Integrand size = 121, antiderivative size = 30 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\left (1+\frac {\left (5+5 e^2\right ) x^2}{\left (-3-e^4\right ) \log \left (x^2\right )}\right )^2 \]
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=-\frac {20 \left (1+e^2\right ) \left (-\frac {5 \left (1+e^2\right ) x^4}{4 \log ^2\left (x^2\right )}+\frac {\left (3+e^4\right ) x^2}{2 \log \left (x^2\right )}\right )}{\left (3+e^4\right )^2} \]
Integrate[(-100*x^3 - 200*E^2*x^3 - 100*E^4*x^3 + (60*x + 100*x^3 + 100*E^ 4*x^3 + E^4*(20*x + 20*E^2*x) + E^2*(60*x + 200*x^3))*Log[x^2] + (-60*x - 60*E^2*x + E^4*(-20*x - 20*E^2*x))*Log[x^2]^2)/((9 + 6*E^4 + E^8)*Log[x^2] ^3),x]
(-20*(1 + E^2)*((-5*(1 + E^2)*x^4)/(4*Log[x^2]^2) + ((3 + E^4)*x^2)/(2*Log [x^2])))/(3 + E^4)^2
Time = 0.47 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {6, 6, 27, 27, 7239, 27, 7263, 17}
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.
| \(\displaystyle \int \frac {-100 e^4 x^3-200 e^2 x^3-100 x^3+\left (-60 e^2 x-60 x+e^4 \left (-20 e^2 x-20 x\right )\right ) \log ^2\left (x^2\right )+\left (100 e^4 x^3+100 x^3+e^2 \left (200 x^3+60 x\right )+60 x+e^4 \left (20 e^2 x+20 x\right )\right ) \log \left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (-100-200 e^2\right ) x^3-100 e^4 x^3+\left (-60 e^2 x-60 x+e^4 \left (-20 e^2 x-20 x\right )\right ) \log ^2\left (x^2\right )+\left (100 e^4 x^3+100 x^3+e^2 \left (200 x^3+60 x\right )+60 x+e^4 \left (20 e^2 x+20 x\right )\right ) \log \left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (-100-200 e^2-100 e^4\right ) x^3+\left (-60 e^2 x-60 x+e^4 \left (-20 e^2 x-20 x\right )\right ) \log ^2\left (x^2\right )+\left (100 e^4 x^3+100 x^3+e^2 \left (200 x^3+60 x\right )+60 x+e^4 \left (20 e^2 x+20 x\right )\right ) \log \left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {20 \left (5 \left (1+e^2\right )^2 x^3+\left (1+e^2\right ) \left (3+e^4\right ) \log ^2\left (x^2\right ) x-\left (5 e^4 x^3+5 x^3+e^4 \left (1+e^2\right ) x+3 x+e^2 \left (10 x^3+3 x\right )\right ) \log \left (x^2\right )\right )}{\log ^3\left (x^2\right )}dx}{\left (3+e^4\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {20 \int \frac {5 \left (1+e^2\right )^2 x^3+\left (1+e^2\right ) \left (3+e^4\right ) \log ^2\left (x^2\right ) x-\left (5 e^4 x^3+5 x^3+e^4 \left (1+e^2\right ) x+3 x+e^2 \left (10 x^3+3 x\right )\right ) \log \left (x^2\right )}{\log ^3\left (x^2\right )}dx}{\left (3+e^4\right )^2}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -\frac {20 \int \frac {\left (1+e^2\right ) x \left (1-\log \left (x^2\right )\right ) \left (5 \left (1+e^2\right ) x^2-\left (3+e^4\right ) \log \left (x^2\right )\right )}{\log ^3\left (x^2\right )}dx}{\left (3+e^4\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {20 \left (1+e^2\right ) \int \frac {x \left (1-\log \left (x^2\right )\right ) \left (5 \left (1+e^2\right ) x^2-\left (3+e^4\right ) \log \left (x^2\right )\right )}{\log ^3\left (x^2\right )}dx}{\left (3+e^4\right )^2}\) |
| \(\Big \downarrow \) 7263 |
| \(\displaystyle \frac {10 \left (1+e^2\right ) \int \left (\frac {5 \left (1+e^2\right ) x^2}{\log \left (x^2\right )}-e^4-3\right )d\frac {x^2}{\log \left (x^2\right )}}{\left (3+e^4\right )^2}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle \frac {\left (-\frac {5 \left (1+e^2\right ) x^2}{\log \left (x^2\right )}+e^4+3\right )^2}{\left (3+e^4\right )^2}\) |
Int[(-100*x^3 - 200*E^2*x^3 - 100*E^4*x^3 + (60*x + 100*x^3 + 100*E^4*x^3 + E^4*(20*x + 20*E^2*x) + E^2*(60*x + 200*x^3))*Log[x^2] + (-60*x - 60*E^2 *x + E^4*(-20*x - 20*E^2*x))*Log[x^2]^2)/((9 + 6*E^4 + E^8)*Log[x^2]^3),x]
3.21.16.3.1 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[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w, x])]}, Simp[(-c)*q Subst[ Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ[{ a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && Inte gerQ[m]
Time = 0.77 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67
| method | result | size |
| norman | \(\frac {\left (-10 \,{\mathrm e}^{2}-10\right ) x^{2} \ln \left (x^{2}\right )+\frac {25 \left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2}+1\right ) x^{4}}{{\mathrm e}^{4}+3}}{\left ({\mathrm e}^{4}+3\right ) \ln \left (x^{2}\right )^{2}}\) | \(50\) |
