Integrand size = 83, antiderivative size = 24 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=-5+\frac {9 \left (e^x+3 \left (3+e^{\log ^2(x)}\right )+x\right )^2}{x^2} \] Output:
3*(exp(x)+x+3*exp(ln(x)^2)+9)/x^2*(3*exp(x)+3*x+9*exp(ln(x)^2)+27)-5
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\frac {9 \left (9+e^x+3 e^{\log ^2(x)}\right ) \left (9+e^x+3 e^{\log ^2(x)}+2 x\right )}{x^2} \] Input:
Integrate[(-1458 - 162*x + E^(2*x)*(-18 + 18*x) + E^x*(-324 + 144*x + 18*x ^2) + E^(2*Log[x]^2)*(-162 + 324*Log[x]) + E^Log[x]^2*(-972 - 54*x + E^x*( -108 + 54*x) + (972 + 108*E^x + 108*x)*Log[x]))/x^3,x]
Output:
(9*(9 + E^x + 3*E^Log[x]^2)*(9 + E^x + 3*E^Log[x]^2 + 2*x))/x^2
Leaf count is larger than twice the leaf count of optimal. \(80\) vs. \(2(24)=48\).
Time = 0.75 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2010, 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.
\(\displaystyle \int \frac {e^x \left (18 x^2+144 x-324\right )-162 x+e^{2 x} (18 x-18)+e^{2 \log ^2(x)} (324 \log (x)-162)+e^{\log ^2(x)} \left (-54 x+e^x (54 x-108)+\left (108 x+108 e^x+972\right ) \log (x)-972\right )-1458}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {162 e^{2 \log ^2(x)} (2 \log (x)-1)}{x^3}+\frac {54 e^{\log ^2(x)} \left (e^x x-x-2 e^x+2 x \log (x)+2 e^x \log (x)+18 \log (x)-18\right )}{x^3}+\frac {18 \left (e^x x^2+8 e^x x+e^{2 x} x-9 x-18 e^x-e^{2 x}-81\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 (x+9)^2}{x^2}+\frac {162 e^x}{x^2}+\frac {9 e^{2 x}}{x^2}+\frac {54 e^{\log ^2(x)} \left (e^x \log (x)+x \log (x)+9 \log (x)\right )}{x^2 \log (x)}+\frac {81 e^{2 \log ^2(x)}}{x^2}+\frac {18 e^x}{x}\) |
Input:
Int[(-1458 - 162*x + E^(2*x)*(-18 + 18*x) + E^x*(-324 + 144*x + 18*x^2) + E^(2*Log[x]^2)*(-162 + 324*Log[x]) + E^Log[x]^2*(-972 - 54*x + E^x*(-108 + 54*x) + (972 + 108*E^x + 108*x)*Log[x]))/x^3,x]
Output:
(162*E^x)/x^2 + (9*E^(2*x))/x^2 + (81*E^(2*Log[x]^2))/x^2 + (18*E^x)/x + ( 9*(9 + x)^2)/x^2 + (54*E^Log[x]^2*(9*Log[x] + E^x*Log[x] + x*Log[x]))/(x^2 *Log[x])
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 = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17
method | result | size |
risch | \(\frac {18 \,{\mathrm e}^{x} x +9 \,{\mathrm e}^{2 x}+162 x +162 \,{\mathrm e}^{x}+729}{x^{2}}+\frac {81 \,{\mathrm e}^{2 \ln \left (x \right )^{2}}}{x^{2}}+\frac {54 \left (x +{\mathrm e}^{x}+9\right ) {\mathrm e}^{\ln \left (x \right )^{2}}}{x^{2}}\) | \(52\) |
parallelrisch | \(\frac {729+18 \,{\mathrm e}^{x} x +54 \,{\mathrm e}^{\ln \left (x \right )^{2}} x +9 \,{\mathrm e}^{2 x}+54 \,{\mathrm e}^{\ln \left (x \right )^{2}} {\mathrm e}^{x}+81 \,{\mathrm e}^{2 \ln \left (x \right )^{2}}+162 x +162 \,{\mathrm e}^{x}+486 \,{\mathrm e}^{\ln \left (x \right )^{2}}}{x^{2}}\) | \(58\) |
Input:
