Integrand size = 133, antiderivative size = 28 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {3+\frac {25 e^{e^{\left (2+e^2 (3+x)^2\right )^2}}}{x^2}}{x} \]
Time = 4.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {25 e^{e^{\left (2+e^2 (3+x)^2\right )^2}}+3 x^2}{x^3} \]
Integrate[(-3*x^2 + E^E^(4 + E^2*(36 + 24*x + 4*x^2) + E^4*(81 + 108*x + 5 4*x^2 + 12*x^3 + x^4))*(-75 + E^(4 + E^2*(36 + 24*x + 4*x^2) + E^4*(81 + 1 08*x + 54*x^2 + 12*x^3 + x^4))*(E^2*(600*x + 200*x^2) + E^4*(2700*x + 2700 *x^2 + 900*x^3 + 100*x^4))))/x^4,x]
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.
\(\displaystyle \int \frac {\exp \left (\exp \left (e^2 \left (4 x^2+24 x+36\right )+e^4 \left (x^4+12 x^3+54 x^2+108 x+81\right )+4\right )\right ) \left (\left (e^2 \left (200 x^2+600 x\right )+e^4 \left (100 x^4+900 x^3+2700 x^2+2700 x\right )\right ) \exp \left (e^2 \left (4 x^2+24 x+36\right )+e^4 \left (x^4+12 x^3+54 x^2+108 x+81\right )+4\right )-75\right )-3 x^2}{x^4} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {100 (x+3) \left (e^2 x^2+6 e^2 x+9 e^2+2\right ) \exp \left (e^4 (x+3)^4+4 e^2 (x+3)^2+e^{\left (e^2 (x+3)^2+2\right )^2}+6\right )}{x^3}-\frac {3 \left (x^2+25 e^{e^{\left (e^2 (x+3)^2+2\right )^2}}\right )}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 300 \left (2+9 e^2\right ) \int \frac {\exp \left (e^4 (x+3)^4+4 e^2 (x+3)^2+e^{\left (e^2 (x+3)^2+2\right )^2}+6\right )}{x^3}dx+100 \left (2+27 e^2\right ) \int \frac {\exp \left (e^4 (x+3)^4+4 e^2 (x+3)^2+e^{\left (e^2 (x+3)^2+2\right )^2}+6\right )}{x^2}dx+900 \int \frac {\exp \left (e^4 (x+3)^4+4 e^2 (x+3)^2+e^{\left (e^2 (x+3)^2+2\right )^2}+8\right )}{x}dx+100 \text {Subst}\left (\int e^{e^4 x^4+4 e^2 x^2+e^{\left (e^2 x^2+2\right )^2}+8}dx,x,x+3\right )-75 \int \frac {e^{e^{\left (e^2 x^2+6 e^2 x+9 e^2+2\right )^2}}}{x^4}dx+\frac {3}{x}\) |
Int[(-3*x^2 + E^E^(4 + E^2*(36 + 24*x + 4*x^2) + E^4*(81 + 108*x + 54*x^2 + 12*x^3 + x^4))*(-75 + E^(4 + E^2*(36 + 24*x + 4*x^2) + E^4*(81 + 108*x + 54*x^2 + 12*x^3 + x^4))*(E^2*(600*x + 200*x^2) + E^4*(2700*x + 2700*x^2 + 900*x^3 + 100*x^4))))/x^4,x]
3.5.89.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96
method | result | size |
norman | \(\frac {3 x^{2}+25 \,{\mathrm e}^{{\mathrm e}^{\left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right ) {\mathrm e}^{4}+\left (4 x^{2}+24 x +36\right ) {\mathrm e}^{2}+4}}}{x^{3}}\) | \(55\) |
parallelrisch | \(\frac {3 x^{2}+25 \,{\mathrm e}^{{\mathrm e}^{\left (x^{4}+12 x^{3}+54 x^{2}+108 x +81\right ) {\mathrm e}^{4}+\left (4 x^{2}+24 x +36\right ) {\mathrm e}^{2}+4}}}{x^{3}}\) | \(55\) |
risch | \(\frac {3}{x}+\frac {25 \,{\mathrm e}^{{\mathrm e}^{x^{4} {\mathrm e}^{4}+12 x^{3} {\mathrm e}^{4}+4 x^{2} {\mathrm e}^{2}+54 x^{2} {\mathrm e}^{4}+24 \,{\mathrm e}^{2} x +108 x \,{\mathrm e}^{4}+36 \,{\mathrm e}^{2}+81 \,{\mathrm e}^{4}+4}}}{x^{3}}\) | \(61\) |
int(((((100*x^4+900*x^3+2700*x^2+2700*x)*exp(1)^4+(200*x^2+600*x)*exp(1)^2 )*exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp(1)^2+4)-75 )*exp(exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp(1)^2+4 ))-3*x^2)/x^4,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {3 \, x^{2} + 25 \, e^{\left (e^{\left ({\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )} e^{4} + 4 \, {\left (x^{2} + 6 \, x + 9\right )} e^{2} + 4\right )}\right )}}{x^{3}} \]
integrate(((((100*x^4+900*x^3+2700*x^2+2700*x)*exp(1)^4+(200*x^2+600*x)*ex p(1)^2)*exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp(1)^2 +4)-75)*exp(exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp( 1)^2+4))-3*x^2)/x^4,x, algorithm=\
