Integrand size = 134, antiderivative size = 28 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3 \left (x-(x-x \log (5+x))^2\right )}{-2+e^{2-x}} \]
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3 e^x x \left (-1+x-2 x \log (5+x)+x \log ^2(5+x)\right )}{-e^2+2 e^x} \]
Integrate[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 12*x^2 + E^(2 - x)*(60*x + 36*x^2 + 6*x^3))*Log[5 + x] + (60*x + 12*x^2 + E^(2 - x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20 - 4*x) + 4*x + E^(4 - 2*x)*(5 + x)),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.
\(\displaystyle \int \frac {e^{2-x} \left (-3 x^3-12 x^2-12 x+15\right )+\left (12 x^2+e^{2-x} \left (-3 x^3-21 x^2-30 x\right )+60 x\right ) \log ^2(x+5)+\left (-12 x^2+e^{2-x} \left (6 x^3+36 x^2+60 x\right )-120 x\right ) \log (x+5)+54 x-30}{e^{2-x} (-4 x-20)+4 x+e^{4-2 x} (x+5)+20} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} \left (e^{2-x} \left (-3 x^3-12 x^2-12 x+15\right )+\left (12 x^2+e^{2-x} \left (-3 x^3-21 x^2-30 x\right )+60 x\right ) \log ^2(x+5)+\left (-12 x^2+e^{2-x} \left (6 x^3+36 x^2+60 x\right )-120 x\right ) \log (x+5)+54 x-30\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {3 e^{x-2} \left (x^3+x^3 \log ^2(x+5)-2 x^3 \log (x+5)+4 x^2+7 x^2 \log ^2(x+5)-12 x^2 \log (x+5)+4 x+10 x \log ^2(x+5)-20 x \log (x+5)-5\right )}{x+5}-\frac {6 e^{2 x-2} \left (x^3+x^3 \log ^2(x+5)-2 x^3 \log (x+5)+4 x^2+7 x^2 \log ^2(x+5)-12 x^2 \log (x+5)+4 x+10 x \log ^2(x+5)-20 x \log (x+5)-5\right )}{\left (e^2-2 e^x\right ) (x+5)}-\frac {6 e^{2 x} x \left (x+x \log ^2(x+5)-2 x \log (x+5)-1\right )}{\left (2 e^x-e^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 e^x \left (2 x \left (e^2 \left (x^2+6 x+10\right )-2 e^x (x+10)\right ) \log (x+5)-e^2 \left (x^3+4 x^2+4 x-5\right )+2 e^x (9 x-5)-x (x+5) \left (e^2 (x+2)-4 e^x\right ) \log ^2(x+5)\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int -\frac {e^x \left (-x (x+5) \left (4 e^x-e^2 (x+2)\right ) \log ^2(x+5)+2 x \left (2 e^x (x+10)-e^2 \left (x^2+6 x+10\right )\right ) \log (x+5)+2 e^x (5-9 x)-e^2 \left (-x^3-4 x^2-4 x+5\right )\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -3 \int \frac {e^x \left (-x (x+5) \left (4 e^x-e^2 (x+2)\right ) \log ^2(x+5)+2 x \left (2 e^x (x+10)-e^2 \left (x^2+6 x+10\right )\right ) \log (x+5)+2 e^x (5-9 x)-e^2 \left (-x^3-4 x^2-4 x+5\right )\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -3 \int \left (\frac {e^{x+2} x \left (x \log ^2(x+5)-2 x \log (x+5)+x-1\right )}{\left (e^2-2 e^x\right )^2}-\frac {e^x \left (2 \log ^2(x+5) x^2-2 \log (x+5) x^2+10 \log ^2(x+5) x-20 \log (x+5) x+9 x-5\right )}{\left (-e^2+2 e^x\right ) (x+5)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\int \frac {e^{x+2} x^2 \log ^2(x+5)}{\left (-e^2+2 e^x\right )^2}dx-2 \int \frac {e^{x+2} x^2 \log (x+5)}{\left (-e^2+2 e^x\right )^2}dx+50 \int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx+50 \int \frac {\int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx}{x+5}dx-2 \int \frac {e^x x \log ^2(x+5)}{-e^2+2 e^x}dx-50 \log (x+5) \int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx+10 \int \frac {e^x \log (x+5)}{-e^2+2 e^x}dx+2 \int \frac {e^x x \log (x+5)}{-e^2+2 e^x}dx+\operatorname {PolyLog}\left (2,2 e^{x-2}\right )+\frac {e^2 x^2}{2 \left (e^2-2 e^x\right )}-\frac {x^2}{2}-\frac {e^2 x}{2 \left (e^2-2 e^x\right )}+\frac {x}{2}+x \log \left (1-2 e^{x-2}\right )-5 \log \left (e^2-2 e^x\right )\right )\) |
Int[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 12*x^ 2 + E^(2 - x)*(60*x + 36*x^2 + 6*x^3))*Log[5 + x] + (60*x + 12*x^2 + E^(2 - x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20 - 4*x) + 4*x + E^(4 - 2*x)*(5 + x)),x]
3.5.19.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50
method | result | size |
parallelrisch | \(-\frac {30 \ln \left (5+x \right )^{2} x^{2}-60 \ln \left (5+x \right ) x^{2}+30 x^{2}-30 x}{10 \left ({\mathrm e}^{2-x}-2\right )}\) | \(42\) |
risch | \(-\frac {3 x^{2} \ln \left (5+x \right )^{2}}{{\mathrm e}^{2-x}-2}+\frac {6 x^{2} \ln \left (5+x \right )}{{\mathrm e}^{2-x}-2}-\frac {3 x \left (-1+x \right )}{{\mathrm e}^{2-x}-2}\) | \(58\) |
int((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*ln(5+x)^2+((6*x^3+36*x^2+ 60*x)*exp(2-x)-12*x^2-120*x)*ln(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x -30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} \log \left (x + 5\right )^{2} - 2 \, x^{2} \log \left (x + 5\right ) + x^{2} - x\right )}}{e^{\left (-x + 2\right )} - 2} \]
integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 -x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {- 3 x^{2} \log {\left (x + 5 \right )}^{2} + 6 x^{2} \log {\left (x + 5 \right )} - 3 x^{2} + 3 x}{e^{2 - x} - 2} \]
integrate((((-3*x**3-21*x**2-30*x)*exp(2-x)+12*x**2+60*x)*ln(5+x)**2+((6*x **3+36*x**2+60*x)*exp(2-x)-12*x**2-120*x)*ln(5+x)+(-3*x**3-12*x**2-12*x+15 )*exp(2-x)+54*x-30)/((5+x)*exp(2-x)**2+(-4*x-20)*exp(2-x)+20+4*x),x)
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} e^{x} \log \left (x + 5\right )^{2} - 2 \, x^{2} e^{x} \log \left (x + 5\right ) + {\left (x^{2} - x\right )} e^{x}\right )}}{e^{2} - 2 \, e^{x}} \]
integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 -x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} \log \left (x + 5\right )^{2} - 2 \, x^{2} \log \left (x + 5\right ) + x^{2} - x\right )}}{e^{\left (-x + 2\right )} - 2} \]
integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 -x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3\,x\,{\mathrm {e}}^{x-2}\,\left (x\,{\ln \left (x+5\right )}^2-2\,x\,\ln \left (x+5\right )+x-1\right )}{2\,{\mathrm {e}}^{x-2}-1} \]