Integrand size = 150, antiderivative size = 26 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {1+x}{-x-\frac {5}{\log (2)}+\log (11+x)}\right ) \]
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right ) \]
Integrate[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + x] + (E^x*(-220 - 240*x - 20*x^2 + (-44*x - 48*x^2 - 4*x^3)*Log[2]) + E^x* (44 + 48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log[2 ] + Log[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log [2] + (11 + 12*x + x^2)*Log[2]*Log[11 + x]),x]
Time = 17.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7292, 7279, 7239, 27, 7293, 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.
| \(\displaystyle \int \frac {\left (e^x \left (4 x^2+48 x+44\right ) \log (2) \log (x+11)+e^x \left (-20 x^2+\left (-4 x^3-48 x^2-44 x\right ) \log (2)-240 x-220\right )\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+e^x (-20 x-220+40 \log (2))+e^x (4 x+44) \log (2) \log (x+11)}{-5 x^2+\left (x^2+12 x+11\right ) \log (2) \log (x+11)+\left (-x^3-12 x^2-11 x\right ) \log (2)-60 x-55} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (e^x \left (4 x^2+48 x+44\right ) \log (2) \log (x+11)+e^x \left (-20 x^2+\left (-4 x^3-48 x^2-44 x\right ) \log (2)-240 x-220\right )\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-e^x (-20 x-220+40 \log (2))-e^x (4 x+44) \log (2) \log (x+11)}{\left (x^2+12 x+11\right ) (x \log (2)-\log (2) \log (x+11)+5)}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {2 e^x \left (x^3 \log (2) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-x^2 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 x^2 \left (1+\frac {12 \log (2)}{5}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 x-x \log (2) \log (x+11)-12 x \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+60 x \left (1+\frac {11 \log (2)}{60}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-11 \log (2) \log (x+11)-11 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+55 \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{5 (x+1) (x \log (2)-\log (2) \log (x+11)+5)}+\frac {2 e^x \left (x^3 (-\log (2)) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+x^2 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-5 x^2 \left (1+\frac {12 \log (2)}{5}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-5 x+x \log (2) \log (x+11)+12 x \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-60 x \left (1+\frac {11 \log (2)}{60}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+11 \log (2) \log (x+11)+11 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-55 \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{5 (x+11) (x \log (2)-\log (2) \log (x+11)+5)}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {4 e^x \left (\left (x^2+12 x+11\right ) (x \log (2)+5) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 (x+11-\log (4))-(x+11) \log (2) \log (x+11) \left ((x+1) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+1\right )\right )}{(x+1) (x+11) (x \log (2)-\log (2) \log (x+11)+5)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int \frac {e^x \left (5 (x-\log (4)+11)+\left (x^2+12 x+11\right ) (\log (2) x+5) \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )-(x+11) \log (2) \log (x+11) \left ((x+1) \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )+1\right )\right )}{(x+1) (x+11) (\log (2) x-\log (2) \log (x+11)+5)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 4 \int \left (\frac {e^x \left (-\log (2) \log (x+11) x+5 x-11 \log (2) \log (x+11)+55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{(x+1) (x+11) (\log (2) x-\log (2) \log (x+11)+5)}+e^x \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 e^x \log \left (-\frac {(x+1) \log (2)}{x \log (2)-\log (2) \log (x+11)+5}\right )\) |
Int[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + x] + ( E^x*(-220 - 240*x - 20*x^2 + (-44*x - 48*x^2 - 4*x^3)*Log[2]) + E^x*(44 + 48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log[2] + Lo g[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log[2] + (11 + 12*x + x^2)*Log[2]*Log[11 + x]),x]
3.5.29.3.1 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]]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 32.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
| method | result | size |
| parallelrisch | \(4 \,{\mathrm e}^{x} \ln \left (\frac {\left (1+x \right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (11+x \right )-x \ln \left (2\right )-5}\right )\) | \(28\) |
| risch | \(-4 \,{\mathrm e}^{x} \ln \left (5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )\right )-2 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}-4 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}+2 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{3} {\mathrm e}^{x}+4 i \pi \,{\mathrm e}^{x}+4 \ln \left (\ln \left (2\right )\right ) {\mathrm e}^{x}+4 \ln \left (1+x \right ) {\mathrm e}^{x}\) | \(241\) |
int((((4*x^2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x^3-48*x^2-44*x)*ln(2)-20 *x^2-240*x-220)*exp(x))*ln((1+x)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5))+(4*x+44 )*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x^2+12*x+11)*ln(2)*l n(11+x)+(-x^3-12*x^2-11*x)*ln(2)-5*x^2-60*x-55),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 \, e^{x} \log \left (-\frac {{\left (x + 1\right )} \log \left (2\right )}{x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5}\right ) \]
integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 +12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg orithm=\
Time = 14.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^{x} \log {\left (\frac {\left (x + 1\right ) \log {\left (2 \right )}}{- x \log {\left (2 \right )} + \log {\left (2 \right )} \log {\left (x + 11 \right )} - 5} \right )} \]
integrate((((4*x**2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x**3-48*x**2-44*x) *ln(2)-20*x**2-240*x-220)*exp(x))*ln((1+x)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5 ))+(4*x+44)*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x**2+12*x+ 11)*ln(2)*ln(11+x)+(-x**3-12*x**2-11*x)*ln(2)-5*x**2-60*x-55),x)
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (-x \log \left (2\right ) + \log \left (2\right ) \log \left (x + 11\right ) - 5\right ) + 4 \, e^{x} \log \left (x + 1\right ) + 4 \, e^{x} \log \left (\log \left (2\right )\right ) \]
integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 +12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg orithm=\
Time = 0.59 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5\right ) + 4 \, e^{x} \log \left (-x \log \left (2\right ) - \log \left (2\right )\right ) \]
integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 +12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg orithm=\
Time = 0.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4\,\ln \left (-\frac {\ln \left (2\right )\,\left (x+1\right )}{x\,\ln \left (2\right )-\ln \left (x+11\right )\,\ln \left (2\right )+5}\right )\,{\mathrm {e}}^x \]
int((exp(x)*(20*x - 40*log(2) + 220) + log(-(log(2)*(x + 1))/(x*log(2) - l og(x + 11)*log(2) + 5))*(exp(x)*(240*x + log(2)*(44*x + 48*x^2 + 4*x^3) + 20*x^2 + 220) - log(x + 11)*exp(x)*log(2)*(48*x + 4*x^2 + 44)) - log(x + 1 1)*exp(x)*log(2)*(4*x + 44))/(60*x + log(2)*(11*x + 12*x^2 + x^3) + 5*x^2 - log(x + 11)*log(2)*(12*x + x^2 + 11) + 55),x)