(3*x^2 + 25*e^(e^((x^4 + 12*x^3 + 54*x^2 + 108*x + 81)*e^4 + 4*(x^2 + 6*x + 9)*e^2 + 4)))/x^3
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {3}{x} + \frac {25 e^{e^{\left (4 x^{2} + 24 x + 36\right ) e^{2} + \left (x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81\right ) e^{4} + 4}}}{x^{3}} \]
integrate(((((100*x**4+900*x**3+2700*x**2+2700*x)*exp(1)**4+(200*x**2+600* x)*exp(1)**2)*exp((x**4+12*x**3+54*x**2+108*x+81)*exp(1)**4+(4*x**2+24*x+3 6)*exp(1)**2+4)-75)*exp(exp((x**4+12*x**3+54*x**2+108*x+81)*exp(1)**4+(4*x **2+24*x+36)*exp(1)**2+4))-3*x**2)/x**4,x)
3/x + 25*exp(exp((4*x**2 + 24*x + 36)*exp(2) + (x**4 + 12*x**3 + 54*x**2 + 108*x + 81)*exp(4) + 4))/x**3
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {3}{x} + \frac {25 \, e^{\left (e^{\left (x^{4} e^{4} + 12 \, x^{3} e^{4} + 54 \, x^{2} e^{4} + 4 \, x^{2} e^{2} + 108 \, x e^{4} + 24 \, x e^{2} + 81 \, e^{4} + 36 \, e^{2} + 4\right )}\right )}}{x^{3}} \]
integrate(((((100*x^4+900*x^3+2700*x^2+2700*x)*exp(1)^4+(200*x^2+600*x)*ex p(1)^2)*exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp(1)^2 +4)-75)*exp(exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp( 1)^2+4))-3*x^2)/x^4,x, algorithm=\
3/x + 25*e^(e^(x^4*e^4 + 12*x^3*e^4 + 54*x^2*e^4 + 4*x^2*e^2 + 108*x*e^4 + 24*x*e^2 + 81*e^4 + 36*e^2 + 4))/x^3
\[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\int { -\frac {3 \, x^{2} - 25 \, {\left (4 \, {\left ({\left (x^{4} + 9 \, x^{3} + 27 \, x^{2} + 27 \, x\right )} e^{4} + 2 \, {\left (x^{2} + 3 \, x\right )} e^{2}\right )} e^{\left ({\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )} e^{4} + 4 \, {\left (x^{2} + 6 \, x + 9\right )} e^{2} + 4\right )} - 3\right )} e^{\left (e^{\left ({\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )} e^{4} + 4 \, {\left (x^{2} + 6 \, x + 9\right )} e^{2} + 4\right )}\right )}}{x^{4}} \,d x } \]
integrate(((((100*x^4+900*x^3+2700*x^2+2700*x)*exp(1)^4+(200*x^2+600*x)*ex p(1)^2)*exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp(1)^2 +4)-75)*exp(exp((x^4+12*x^3+54*x^2+108*x+81)*exp(1)^4+(4*x^2+24*x+36)*exp( 1)^2+4))-3*x^2)/x^4,x, algorithm=\
integrate(-(3*x^2 - 25*(4*((x^4 + 9*x^3 + 27*x^2 + 27*x)*e^4 + 2*(x^2 + 3* x)*e^2)*e^((x^4 + 12*x^3 + 54*x^2 + 108*x + 81)*e^4 + 4*(x^2 + 6*x + 9)*e^ 2 + 4) - 3)*e^(e^((x^4 + 12*x^3 + 54*x^2 + 108*x + 81)*e^4 + 4*(x^2 + 6*x + 9)*e^2 + 4)))/x^4, x)
Time = 12.55 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {-3 x^2+e^{e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )}} \left (-75+e^{4+e^2 \left (36+24 x+4 x^2\right )+e^4 \left (81+108 x+54 x^2+12 x^3+x^4\right )} \left (e^2 \left (600 x+200 x^2\right )+e^4 \left (2700 x+2700 x^2+900 x^3+100 x^4\right )\right )\right )}{x^4} \, dx=\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^{x^4\,{\mathrm {e}}^4}\,{\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{12\,x^3\,{\mathrm {e}}^4}\,{\mathrm {e}}^{54\,x^2\,{\mathrm {e}}^4}\,{\mathrm {e}}^{36\,{\mathrm {e}}^2}\,{\mathrm {e}}^{81\,{\mathrm {e}}^4}\,{\mathrm {e}}^4\,{\mathrm {e}}^{24\,x\,{\mathrm {e}}^2}\,{\mathrm {e}}^{108\,x\,{\mathrm {e}}^4}}}{x^3}+\frac {3}{x} \]
int((exp(exp(exp(2)*(24*x + 4*x^2 + 36) + exp(4)*(108*x + 54*x^2 + 12*x^3 + x^4 + 81) + 4))*(exp(exp(2)*(24*x + 4*x^2 + 36) + exp(4)*(108*x + 54*x^2 + 12*x^3 + x^4 + 81) + 4)*(exp(2)*(600*x + 200*x^2) + exp(4)*(2700*x + 27 00*x^2 + 900*x^3 + 100*x^4)) - 75) - 3*x^2)/x^4,x)