| risch | \(-\frac {5 x^{2} \left (2 \,{\mathrm e}^{6} \ln \left (x^{2}\right )-5 x^{2} {\mathrm e}^{4}+2 \ln \left (x^{2}\right ) {\mathrm e}^{4}-10 x^{2} {\mathrm e}^{2}+6 \,{\mathrm e}^{2} \ln \left (x^{2}\right )-5 x^{2}+6 \ln \left (x^{2}\right )\right )}{\left ({\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9\right ) \ln \left (x^{2}\right )^{2}}\) | \(72\) |
| parallelrisch | \(\frac {50 x^{4} {\mathrm e}^{4}-20 \,{\mathrm e}^{2} {\mathrm e}^{4} x^{2} \ln \left (x^{2}\right )+100 x^{4} {\mathrm e}^{2}-60 \,{\mathrm e}^{2} \ln \left (x^{2}\right ) x^{2}-20 \,{\mathrm e}^{4} x^{2} \ln \left (x^{2}\right )+50 x^{4}-60 x^{2} \ln \left (x^{2}\right )}{2 \left ({\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9\right ) \ln \left (x^{2}\right )^{2}}\) | \(87\) |
| default | \(\frac {100 \left (-\frac {{\mathrm e}^{4}}{2}-{\mathrm e}^{2}-\frac {1}{2}\right ) \left (-\frac {x^{4}}{2 \ln \left (x^{2}\right )^{2}}-\frac {x^{4}}{\ln \left (x^{2}\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )-20 \left (-\frac {{\mathrm e}^{2} {\mathrm e}^{4}}{2}-\frac {3 \,{\mathrm e}^{2}}{2}-\frac {{\mathrm e}^{4}}{2}-\frac {3}{2}\right ) \operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )+10 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )+10 \,{\mathrm e}^{2} {\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )-\frac {30 x^{2}}{\ln \left (x^{2}\right )}-30 \,\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )-\frac {50 x^{4}}{\ln \left (x^{2}\right )}-100 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )+30 \,{\mathrm e}^{2} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )+100 \,{\mathrm e}^{2} \left (-\frac {x^{4}}{\ln \left (x^{2}\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )+50 \,{\mathrm e}^{4} \left (-\frac {x^{4}}{\ln \left (x^{2}\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )}{{\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9}\) | \(264\) |
| parts | \(\frac {10 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )+10 \,{\mathrm e}^{2} {\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )-\frac {30 x^{2}}{\ln \left (x^{2}\right )}-30 \,\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )-\frac {50 x^{4}}{\ln \left (x^{2}\right )}-100 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )+30 \,{\mathrm e}^{2} \left (-\frac {x^{2}}{\ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )\right )+100 \,{\mathrm e}^{2} \left (-\frac {x^{4}}{\ln \left (x^{2}\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )+50 \,{\mathrm e}^{4} \left (-\frac {x^{4}}{\ln \left (x^{2}\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )}{{\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9}-\frac {100 \left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2}+1\right ) \left (-\frac {x^{4}}{4 \ln \left (x^{2}\right )^{2}}-\frac {x^{4}}{2 \ln \left (x^{2}\right )}-\operatorname {Ei}_{1}\left (-2 \ln \left (x^{2}\right )\right )\right )}{{\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9}+\frac {10 \left ({\mathrm e}^{2}+1\right ) \operatorname {Ei}_{1}\left (-\ln \left (x^{2}\right )\right )}{{\mathrm e}^{4}+3}\) | \(270\) |
int((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*ln(x^2)^2+((20*exp(2)*x +20*x)*exp(4)+100*x^3*exp(2)^2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)*ln(x^2) -100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/ln(x^2)^3, x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {5 \, {\left (5 \, x^{4} e^{4} + 10 \, x^{4} e^{2} + 5 \, x^{4} - 2 \, {\left (x^{2} e^{6} + x^{2} e^{4} + 3 \, x^{2} e^{2} + 3 \, x^{2}\right )} \log \left (x^{2}\right )\right )}}{{\left (e^{8} + 6 \, e^{4} + 9\right )} \log \left (x^{2}\right )^{2}} \]
integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*e xp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)* log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/lo g(x^2)^3,x, algorithm=\
5*(5*x^4*e^4 + 10*x^4*e^2 + 5*x^4 - 2*(x^2*e^6 + x^2*e^4 + 3*x^2*e^2 + 3*x ^2)*log(x^2))/((e^8 + 6*e^4 + 9)*log(x^2)^2)
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (24) = 48\).