int(((324*ln(x)-162)*exp(ln(x)^2)^2+((108*exp(x)+108*x+972)*ln(x)+(54*x-10 8)*exp(x)-54*x-972)*exp(ln(x)^2)+(18*x-18)*exp(x)^2+(18*x^2+144*x-324)*exp (x)-162*x-1458)/x^3,x,method=_RETURNVERBOSE)
Output:
9*(2*exp(x)*x+exp(x)^2+18*x+18*exp(x)+81)/x^2+81/x^2*exp(ln(x)^2)^2+54*(x+ exp(x)+9)/x^2*exp(ln(x)^2)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\frac {9 \, {\left (6 \, {\left (x + e^{x} + 9\right )} e^{\left (\log \left (x\right )^{2}\right )} + 2 \, {\left (x + 9\right )} e^{x} + 18 \, x + 9 \, e^{\left (2 \, \log \left (x\right )^{2}\right )} + e^{\left (2 \, x\right )} + 81\right )}}{x^{2}} \] Input:
integrate(((324*log(x)-162)*exp(log(x)^2)^2+((108*exp(x)+108*x+972)*log(x) +(54*x-108)*exp(x)-54*x-972)*exp(log(x)^2)+(18*x-18)*exp(x)^2+(18*x^2+144* x-324)*exp(x)-162*x-1458)/x^3,x, algorithm="fricas")
Output:
9*(6*(x + e^x + 9)*e^(log(x)^2) + 2*(x + 9)*e^x + 18*x + 9*e^(2*log(x)^2) + e^(2*x) + 81)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (24) = 48\).
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.42 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=- \frac {- 162 x - 729}{x^{2}} + \frac {9 x^{2} e^{2 x} + \left (18 x^{3} + 54 x^{2} e^{\log {\left (x \right )}^{2}} + 162 x^{2}\right ) e^{x}}{x^{4}} + \frac {81 x^{2} e^{2 \log {\left (x \right )}^{2}} + \left (54 x^{3} + 486 x^{2}\right ) e^{\log {\left (x \right )}^{2}}}{x^{4}} \] Input:
integrate(((324*ln(x)-162)*exp(ln(x)**2)**2+((108*exp(x)+108*x+972)*ln(x)+ (54*x-108)*exp(x)-54*x-972)*exp(ln(x)**2)+(18*x-18)*exp(x)**2+(18*x**2+144 *x-324)*exp(x)-162*x-1458)/x**3,x)
Output:
-(-162*x - 729)/x**2 + (9*x**2*exp(2*x) + (18*x**3 + 54*x**2*exp(log(x)**2 ) + 162*x**2)*exp(x))/x**4 + (81*x**2*exp(2*log(x)**2) + (54*x**3 + 486*x* *2)*exp(log(x)**2))/x**4
\[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\int { \frac {18 \, {\left (9 \, {\left (2 \, \log \left (x\right ) - 1\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} + 3 \, {\left ({\left (x - 2\right )} e^{x} + 2 \, {\left (x + e^{x} + 9\right )} \log \left (x\right ) - x - 18\right )} e^{\left (\log \left (x\right )^{2}\right )} + {\left (x - 1\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 8 \, x - 18\right )} e^{x} - 9 \, x - 81\right )}}{x^{3}} \,d x } \] Input:
integrate(((324*log(x)-162)*exp(log(x)^2)^2+((108*exp(x)+108*x+972)*log(x) +(54*x-108)*exp(x)-54*x-972)*exp(log(x)^2)+(18*x-18)*exp(x)^2+(18*x^2+144* x-324)*exp(x)-162*x-1458)/x^3,x, algorithm="maxima")
Output:
81/2*I*sqrt(2)*sqrt(pi)*erf(I*sqrt(2)*log(x) - 1/2*I*sqrt(2))*e^(-1/2) + 2 7*I*sqrt(pi)*erf(I*log(x) - 1/2*I)*e^(-1/4) + 486*I*sqrt(pi)*erf(I*log(x) - I)*e^(-1) + 162/x + 54*e^(log(x)^2 + x)/x^2 + 729/x^2 + 18*Ei(x) + 144*g amma(-1, -x) + 36*gamma(-1, -2*x) + 324*gamma(-2, -x) + 72*gamma(-2, -2*x) + 18*integrate(6*(x + 9)*e^(log(x)^2)*log(x)/x^3, x) + 324*integrate(e^(2 *log(x)^2)*log(x)/x^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (21) = 42\).
