Integrand size = 115, antiderivative size = 26 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=1+\left (-x+e \log \left (-x (-16+3 x)+\log \left (e^2 x\right )\right )\right )^2 \] Output:
(ln(ln(exp(2)*x)-x*(3*x-16))*exp(1)-x)^2+1
Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=\left (-x+e \log \left (2+16 x-3 x^2+\log (x)\right )\right )^2 \] Input:
Integrate[(32*x^3 - 6*x^4 + E*(-2*x - 32*x^2 + 12*x^3) + 2*x^2*Log[E^2*x] + (E^2*(2 + 32*x - 12*x^2) + E*(-32*x^2 + 6*x^3) - 2*E*x*Log[E^2*x])*Log[1 6*x - 3*x^2 + Log[E^2*x]])/(16*x^2 - 3*x^3 + x*Log[E^2*x]),x]
Output:
(-x + E*Log[2 + 16*x - 3*x^2 + Log[x]])^2
Time = 0.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {-6 x^4+32 x^3+2 x^2 \log \left (e^2 x\right )+e \left (12 x^3-32 x^2-2 x\right )+\left (e^2 \left (-12 x^2+32 x+2\right )+e \left (6 x^3-32 x^2\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (-3 x^2+16 x+\log \left (e^2 x\right )\right )}{-3 x^3+16 x^2+x \log \left (e^2 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (x \left (-3 x^2+16 x+2\right )+e \left (6 x^2-16 x-1\right )+x \log (x)\right ) \left (x-e \log \left (-3 x^2+16 x+\log (x)+2\right )\right )}{x \left (-3 x^2+16 x+\log (x)+2\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\left (e \left (-6 x^2+16 x+1\right )-x \left (-3 x^2+16 x+2\right )-x \log (x)\right ) \left (x-e \log \left (-3 x^2+16 x+\log (x)+2\right )\right )}{x \left (-3 x^2+16 x+\log (x)+2\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\left (e \left (-6 x^2+16 x+1\right )-x \left (-3 x^2+16 x+2\right )-x \log (x)\right ) \left (x-e \log \left (-3 x^2+16 x+\log (x)+2\right )\right )}{x \left (-3 x^2+16 x+\log (x)+2\right )}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (x-e \log \left (-3 x^2+16 x+\log (x)+2\right )\right )^2\) |
Input:
Int[(32*x^3 - 6*x^4 + E*(-2*x - 32*x^2 + 12*x^3) + 2*x^2*Log[E^2*x] + (E^2 *(2 + 32*x - 12*x^2) + E*(-32*x^2 + 6*x^3) - 2*E*x*Log[E^2*x])*Log[16*x - 3*x^2 + Log[E^2*x]])/(16*x^2 - 3*x^3 + x*Log[E^2*x]),x]
Output:
(x - E*Log[2 + 16*x - 3*x^2 + Log[x]])^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 8.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73
method | result | size |
risch | \({\mathrm e}^{2} {\ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right )}^{2}-2 \ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right ) {\mathrm e} x +x^{2}\) | \(45\) |
parallelrisch | \({\mathrm e}^{2} {\ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right )}^{2}-2 \ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right ) {\mathrm e} x +x^{2}\) | \(47\) |
Input:
int(((-2*x*exp(1)*ln(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*ex p(1))*ln(ln(exp(2)*x)-3*x^2+16*x)+2*x^2*ln(exp(2)*x)+(12*x^3-32*x^2-2*x)*e xp(1)-6*x^4+32*x^3)/(x*ln(exp(2)*x)-3*x^3+16*x^2),x,method=_RETURNVERBOSE)
Output:
exp(2)*ln(ln(exp(2)*x)-3*x^2+16*x)^2-2*ln(ln(exp(2)*x)-3*x^2+16*x)*exp(1)* x+x^2
Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=-2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \left (x e^{2}\right )\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \left (x e^{2}\right )\right )^{2} + x^{2} \] Input:
integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32* x^2)*exp(1))*log(log(exp(2)*x)-3*x^2+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32* x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorithm= "fricas")
Output:
-2*x*e*log(-3*x^2 + 16*x + log(x*e^2)) + e^2*log(-3*x^2 + 16*x + log(x*e^2 ))^2 + x^2
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=x^{2} - 2 e x \log {\left (- 3 x^{2} + 16 x + \log {\left (x e^{2} \right )} \right )} + e^{2} \log {\left (- 3 x^{2} + 16 x + \log {\left (x e^{2} \right )} \right )}^{2} \] Input:
integrate(((-2*x*exp(1)*ln(exp(2)*x)+(-12*x**2+32*x+2)*exp(1)**2+(6*x**3-3 2*x**2)*exp(1))*ln(ln(exp(2)*x)-3*x**2+16*x)+2*x**2*ln(exp(2)*x)+(12*x**3- 32*x**2-2*x)*exp(1)-6*x**4+32*x**3)/(x*ln(exp(2)*x)-3*x**3+16*x**2),x)
Output:
x**2 - 2*E*x*log(-3*x**2 + 16*x + log(x*exp(2))) + exp(2)*log(-3*x**2 + 16 *x + log(x*exp(2)))**2
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=-2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \left (x\right ) + 2\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \left (x\right ) + 2\right )^{2} + x^{2} \] Input:
integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32* x^2)*exp(1))*log(log(exp(2)*x)-3*x^2+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32* x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorithm= "maxima")
Output:
-2*x*e*log(-3*x^2 + 16*x + log(x) + 2) + e^2*log(-3*x^2 + 16*x + log(x) + 2)^2 + x^2
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx=-2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \left (x\right ) + 2\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \left (x\right ) + 2\right )^{2} + x^{2} \] Input:
integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32* x^2)*exp(1))*log(log(exp(2)*x)-3*x^2+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32* x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorithm= "giac")
Output:
-2*x*e*log(-3*x^2 + 16*x + log(x) + 2) + e^2*log(-3*x^2 + 16*x + log(x) + 2)^2 + x^2
Time = 2.94 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx={\left (x-\mathrm {e}\,\ln \left (16\,x+\ln \left (x\right )-3\,x^2+2\right )\right )}^2 \] Input:
int(-(exp(1)*(2*x + 32*x^2 - 12*x^3) - 32*x^3 + 6*x^4 - 2*x^2*log(x*exp(2) ) + log(16*x + log(x*exp(2)) - 3*x^2)*(exp(1)*(32*x^2 - 6*x^3) - exp(2)*(3 2*x - 12*x^2 + 2) + 2*x*exp(1)*log(x*exp(2))))/(x*log(x*exp(2)) + 16*x^2 - 3*x^3),x)
Output:
(x - exp(1)*log(16*x + log(x) - 3*x^2 + 2))^2
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {32 x^3-6 x^4+e \left (-2 x-32 x^2+12 x^3\right )+2 x^2 \log \left (e^2 x\right )+\left (e^2 \left (2+32 x-12 x^2\right )+e \left (-32 x^2+6 x^3\right )-2 e x \log \left (e^2 x\right )\right ) \log \left (16 x-3 x^2+\log \left (e^2 x\right )\right )}{16 x^2-3 x^3+x \log \left (e^2 x\right )} \, dx={\mathrm {log}\left (\mathrm {log}\left (e^{2} x \right )-3 x^{2}+16 x \right )}^{2} e^{2}-2 \,\mathrm {log}\left (\mathrm {log}\left (e^{2} x \right )-3 x^{2}+16 x \right ) e x +x^{2} \] Input:
int(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*e xp(1))*log(log(exp(2)*x)-3*x^2+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32*x^2-2* x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x)
Output:
log(log(e**2*x) - 3*x**2 + 16*x)**2*e**2 - 2*log(log(e**2*x) - 3*x**2 + 16 *x)*e*x + x**2