x^2)*log(3))*exp(x*log(3))+(x^2+2*x-4)*exp(1+x)^2+(-10*x^3-20* x^2+40*x)*exp(1+x)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(1+x)^2+(-50* x^3-100*x^2-50*x)*exp(1+x)+125*x^4+250*x^3+125*x^2),x, algorithm=\
1/5*(5*x^3 - 5*(x^2 + x)*3^x + 5*x^2 - (x^2 + x + 5)*e^(x + 1) + 25*x)/(5* x^2 - (x + 1)*e^(x + 1) + 5*x)
Exception generated. \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
integrate(((((5*x**3+10*x**2+5*x)*ln(3)-5*x**3-5*x**2+5*x+5)*exp(1+x)+(-25 *x**4-50*x**3-25*x**2)*ln(3))*exp(x*ln(3))+(x**2+2*x-4)*exp(1+x)**2+(-10*x **3-20*x**2+40*x)*exp(1+x)+25*x**4+50*x**3-100*x**2)/((5*x**2+10*x+5)*exp( 1+x)**2+(-50*x**3-100*x**2-50*x)*exp(1+x)+125*x**4+250*x**3+125*x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (27) = 54\).
Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} e + x e + 5 \, e\right )} e^{x} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x e + e\right )} e^{x} + 5 \, x\right )}} \]
integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(1+x)+(-25*x^ 4-50*x^3-25*x^2)*log(3))*exp(x*log(3))+(x^2+2*x-4)*exp(1+x)^2+(-10*x^3-20* x^2+40*x)*exp(1+x)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(1+x)^2+(-50* x^3-100*x^2-50*x)*exp(1+x)+125*x^4+250*x^3+125*x^2),x, algorithm=\
1/5*(5*x^3 - 5*(x^2 + x)*3^x + 5*x^2 - (x^2*e + x*e + 5*e)*e^x + 25*x)/(5* x^2 - (x*e + e)*e^x + 5*x)
\[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\int { \frac {25 \, x^{4} + 50 \, x^{3} - 5 \, {\left ({\left (x^{3} + x^{2} - {\left (x^{3} + 2 \, x^{2} + x\right )} \log \left (3\right ) - x - 1\right )} e^{\left (x + 1\right )} + 5 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \left (3\right )\right )} 3^{x} - 100 \, x^{2} + {\left (x^{2} + 2 \, x - 4\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} - 4 \, x\right )} e^{\left (x + 1\right )}}{5 \, {\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{\left (x + 1\right )}\right )}} \,d x } \]
integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(1+x)+(-25*x^ 4-50*x^3-25*x^2)*log(3))*exp(x*log(3))+(x^2+2*x-4)*exp(1+x)^2+(-10*x^3-20* x^2+40*x)*exp(1+x)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(1+x)^2+(-50* x^3-100*x^2-50*x)*exp(1+x)+125*x^4+250*x^3+125*x^2),x, algorithm=\
integrate(1/5*(25*x^4 + 50*x^3 - 5*((x^3 + x^2 - (x^3 + 2*x^2 + x)*log(3) - x - 1)*e^(x + 1) + 5*(x^4 + 2*x^3 + x^2)*log(3))*3^x - 100*x^2 + (x^2 + 2*x - 4)*e^(2*x + 2) - 10*(x^3 + 2*x^2 - 4*x)*e^(x + 1))/(25*x^4 + 50*x^3 + 25*x^2 + (x^2 + 2*x + 1)*e^(2*x + 2) - 10*(x^3 + 2*x^2 + x)*e^(x + 1)), x)
Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx=\frac {x}{5}+\frac {1}{x+1}-\frac {3^x\,x}{5\,\left (x-\frac {{\mathrm {e}}^{x+1}}{5}\right )} \]
int((exp(2*x + 2)*(2*x + x^2 - 4) - exp(x + 1)*(20*x^2 - 40*x + 10*x^3) - 100*x^2 + 50*x^3 + 25*x^4 + exp(x*log(3))*(exp(x + 1)*(5*x + log(3)*(5*x + 10*x^2 + 5*x^3) - 5*x^2 - 5*x^3 + 5) - log(3)*(25*x^2 + 50*x^3 + 25*x^4)) )/(exp(2*x + 2)*(10*x + 5*x^2 + 5) - exp(x + 1)*(50*x + 100*x^2 + 50*x^3) + 125*x^2 + 250*x^3 + 125*x^4),x)