Integrand size = 69, antiderivative size = 25 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\log (x)+\left (3-\frac {3}{4 \log \left (4+\frac {1}{4} (e+4 x)\right )}\right )^2 \]
Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\log (x)+\frac {9}{16 \log ^2\left (4+\frac {e}{4}+x\right )}-\frac {9}{2 \log \left (4+\frac {e}{4}+x\right )} \]
Integrate[(-9*x + 36*x*Log[(16 + E + 4*x)/4] + (32 + 2*E + 8*x)*Log[(16 + E + 4*x)/4]^3)/((32*x + 2*E*x + 8*x^2)*Log[(16 + E + 4*x)/4]^3),x]
Time = 1.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6, 2026, 7292, 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.
\(\displaystyle \int \frac {-9 x+(8 x+2 e+32) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )+36 x \log \left (\frac {1}{4} (4 x+e+16)\right )}{\left (8 x^2+2 e x+32 x\right ) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-9 x+(8 x+2 e+32) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )+36 x \log \left (\frac {1}{4} (4 x+e+16)\right )}{\left (8 x^2+(32+2 e) x\right ) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-9 x+(8 x+2 e+32) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )+36 x \log \left (\frac {1}{4} (4 x+e+16)\right )}{x (8 x+2 (16+e)) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-9 x+(8 x+2 e+32) \log ^3\left (\frac {1}{4} (4 x+e+16)\right )+36 x \log \left (\frac {1}{4} (4 x+e+16)\right )}{x (8 x+2 (16+e)) \log ^3\left (x+\frac {16+e}{4}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{x}-\frac {9}{2 (4 x+e+16) \log ^3\left (x+\frac {e}{4}+4\right )}+\frac {18}{(4 x+e+16) \log ^2\left (x+\frac {e}{4}+4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9}{16 \log ^2\left (x+\frac {16+e}{4}\right )}+\log (x)-\frac {9}{2 \log \left (x+\frac {16+e}{4}\right )}\) |
Int[(-9*x + 36*x*Log[(16 + E + 4*x)/4] + (32 + 2*E + 8*x)*Log[(16 + E + 4* x)/4]^3)/((32*x + 2*E*x + 8*x^2)*Log[(16 + E + 4*x)/4]^3),x]
3.16.6.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
norman | \(\frac {\frac {9}{16}-\frac {9 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}{2}}{\ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}+\ln \left (x \right )\) | \(27\) |
risch | \(\ln \left (x \right )-\frac {9 \left (-1+8 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )\right )}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}\) | \(28\) |
parts | \(\ln \left (x \right )+\frac {9}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}-\frac {9}{2 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}\) | \(28\) |
derivativedivides | \(\ln \left (-4 x \right )+\frac {9}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}-\frac {9}{2 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}\) | \(30\) |
default | \(\ln \left (-4 x \right )+\frac {9}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}-\frac {9}{2 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}\) | \(30\) |
parallelrisch | \(\frac {32 \ln \left (x \right ) \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}+18-144 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}{32 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}\) | \(39\) |
int(((2*exp(1)+8*x+32)*ln(1/4*exp(1)+x+4)^3+36*x*ln(1/4*exp(1)+x+4)-9*x)/( 2*x*exp(1)+8*x^2+32*x)/ln(1/4*exp(1)+x+4)^3,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\frac {16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2} \log \left (x\right ) - 72 \, \log \left (x + \frac {1}{4} \, e + 4\right ) + 9}{16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}} \]
integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4 )-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1/4*exp(1)+x+4)^3,x, algorithm=\
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\frac {9 - 72 \log {\left (x + \frac {e}{4} + 4 \right )}}{16 \log {\left (x + \frac {e}{4} + 4 \right )}^{2}} + \log {\left (x \right )} \]
integrate(((2*exp(1)+8*x+32)*ln(1/4*exp(1)+x+4)**3+36*x*ln(1/4*exp(1)+x+4) -9*x)/(2*x*exp(1)+8*x**2+32*x)/ln(1/4*exp(1)+x+4)**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (19) = 38\).
Time = 0.38 (sec) , antiderivative size = 287, normalized size of antiderivative = 11.48 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=-{\left (\frac {\log \left (4 \, x + e + 16\right )}{e + 16} - \frac {\log \left (x\right )}{e + 16}\right )} e - \frac {\log \left (x + \frac {1}{4} \, e + 4\right )^{3}}{2 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + 3 \, {\left (2 \, \log \left (2\right ) - \log \left (4 \, x + e + 16\right )\right )} \log \left (-2 \, \log \left (2\right ) + \log \left (4 \, x + e + 16\right )\right ) + 3 \, \log \left (x + \frac {1}{4} \, e + 4\right ) \log \left (-2 \, \log \left (2\right ) + \log \left (4 \, x + e + 16\right )\right ) + \frac {3 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}}{2 \, {\left (2 \, \log \left (2\right ) - \log \left (4 \, x + e + 16\right )\right )}} - \frac {16 \, \log \left (4 \, x + e + 16\right )}{e + 16} - \frac {9 \, \log \left (x + \frac {1}{4} \, e + 4\right )}{4 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + \frac {16 \, \log \left (x\right )}{e + 16} + \frac {9}{16 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + \frac {9}{4 \, {\left (2 \, \log \left (2\right ) - \log \left (4 \, x + e + 16\right )\right )}} + 3 \, \log \left (4 \, x + e + 16\right ) \]
integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4 )-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1/4*exp(1)+x+4)^3,x, algorithm=\
-(log(4*x + e + 16)/(e + 16) - log(x)/(e + 16))*e - 1/2*log(x + 1/4*e + 4) ^3/(4*log(2)^2 - 4*log(2)*log(4*x + e + 16) + log(4*x + e + 16)^2) + 3*(2* log(2) - log(4*x + e + 16))*log(-2*log(2) + log(4*x + e + 16)) + 3*log(x + 1/4*e + 4)*log(-2*log(2) + log(4*x + e + 16)) + 3/2*log(x + 1/4*e + 4)^2/ (2*log(2) - log(4*x + e + 16)) - 16*log(4*x + e + 16)/(e + 16) - 9/4*log(x + 1/4*e + 4)/(4*log(2)^2 - 4*log(2)*log(4*x + e + 16) + log(4*x + e + 16) ^2) + 16*log(x)/(e + 16) + 9/16/(4*log(2)^2 - 4*log(2)*log(4*x + e + 16) + log(4*x + e + 16)^2) + 9/4/(2*log(2) - log(4*x + e + 16)) + 3*log(4*x + e + 16)
Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\frac {16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2} \log \left (x\right ) - 72 \, \log \left (x + \frac {1}{4} \, e + 4\right ) + 9}{16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}} \]
integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4 )-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1/4*exp(1)+x+4)^3,x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left (32 x+2 e x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx=\frac {16\,\ln \left (x\right )\,{\ln \left (x+\frac {\mathrm {e}}{4}+4\right )}^2-72\,\ln \left (x+\frac {\mathrm {e}}{4}+4\right )+9}{16\,{\ln \left (x+\frac {\mathrm {e}}{4}+4\right )}^2} \]
int((log(x + exp(1)/4 + 4)^3*(8*x + 2*exp(1) + 32) - 9*x + 36*x*log(x + ex p(1)/4 + 4))/(log(x + exp(1)/4 + 4)^3*(32*x + 2*x*exp(1) + 8*x^2)),x)