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.50 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {25 x^{4} + 50 x^{4} e^{2} + 25 x^{4} e^{4} + \left (- 10 x^{2} e^{6} - 10 x^{2} e^{4} - 30 x^{2} e^{2} - 30 x^{2}\right ) \log {\left (x^{2} \right )}}{\left (9 + 6 e^{4} + e^{8}\right ) \log {\left (x^{2} \right )}^{2}} \]
integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*ln(x**2)**2+((20* exp(2)*x+20*x)*exp(4)+100*x**3*exp(2)**2+(200*x**3+60*x)*exp(2)+100*x**3+6 0*x)*ln(x**2)-100*x**3*exp(2)**2-200*x**3*exp(2)-100*x**3)/(exp(4)**2+6*ex p(4)+9)/ln(x**2)**3,x)
(25*x**4 + 50*x**4*exp(2) + 25*x**4*exp(4) + (-10*x**2*exp(6) - 10*x**2*ex p(4) - 30*x**2*exp(2) - 30*x**2)*log(x**2))/((9 + 6*exp(4) + exp(8))*log(x **2)**2)
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {5 \, {\left (5 \, x^{4} {\left (e^{4} + 2 \, e^{2} + 1\right )} - 4 \, x^{2} {\left (e^{6} + e^{4} + 3 \, e^{2} + 3\right )} \log \left (x\right )\right )}}{4 \, {\left (e^{8} + 6 \, e^{4} + 9\right )} \log \left (x\right )^{2}} \]
integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*e xp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)* log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/lo g(x^2)^3,x, algorithm=\
5/4*(5*x^4*(e^4 + 2*e^2 + 1) - 4*x^2*(e^6 + e^4 + 3*e^2 + 3)*log(x))/((e^8 + 6*e^4 + 9)*log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (25) = 50\).
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.33 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {5 \, {\left (\frac {5 \, x^{4} e^{4}}{\log \left (x^{2}\right )^{2}} + \frac {10 \, x^{4} e^{2}}{\log \left (x^{2}\right )^{2}} + \frac {5 \, x^{4}}{\log \left (x^{2}\right )^{2}} - \frac {2 \, x^{2} e^{6}}{\log \left (x^{2}\right )} - \frac {2 \, x^{2} e^{4}}{\log \left (x^{2}\right )} - \frac {6 \, x^{2} e^{2}}{\log \left (x^{2}\right )} - \frac {6 \, x^{2}}{\log \left (x^{2}\right )}\right )}}{e^{8} + 6 \, e^{4} + 9} \]
integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*e xp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)* log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/lo g(x^2)^3,x, algorithm=\
5*(5*x^4*e^4/log(x^2)^2 + 10*x^4*e^2/log(x^2)^2 + 5*x^4/log(x^2)^2 - 2*x^2 *e^6/log(x^2) - 2*x^2*e^4/log(x^2) - 6*x^2*e^2/log(x^2) - 6*x^2/log(x^2))/ (e^8 + 6*e^4 + 9)
Time = 12.60 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {25\,x^5\,{\left ({\mathrm {e}}^2+1\right )}^2-x^3\,\ln \left (x^2\right )\,\left (30\,{\mathrm {e}}^2+10\,{\mathrm {e}}^4+10\,{\mathrm {e}}^6+30\right )}{x\,{\ln \left (x^2\right )}^2\,\left (6\,{\mathrm {e}}^4+{\mathrm {e}}^8+9\right )} \]
int(-(200*x^3*exp(2) - log(x^2)*(60*x + exp(2)*(60*x + 200*x^3) + exp(4)*( 20*x + 20*x*exp(2)) + 100*x^3*exp(4) + 100*x^3) + 100*x^3*exp(4) + log(x^2 )^2*(60*x + exp(4)*(20*x + 20*x*exp(2)) + 60*x*exp(2)) + 100*x^3)/(log(x^2 )^3*(6*exp(4) + exp(8) + 9)),x)