Integrand size = 181, antiderivative size = 28 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=\log \left (\frac {(8+x) \left (3+e^{(5+x)^2} \log (5-x)\right ) \log (x)}{4+x}\right ) \]
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=-25-\log (4+x)+\log (8+x)+\log \left (3+e^{(5+x)^2} \log (5-x)\right )+\log (\log (x)) \]
Integrate[(-480 - 84*x + 21*x^2 + 3*x^3 + E^(25 + 10*x + x^2)*(-160 - 28*x + 7*x^2 + x^3)*Log[5 - x] + (60*x - 12*x^2 + E^(25 + 10*x + x^2)*(32*x + 12*x^2 + x^3) + E^(25 + 10*x + x^2)*(-1580*x - 604*x^2 + 14*x^3 + 24*x^4 + 2*x^5)*Log[5 - x])*Log[x])/((-480*x - 84*x^2 + 21*x^3 + 3*x^4 + E^(25 + 1 0*x + x^2)*(-160*x - 28*x^2 + 7*x^3 + x^4)*Log[5 - x])*Log[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.
| \(\displaystyle \int \frac {3 x^3+21 x^2+e^{x^2+10 x+25} \left (x^3+7 x^2-28 x-160\right ) \log (5-x)+\left (-12 x^2+e^{x^2+10 x+25} \left (x^3+12 x^2+32 x\right )+e^{x^2+10 x+25} \left (2 x^5+24 x^4+14 x^3-604 x^2-1580 x\right ) \log (5-x)+60 x\right ) \log (x)-84 x-480}{\left (3 x^4+21 x^3-84 x^2+e^{x^2+10 x+25} \left (x^4+7 x^3-28 x^2-160 x\right ) \log (5-x)-480 x\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-3 x^3-21 x^2-e^{x^2+10 x+25} \left (x^3+7 x^2-28 x-160\right ) \log (5-x)-\left (-12 x^2+e^{x^2+10 x+25} \left (x^3+12 x^2+32 x\right )+e^{x^2+10 x+25} \left (2 x^5+24 x^4+14 x^3-604 x^2-1580 x\right ) \log (5-x)+60 x\right ) \log (x)+84 x+480}{x \left (-x^3-7 x^2+28 x+160\right ) \left (e^{(x+5)^2} \log (5-x)+3\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (-\frac {-3 x^3-21 x^2-e^{x^2+10 x+25} \left (x^3+7 x^2-28 x-160\right ) \log (5-x)-\left (-12 x^2+e^{x^2+10 x+25} \left (x^3+12 x^2+32 x\right )+e^{x^2+10 x+25} \left (2 x^5+24 x^4+14 x^3-604 x^2-1580 x\right ) \log (5-x)+60 x\right ) \log (x)+84 x+480}{117 (x-5) x \left (e^{(x+5)^2} \log (5-x)+3\right ) \log (x)}+\frac {-3 x^3-21 x^2-e^{x^2+10 x+25} \left (x^3+7 x^2-28 x-160\right ) \log (5-x)-\left (-12 x^2+e^{x^2+10 x+25} \left (x^3+12 x^2+32 x\right )+e^{x^2+10 x+25} \left (2 x^5+24 x^4+14 x^3-604 x^2-1580 x\right ) \log (5-x)+60 x\right ) \log (x)+84 x+480}{36 x (x+4) \left (e^{(x+5)^2} \log (5-x)+3\right ) \log (x)}-\frac {-3 x^3-21 x^2-e^{x^2+10 x+25} \left (x^3+7 x^2-28 x-160\right ) \log (5-x)-\left (-12 x^2+e^{x^2+10 x+25} \left (x^3+12 x^2+32 x\right )+e^{x^2+10 x+25} \left (2 x^5+24 x^4+14 x^3-604 x^2-1580 x\right ) \log (5-x)+60 x\right ) \log (x)+84 x+480}{52 x (x+8) \left (e^{(x+5)^2} \log (5-x)+3\right ) \log (x)}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-x \left (e^{(x+5)^2} \left (x^2+12 x+32\right )-12 (x-5)\right ) \log (x)-3 \left (x^3+7 x^2-28 x-160\right )-e^{(x+5)^2} (x-5) \log (5-x) \left (x^2+2 \left (x^3+17 x^2+92 x+158\right ) x \log (x)+12 x+32\right )}{(5-x) x (x+4) (x+8) \left (e^{(x+5)^2} \log (5-x)+3\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x^5 \log (5-x) \log (x)+24 x^4 \log (5-x) \log (x)+x^3 \log (5-x)+14 x^3 \log (5-x) \log (x)+x^3 \log (x)+7 x^2 \log (5-x)-604 x^2 \log (5-x) \log (x)+12 x^2 \log (x)-28 x \log (5-x)-1580 x \log (5-x) \log (x)+32 x \log (x)-160 \log (5-x)}{(x-5) x (x+4) (x+8) \log (5-x) \log (x)}-\frac {3 \left (2 x^2 \log (5-x)-50 \log (5-x)+1\right )}{(x-5) \log (5-x) \left (e^{(x+5)^2} \log (5-x)+3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -30 \int \frac {1}{e^{(x+5)^2} \log (5-x)+3}dx-6 \int \frac {x}{e^{(x+5)^2} \log (5-x)+3}dx-3 \int \frac {1}{(x-5) \log (5-x) \left (e^{(x+5)^2} \log (5-x)+3\right )}dx+x^2+10 x-\log (x+4)+\log (x+8)+\log (\log (5-x))+\log (\log (x))\) |
Int[(-480 - 84*x + 21*x^2 + 3*x^3 + E^(25 + 10*x + x^2)*(-160 - 28*x + 7*x ^2 + x^3)*Log[5 - x] + (60*x - 12*x^2 + E^(25 + 10*x + x^2)*(32*x + 12*x^2 + x^3) + E^(25 + 10*x + x^2)*(-1580*x - 604*x^2 + 14*x^3 + 24*x^4 + 2*x^5 )*Log[5 - x])*Log[x])/((-480*x - 84*x^2 + 21*x^3 + 3*x^4 + E^(25 + 10*x + x^2)*(-160*x - 28*x^2 + 7*x^3 + x^4)*Log[5 - x])*Log[x]),x]
3.14.95.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 25.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
| method | result | size |
| parallelrisch | \(\ln \left (\ln \left (x \right )\right )-\ln \left (4+x \right )+\ln \left (x +8\right )+\ln \left ({\mathrm e}^{x^{2}+10 x +25} \ln \left (5-x \right )+3\right )\) | \(34\) |
| risch | \(x^{2}+10 x -\ln \left (4+x \right )+\ln \left (x +8\right )+\ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (5-x \right )+3 \,{\mathrm e}^{-\left (5+x \right )^{2}}\right )\) | \(39\) |
int((((2*x^5+24*x^4+14*x^3-604*x^2-1580*x)*exp(x^2+10*x+25)*ln(5-x)+(x^3+1 2*x^2+32*x)*exp(x^2+10*x+25)-12*x^2+60*x)*ln(x)+(x^3+7*x^2-28*x-160)*exp(x ^2+10*x+25)*ln(5-x)+3*x^3+21*x^2-84*x-480)/((x^4+7*x^3-28*x^2-160*x)*exp(x ^2+10*x+25)*ln(5-x)+3*x^4+21*x^3-84*x^2-480*x)/ln(x),x,method=_RETURNVERBO SE)
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=x^{2} + 10 \, x + \log \left ({\left (e^{\left (x^{2} + 10 \, x + 25\right )} \log \left (-x + 5\right ) + 3\right )} e^{\left (-x^{2} - 10 \, x - 25\right )}\right ) + \log \left (x + 8\right ) - \log \left (x + 4\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^5+24*x^4+14*x^3-604*x^2-1580*x)*exp(x^2+10*x+25)*log(5-x) +(x^3+12*x^2+32*x)*exp(x^2+10*x+25)-12*x^2+60*x)*log(x)+(x^3+7*x^2-28*x-16 0)*exp(x^2+10*x+25)*log(5-x)+3*x^3+21*x^2-84*x-480)/((x^4+7*x^3-28*x^2-160 *x)*exp(x^2+10*x+25)*log(5-x)+3*x^4+21*x^3-84*x^2-480*x)/log(x),x, algorit hm=\
x^2 + 10*x + log((e^(x^2 + 10*x + 25)*log(-x + 5) + 3)*e^(-x^2 - 10*x - 25 )) + log(x + 8) - log(x + 4) + log(log(x))
Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=- \log {\left (x + 4 \right )} + \log {\left (x + 8 \right )} + \log {\left (e^{x^{2} + 10 x + 25} + \frac {3}{\log {\left (5 - x \right )}} \right )} + \log {\left (\log {\left (x \right )} \right )} + \log {\left (\log {\left (5 - x \right )} \right )} \]
