Integrand size = 80, antiderivative size = 22 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=e^x \left (2 x-4 x (2+x)-\frac {4}{\log (3)}\right ) \log (x) \]
Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=-\frac {2 e^x \left (2+x^2 \log (9)+x \log (27)\right ) \log (x)}{\log (3)} \]
Integrate[(E^x*(-4 + (-6*x - 4*x^2)*Log[3])*(2 + (3*x + 2*x^2)*Log[3] + (2 *x + (3*x + 7*x^2 + 2*x^3)*Log[3])*Log[x]))/(Log[3]*(2*x + (3*x^2 + 2*x^3) *Log[3])),x]
Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(22)=44\).
Time = 0.58 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {27, 27, 2019, 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.
\(\displaystyle \int \frac {e^x \left (\left (-4 x^2-6 x\right ) \log (3)-4\right ) \left (\left (2 x^2+3 x\right ) \log (3)+\left (\left (2 x^3+7 x^2+3 x\right ) \log (3)+2 x\right ) \log (x)+2\right )}{\log (3) \left (\left (2 x^3+3 x^2\right ) \log (3)+2 x\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {2 e^x \left (\log (3) \left (2 x^2+3 x\right )+2\right ) \left (\log (3) \left (2 x^2+3 x\right )+\left (2 x+\left (2 x^3+7 x^2+3 x\right ) \log (3)\right ) \log (x)+2\right )}{2 x+\left (2 x^3+3 x^2\right ) \log (3)}dx}{\log (3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \int \frac {e^x \left (\log (3) \left (2 x^2+3 x\right )+2\right ) \left (\log (3) \left (2 x^2+3 x\right )+\left (2 x+\left (2 x^3+7 x^2+3 x\right ) \log (3)\right ) \log (x)+2\right )}{2 x+\left (2 x^3+3 x^2\right ) \log (3)}dx}{\log (3)}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\frac {2 \int \frac {e^x \left (\log (3) \left (2 x^2+3 x\right )+\left (2 x+\left (2 x^3+7 x^2+3 x\right ) \log (3)\right ) \log (x)+2\right )}{x}dx}{\log (3)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \int \left (\frac {e^x \left (\log (9) x^2+\log (27) x+2\right )}{x}+e^x \left (\log (9) x^2+7 \log (3) x+\log (27)+2\right ) \log (x)\right )dx}{\log (3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (e^x x^2 \log (9) \log (x)-2 e^x x \log (9) \log (x)+7 e^x x \log (3) \log (x)+e^x (2+\log (27)) \log (x)+2 e^x \log (9) \log (x)-7 e^x \log (3) \log (x)\right )}{\log (3)}\) |
Int[(E^x*(-4 + (-6*x - 4*x^2)*Log[3])*(2 + (3*x + 2*x^2)*Log[3] + (2*x + ( 3*x + 7*x^2 + 2*x^3)*Log[3])*Log[x]))/(Log[3]*(2*x + (3*x^2 + 2*x^3)*Log[3 ])),x]
(-2*(-7*E^x*Log[3]*Log[x] + 7*E^x*x*Log[3]*Log[x] + 2*E^x*Log[9]*Log[x] - 2*E^x*x*Log[9]*Log[x] + E^x*x^2*Log[9]*Log[x] + E^x*(2 + Log[27])*Log[x])) /Log[3]
3.22.16.3.1 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_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Time = 23.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18
method | result | size |
parallelrisch | \({\mathrm e}^{\ln \left (\frac {\left (\left (-4 x^{2}-6 x \right ) \ln \left (3\right )-4\right ) \ln \left (x \right )}{\ln \left (3\right )}\right )+x}\) | \(26\) |
risch | \(-\frac {4 \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right ) \ln \left (x \right ) {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )}{2}+\frac {i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}^{2} \operatorname {csgn}\left (i \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}{2}-i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3\right )\right )\right )}^{2}+x}}{\ln \left (3\right )}\) | \(186\) |
int((((2*x^3+7*x^2+3*x)*ln(3)+2*x)*ln(x)+(2*x^2+3*x)*ln(3)+2)*exp(ln(((-4* x^2-6*x)*ln(3)-4)*ln(x)/ln(3))+x)/((2*x^3+3*x^2)*ln(3)+2*x)/ln(x),x,method =_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=e^{\left (x + \log \left (-\frac {2 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \left (3\right ) + 2\right )} \log \left (x\right )}{\log \left (3\right )}\right )\right )} \]
integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp (log(((-4*x^2-6*x)*log(3)-4)*log(x)/log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/ log(x),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=\frac {\left (- 4 x^{2} \log {\left (3 \right )} \log {\left (x \right )} - 6 x \log {\left (3 \right )} \log {\left (x \right )} - 4 \log {\left (x \right )}\right ) e^{x}}{\log {\left (3 \right )}} \]
integrate((((2*x**3+7*x**2+3*x)*ln(3)+2*x)*ln(x)+(2*x**2+3*x)*ln(3)+2)*exp (ln(((-4*x**2-6*x)*ln(3)-4)*ln(x)/ln(3))+x)/((2*x**3+3*x**2)*ln(3)+2*x)/ln (x),x)
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=-\frac {2 \, {\left (2 \, x^{2} \log \left (3\right ) + 3 \, x \log \left (3\right ) + 2\right )} e^{x} \log \left (x\right )}{\log \left (3\right )} \]
integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp (log(((-4*x^2-6*x)*log(3)-4)*log(x)/log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/ log(x),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=-\frac {2 \, {\left (2 \, x^{2} e^{x} \log \left (3\right ) \log \left (x\right ) + 3 \, x e^{x} \log \left (3\right ) \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )\right )}}{\log \left (3\right )} \]
integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp (log(((-4*x^2-6*x)*log(3)-4)*log(x)/log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/ log(x),x, algorithm=\
Time = 16.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{\log (3) \left (2 x+\left (3 x^2+2 x^3\right ) \log (3)\right )} \, dx=-\frac {2\,{\mathrm {e}}^x\,\ln \left (x\right )\,\left (2\,\ln \left (3\right )\,x^2+3\,\ln \left (3\right )\,x+2\right )}{\ln \left (3\right )} \]