Integrand size = 97, antiderivative size = 24 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=\log \left (\left (-2+2 x-(4+2 x-e (1+\log (x)))^2\right )^2\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).
Time = 0.66 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2 \log \left (18-8 e+e^2+14 x-4 e x+4 x^2-8 e \log (x)+2 e^2 \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right ) \]
Integrate[(4*E^2 + E*(-16 - 16*x) + 28*x + 16*x^2 + (4*E^2 - 8*E*x)*Log[x] )/(18*x + E^2*x + 14*x^2 + 4*x^3 + E*(-8*x - 4*x^2) + (2*E^2*x + E*(-8*x - 4*x^2))*Log[x] + E^2*x*Log[x]^2),x]
2*Log[18 - 8*E + E^2 + 14*x - 4*E*x + 4*x^2 - 8*E*Log[x] + 2*E^2*Log[x] - 4*E*x*Log[x] + E^2*Log[x]^2]
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.
\(\displaystyle \int \frac {16 x^2+28 x+e (-16 x-16)+\left (4 e^2-8 e x\right ) \log (x)+4 e^2}{4 x^3+14 x^2+e \left (-4 x^2-8 x\right )+\left (e \left (-4 x^2-8 x\right )+2 e^2 x\right ) \log (x)+e^2 x+18 x+e^2 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {16 x^2+28 x+e (-16 x-16)+\left (4 e^2-8 e x\right ) \log (x)+4 e^2}{4 x^3+14 x^2+e \left (-4 x^2-8 x\right )+\left (e \left (-4 x^2-8 x\right )+2 e^2 x\right ) \log (x)+\left (18+e^2\right ) x+e^2 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 \left (4 x^2+7 \left (1-\frac {4 e}{7}\right ) x-2 e x \log (x)+e^2 \log (x)-4 \left (1-\frac {e}{4}\right ) e\right )}{4 x^3+14 x^2+e \left (-4 x^2-8 x\right )+\left (e \left (-4 x^2-8 x\right )+2 e^2 x\right ) \log (x)+\left (18+e^2\right ) x+e^2 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {-4 x^2+2 e \log (x) x-(7-4 e) x-e^2 \log (x)+(4-e) e}{4 x^3+14 x^2+e^2 \log ^2(x) x+\left (18+e^2\right ) x-4 e \left (x^2+2 x\right )+2 \left (e^2 x-2 e \left (x^2+2 x\right )\right ) \log (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {-4 x^2+2 e \log (x) x-(7-4 e) x-e^2 \log (x)+(4-e) e}{4 x^3+14 x^2+e^2 \log ^2(x) x+\left (18+e^2\right ) x-4 e \left (x^2+2 x\right )+2 \left (e^2 x-2 e \left (x^2+2 x\right )\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {4 x}{-4 x^2+4 e \log (x) x-14 \left (1-\frac {2 e}{7}\right ) x-e^2 \log ^2(x)+8 \left (1-\frac {e}{4}\right ) e \log (x)-18 \left (1+\frac {1}{18} (-8+e) e\right )}+\frac {2 e \log (x)}{4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )}+\frac {-7+4 e}{4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )}+\frac {e^2 \log (x)}{\left (-4 x^2+4 e \log (x) x-14 \left (1-\frac {2 e}{7}\right ) x-e^2 \log ^2(x)+8 \left (1-\frac {e}{4}\right ) e \log (x)-18 \left (1+\frac {1}{18} (-8+e) e\right )\right ) x}+\frac {(4-e) e}{\left (4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (4 \int \frac {x}{-4 x^2+4 e \log (x) x-14 \left (1-\frac {2 e}{7}\right ) x-e^2 \log ^2(x)+8 \left (1-\frac {e}{4}\right ) e \log (x)-18 \left (1+\frac {1}{18} (-8+e) e\right )}dx+e^2 \int \frac {\log (x)}{x \left (-4 x^2+4 e \log (x) x-14 \left (1-\frac {2 e}{7}\right ) x-e^2 \log ^2(x)+8 \left (1-\frac {e}{4}\right ) e \log (x)-18 \left (1+\frac {1}{18} (-8+e) e\right )\right )}dx-(7-4 e) \int \frac {1}{4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )}dx+(4-e) e \int \frac {1}{x \left (4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )\right )}dx+2 e \int \frac {\log (x)}{4 x^2-4 e \log (x) x+14 \left (1-\frac {2 e}{7}\right ) x+e^2 \log ^2(x)-8 \left (1-\frac {e}{4}\right ) e \log (x)+18 \left (1+\frac {1}{18} (-8+e) e\right )}dx\right )\) |
Int[(4*E^2 + E*(-16 - 16*x) + 28*x + 16*x^2 + (4*E^2 - 8*E*x)*Log[x])/(18* x + E^2*x + 14*x^2 + 4*x^3 + E*(-8*x - 4*x^2) + (2*E^2*x + E*(-8*x - 4*x^2 ))*Log[x] + E^2*x*Log[x]^2),x]
3.25.25.3.1 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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.74 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92
method | result | size |
risch | \(2 \ln \left (\ln \left (x \right )^{2}+2 \,{\mathrm e}^{-1} \left ({\mathrm e}-2 x -4\right ) \ln \left (x \right )+\left ({\mathrm e}^{2}-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right ) {\mathrm e}^{-2}\right )\) | \(46\) |