integrate((((2*x**5+24*x**4+14*x**3-604*x**2-1580*x)*exp(x**2+10*x+25)*ln( 5-x)+(x**3+12*x**2+32*x)*exp(x**2+10*x+25)-12*x**2+60*x)*ln(x)+(x**3+7*x** 2-28*x-160)*exp(x**2+10*x+25)*ln(5-x)+3*x**3+21*x**2-84*x-480)/((x**4+7*x* *3-28*x**2-160*x)*exp(x**2+10*x+25)*ln(5-x)+3*x**4+21*x**3-84*x**2-480*x)/ ln(x),x)
-log(x + 4) + log(x + 8) + log(exp(x**2 + 10*x + 25) + 3/log(5 - x)) + log (log(x)) + log(log(5 - x))
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=x^{2} + 10 \, x + \log \left ({\left (e^{\left (x^{2} + 10 \, x + 25\right )} \log \left (-x + 5\right ) + 3\right )} e^{\left (-x^{2} - 10 \, x - 25\right )}\right ) + \log \left (x + 8\right ) - \log \left (x + 4\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^5+24*x^4+14*x^3-604*x^2-1580*x)*exp(x^2+10*x+25)*log(5-x) +(x^3+12*x^2+32*x)*exp(x^2+10*x+25)-12*x^2+60*x)*log(x)+(x^3+7*x^2-28*x-16 0)*exp(x^2+10*x+25)*log(5-x)+3*x^3+21*x^2-84*x-480)/((x^4+7*x^3-28*x^2-160 *x)*exp(x^2+10*x+25)*log(5-x)+3*x^4+21*x^3-84*x^2-480*x)/log(x),x, algorit hm=\
x^2 + 10*x + log((e^(x^2 + 10*x + 25)*log(-x + 5) + 3)*e^(-x^2 - 10*x - 25 )) + log(x + 8) - log(x + 4) + log(log(x))
Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=\log \left (e^{\left (x^{2} + 10 \, x + 25\right )} \log \left (-x + 5\right ) + 3\right ) + \log \left (x + 8\right ) - \log \left (x + 4\right ) + \log \left (\log \left (x\right )\right ) \]
integrate((((2*x^5+24*x^4+14*x^3-604*x^2-1580*x)*exp(x^2+10*x+25)*log(5-x) +(x^3+12*x^2+32*x)*exp(x^2+10*x+25)-12*x^2+60*x)*log(x)+(x^3+7*x^2-28*x-16 0)*exp(x^2+10*x+25)*log(5-x)+3*x^3+21*x^2-84*x-480)/((x^4+7*x^3-28*x^2-160 *x)*exp(x^2+10*x+25)*log(5-x)+3*x^4+21*x^3-84*x^2-480*x)/log(x),x, algorit hm=\
Time = 12.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-480-84 x+21 x^2+3 x^3+e^{25+10 x+x^2} \left (-160-28 x+7 x^2+x^3\right ) \log (5-x)+\left (60 x-12 x^2+e^{25+10 x+x^2} \left (32 x+12 x^2+x^3\right )+e^{25+10 x+x^2} \left (-1580 x-604 x^2+14 x^3+24 x^4+2 x^5\right ) \log (5-x)\right ) \log (x)}{\left (-480 x-84 x^2+21 x^3+3 x^4+e^{25+10 x+x^2} \left (-160 x-28 x^2+7 x^3+x^4\right ) \log (5-x)\right ) \log (x)} \, dx=\ln \left (\ln \left (5-x\right )\right )+\ln \left (\ln \left (x\right )\right )+\ln \left (\frac {{\mathrm {e}}^{{\left (x+5\right )}^2}\,\ln \left (5-x\right )+3}{\ln \left (5-x\right )}\right )-\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{2}+3{}\mathrm {i}\right )\,2{}\mathrm {i} \]
int((84*x - log(x)*(60*x + exp(10*x + x^2 + 25)*(32*x + 12*x^2 + x^3) - 12 *x^2 + exp(10*x + x^2 + 25)*log(5 - x)*(14*x^3 - 604*x^2 - 1580*x + 24*x^4 + 2*x^5)) - 21*x^2 - 3*x^3 + exp(10*x + x^2 + 25)*log(5 - x)*(28*x - 7*x^ 2 - x^3 + 160) + 480)/(log(x)*(480*x + 84*x^2 - 21*x^3 - 3*x^4 + exp(10*x + x^2 + 25)*log(5 - x)*(160*x + 28*x^2 - 7*x^3 - x^4))),x)