Integrand size = 149, antiderivative size = 21 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (-5+9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}+x\right ) \end {dmath*}
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(21)=42\).
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.71 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (5-9 e^{\frac {48}{e^{12}}-\frac {8 x}{e^9}+\frac {x^2}{e^6}+\frac {256}{e^{12} \left (4+e^3 x\right )^2}-\frac {256}{e^{12} \left (4+e^3 x\right )}}-x\right ) \end {dmath*}
Integrate[(64 + 48*E^3*x + 12*E^6*x^2 + E^9*x^3 + E^(x^4/(16 + 8*E^3*x + E ^6*x^2))*(144*x^3 + 18*E^3*x^4))/(-320 + 64*x + E^9*(-5 + x)*x^3 + E^6*x^2 *(-60 + 12*x) + E^3*x*(-240 + 48*x) + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(57 6 + 432*E^3*x + 108*E^6*x^2 + 9*E^9*x^3)),x]
Log[5 - 9*E^(48/E^12 - (8*x)/E^9 + x^2/E^6 + 256/(E^12*(4 + E^3*x)^2) - 25 6/(E^12*(4 + E^3*x))) - x]
Time = 0.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7292, 7235}
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.
\(\displaystyle \int \frac {e^9 x^3+12 e^6 x^2+e^{\frac {x^4}{e^6 x^2+8 e^3 x+16}} \left (18 e^3 x^4+144 x^3\right )+48 e^3 x+64}{e^9 (x-5) x^3+e^6 (12 x-60) x^2+e^{\frac {x^4}{e^6 x^2+8 e^3 x+16}} \left (9 e^9 x^3+108 e^6 x^2+432 e^3 x+576\right )+e^3 (48 x-240) x+64 x-320} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^9 x^3-12 e^6 x^2-e^{\frac {x^4}{e^6 x^2+8 e^3 x+16}} \left (18 e^3 x^4+144 x^3\right )-48 e^3 x-64}{\left (-9 e^{\frac {x^4}{\left (e^3 x+4\right )^2}}-x+5\right ) \left (e^3 x+4\right )^3}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (-9 e^{\frac {x^4}{\left (e^3 x+4\right )^2}}-x+5\right )\) |
Int[(64 + 48*E^3*x + 12*E^6*x^2 + E^9*x^3 + E^(x^4/(16 + 8*E^3*x + E^6*x^2 ))*(144*x^3 + 18*E^3*x^4))/(-320 + 64*x + E^9*(-5 + x)*x^3 + E^6*x^2*(-60 + 12*x) + E^3*x*(-240 + 48*x) + E^(x^4/(16 + 8*E^3*x + E^6*x^2))*(576 + 43 2*E^3*x + 108*E^6*x^2 + 9*E^9*x^3)),x]
3.6.46.3.1 Defintions of rubi rules used
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 20.81 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38
method | result | size |
norman | \(\ln \left (x +9 \,{\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}-5\right )\) | \(29\) |
parallelrisch | \(\ln \left (9 \,{\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 \,{\mathrm e}^{3+\ln \left (x \right )}+16}}+x -5\right )\) | \(30\) |
risch | \({\mathrm e}^{3} {\mathrm e}^{-9} x^{2}-8 x \,{\mathrm e}^{-9}+\frac {\left (-256 x -768 \,{\mathrm e}^{-3}\right ) {\mathrm e}^{-9}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}-\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}+\ln \left ({\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}-\frac {5}{9}+\frac {x}{9}\right )\) | \(87\) |
int(((18*x^3*exp(3+ln(x))+144*x^3)*exp(x^4/(exp(3+ln(x))^2+8*exp(3+ln(x))+ 16))+exp(3+ln(x))^3+12*exp(3+ln(x))^2+48*exp(3+ln(x))+64)/((9*exp(3+ln(x)) ^3+108*exp(3+ln(x))^2+432*exp(3+ln(x))+576)*exp(x^4/(exp(3+ln(x))^2+8*exp( 3+ln(x))+16))+(-5+x)*exp(3+ln(x))^3+(12*x-60)*exp(3+ln(x))^2+(48*x-240)*ex p(3+ln(x))+64*x-320),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \end {dmath*}
integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3 +log(x))+16))+exp(3+log(x))^3+12*exp(3+log(x))^2+48*exp(3+log(x))+64)/((9* exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(3+ log(x))^2+8*exp(3+log(x))+16))+(-5+x)*exp(3+log(x))^3+(12*x-60)*exp(3+log( x))^2+(48*x-240)*exp(3+log(x))+64*x-320),x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log {\left (\frac {x}{9} + e^{\frac {x^{4}}{x^{2} e^{6} + 8 x e^{3} + 16}} - \frac {5}{9} \right )} \end {dmath*}
integrate(((18*x**3*exp(3+ln(x))+144*x**3)*exp(x**4/(exp(3+ln(x))**2+8*exp (3+ln(x))+16))+exp(3+ln(x))**3+12*exp(3+ln(x))**2+48*exp(3+ln(x))+64)/((9* exp(3+ln(x))**3+108*exp(3+ln(x))**2+432*exp(3+ln(x))+576)*exp(x**4/(exp(3+ ln(x))**2+8*exp(3+ln(x))+16))+(-5+x)*exp(3+ln(x))**3+(12*x-60)*exp(3+ln(x) )**2+(48*x-240)*exp(3+ln(x))+64*x-320),x)
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 5.14 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\frac {x^{3} e^{9} - 4 \, x^{2} e^{6} - 32 \, x e^{3} - 256}{x e^{15} + 4 \, e^{12}} + \log \left (\frac {1}{9} \, {\left ({\left (x - 5\right )} e^{\left (8 \, x e^{\left (-9\right )} + \frac {256}{x e^{15} + 4 \, e^{12}}\right )} + 9 \, e^{\left (x^{2} e^{\left (-6\right )} + \frac {256}{x^{2} e^{18} + 8 \, x e^{15} + 16 \, e^{12}} + 48 \, e^{\left (-12\right )}\right )}\right )} e^{\left (-x^{2} e^{\left (-6\right )} - 48 \, e^{\left (-12\right )}\right )}\right ) \end {dmath*}
integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3 +log(x))+16))+exp(3+log(x))^3+12*exp(3+log(x))^2+48*exp(3+log(x))+64)/((9* exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(3+ log(x))^2+8*exp(3+log(x))+16))+(-5+x)*exp(3+log(x))^3+(12*x-60)*exp(3+log( x))^2+(48*x-240)*exp(3+log(x))+64*x-320),x, algorithm=\
(x^3*e^9 - 4*x^2*e^6 - 32*x*e^3 - 256)/(x*e^15 + 4*e^12) + log(1/9*((x - 5 )*e^(8*x*e^(-9) + 256/(x*e^15 + 4*e^12)) + 9*e^(x^2*e^(-6) + 256/(x^2*e^18 + 8*x*e^15 + 16*e^12) + 48*e^(-12)))*e^(-x^2*e^(-6) - 48*e^(-12)))
Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \end {dmath*}
integrate(((18*x^3*exp(3+log(x))+144*x^3)*exp(x^4/(exp(3+log(x))^2+8*exp(3 +log(x))+16))+exp(3+log(x))^3+12*exp(3+log(x))^2+48*exp(3+log(x))+64)/((9* exp(3+log(x))^3+108*exp(3+log(x))^2+432*exp(3+log(x))+576)*exp(x^4/(exp(3+ log(x))^2+8*exp(3+log(x))+16))+(-5+x)*exp(3+log(x))^3+(12*x-60)*exp(3+log( x))^2+(48*x-240)*exp(3+log(x))+64*x-320),x, algorithm=\
Time = 14.67 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \begin {dmath*} \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\ln \left (x+9\,{\mathrm {e}}^{\frac {x^4}{{\left (x\,{\mathrm {e}}^3+4\right )}^2}}-5\right ) \end {dmath*}
int((12*exp(2*log(x) + 6) + exp(3*log(x) + 9) + 48*exp(log(x) + 3) + exp(x ^4/(exp(2*log(x) + 6) + 8*exp(log(x) + 3) + 16))*(144*x^3 + 18*x^3*exp(log (x) + 3)) + 64)/(64*x + exp(x^4/(exp(2*log(x) + 6) + 8*exp(log(x) + 3) + 1 6))*(108*exp(2*log(x) + 6) + 9*exp(3*log(x) + 9) + 432*exp(log(x) + 3) + 5 76) + exp(3*log(x) + 9)*(x - 5) + exp(log(x) + 3)*(48*x - 240) + exp(2*log (x) + 6)*(12*x - 60) - 320),x)