default | \(2 \ln \left ({\mathrm e}^{2} \ln \left (x \right )^{2}+2 \,{\mathrm e}^{2} \ln \left (x \right )-4 x \,{\mathrm e} \ln \left (x \right )+{\mathrm e}^{2}-8 \,{\mathrm e} \ln \left (x \right )-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right )\) | \(57\) |
norman | \(2 \ln \left ({\mathrm e}^{2} \ln \left (x \right )^{2}+2 \,{\mathrm e}^{2} \ln \left (x \right )-4 x \,{\mathrm e} \ln \left (x \right )+{\mathrm e}^{2}-8 \,{\mathrm e} \ln \left (x \right )-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right )\) | \(57\) |
parallelrisch | \(2 \ln \left (\frac {{\mathrm e}^{2} \ln \left (x \right )^{2}}{4}+\frac {{\mathrm e}^{2} \ln \left (x \right )}{2}-x \,{\mathrm e} \ln \left (x \right )+\frac {{\mathrm e}^{2}}{4}-2 \,{\mathrm e} \ln \left (x \right )-x \,{\mathrm e}+x^{2}-2 \,{\mathrm e}+\frac {7 x}{2}+\frac {9}{2}\right )\) | \(58\) |
int(((4*exp(1)^2-8*x*exp(1))*ln(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16*x^2+28* x)/(x*exp(1)^2*ln(x)^2+(2*x*exp(1)^2+(-4*x^2-8*x)*exp(1))*ln(x)+x*exp(1)^2 +(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x,method=_RETURNVERBOSE)
2*ln(ln(x)^2+2*exp(-1)*(exp(1)-2*x-4)*ln(x)+(exp(2)-4*x*exp(1)+4*x^2-8*exp (1)+14*x+18)*exp(-2))
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2 \, \log \left (e^{2} \log \left (x\right )^{2} + 4 \, x^{2} - 4 \, {\left (x + 2\right )} e - 2 \, {\left (2 \, {\left (x + 2\right )} e - e^{2}\right )} \log \left (x\right ) + 14 \, x + e^{2} + 18\right ) \]
integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16* x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x *exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2 \log {\left (\frac {\left (- 4 x - 8 + 2 e\right ) \log {\left (x \right )}}{e} + \frac {4 x^{2} - 4 e x + 14 x - 8 e + e^{2} + 18}{e^{2}} + \log {\left (x \right )}^{2} \right )} \]
integrate(((4*exp(1)**2-8*x*exp(1))*ln(x)+4*exp(1)**2+(-16*x-16)*exp(1)+16 *x**2+28*x)/(x*exp(1)**2*ln(x)**2+(2*x*exp(1)**2+(-4*x**2-8*x)*exp(1))*ln( x)+x*exp(1)**2+(-4*x**2-8*x)*exp(1)+4*x**3+14*x**2+18*x),x)
2*log((-4*x - 8 + 2*E)*exp(-1)*log(x) + (4*x**2 - 4*E*x + 14*x - 8*E + exp (2) + 18)*exp(-2) + log(x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2 \, \log \left ({\left (e^{2} \log \left (x\right )^{2} + 4 \, x^{2} - 2 \, x {\left (2 \, e - 7\right )} - 2 \, {\left (2 \, x e - e^{2} + 4 \, e\right )} \log \left (x\right ) + e^{2} - 8 \, e + 18\right )} e^{\left (-2\right )}\right ) \]
integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16* x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x *exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm=\
2*log((e^2*log(x)^2 + 4*x^2 - 2*x*(2*e - 7) - 2*(2*x*e - e^2 + 4*e)*log(x) + e^2 - 8*e + 18)*e^(-2))
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2 \, \log \left (-4 \, x e \log \left (x\right ) + e^{2} \log \left (x\right )^{2} + 4 \, x^{2} - 4 \, x e + 2 \, e^{2} \log \left (x\right ) - 8 \, e \log \left (x\right ) + 14 \, x + e^{2} - 8 \, e + 18\right ) \]
integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16* x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x *exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm=\
2*log(-4*x*e*log(x) + e^2*log(x)^2 + 4*x^2 - 4*x*e + 2*e^2*log(x) - 8*e*lo g(x) + 14*x + e^2 - 8*e + 18)
Time = 14.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx=2\,\ln \left (14\,x-8\,\mathrm {e}+{\mathrm {e}}^2+{\mathrm {e}}^2\,{\ln \left (x\right )}^2-4\,x\,\mathrm {e}-8\,\mathrm {e}\,\ln \left (x\right )+2\,{\mathrm {e}}^2\,\ln \left (x\right )+4\,x^2-4\,x\,\mathrm {e}\,\ln \left (x\right )+18\right ) \]
int((28*x + 4*exp(2) + log(x)*(4*exp(2) - 8*x*exp(1)) + 16*x^2 - exp(1)*(1 6*x + 16))/(18*x - log(x)*(exp(1)*(8*x + 4*x^2) - 2*x*exp(2)) - exp(1)*(8* x + 4*x^2) + x*exp(2) + 14*x^2 + 4*x^3 + x*exp(2)*log(x)^2),x)