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\frac {9 \, {\left (6 \, x e^{\left (\log \left (x\right )^{2}\right )} + 2 \, x e^{x} + 18 \, x + 9 \, e^{\left (2 \, \log \left (x\right )^{2}\right )} + 6 \, e^{\left (\log \left (x\right )^{2} + x\right )} + 54 \, e^{\left (\log \left (x\right )^{2}\right )} + e^{\left (2 \, x\right )} + 18 \, e^{x} + 81\right )}}{x^{2}} \] Input:
integrate(((324*log(x)-162)*exp(log(x)^2)^2+((108*exp(x)+108*x+972)*log(x) +(54*x-108)*exp(x)-54*x-972)*exp(log(x)^2)+(18*x-18)*exp(x)^2+(18*x^2+144* x-324)*exp(x)-162*x-1458)/x^3,x, algorithm="giac")
Output:
9*(6*x*e^(log(x)^2) + 2*x*e^x + 18*x + 9*e^(2*log(x)^2) + 6*e^(log(x)^2 + x) + 54*e^(log(x)^2) + e^(2*x) + 18*e^x + 81)/x^2
Time = 3.80 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\frac {9\,\left (3\,{\mathrm {e}}^{{\ln \left (x\right )}^2}+{\mathrm {e}}^x+9\right )\,\left (2\,x+3\,{\mathrm {e}}^{{\ln \left (x\right )}^2}+{\mathrm {e}}^x+9\right )}{x^2} \] Input:
int(-(162*x - exp(2*log(x)^2)*(324*log(x) - 162) + exp(log(x)^2)*(54*x - e xp(x)*(54*x - 108) - log(x)*(108*x + 108*exp(x) + 972) + 972) - exp(x)*(14 4*x + 18*x^2 - 324) - exp(2*x)*(18*x - 18) + 1458)/x^3,x)
Output:
(9*(3*exp(log(x)^2) + exp(x) + 9)*(2*x + 3*exp(log(x)^2) + exp(x) + 9))/x^ 2
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.62 \[ \int \frac {-1458-162 x+e^{2 x} (-18+18 x)+e^x \left (-324+144 x+18 x^2\right )+e^{2 \log ^2(x)} (-162+324 \log (x))+e^{\log ^2(x)} \left (-972-54 x+e^x (-108+54 x)+\left (972+108 e^x+108 x\right ) \log (x)\right )}{x^3} \, dx=\frac {81 e^{2 \mathrm {log}\left (x \right )^{2}}+54 e^{\mathrm {log}\left (x \right )^{2}+x}+54 e^{\mathrm {log}\left (x \right )^{2}} x +486 e^{\mathrm {log}\left (x \right )^{2}}+9 e^{2 x}+18 e^{x} x +162 e^{x}+162 x +729}{x^{2}} \] Input:
int(((324*log(x)-162)*exp(log(x)^2)^2+((108*exp(x)+108*x+972)*log(x)+(54*x -108)*exp(x)-54*x-972)*exp(log(x)^2)+(18*x-18)*exp(x)^2+(18*x^2+144*x-324) *exp(x)-162*x-1458)/x^3,x)
Output:
(9*(9*e**(2*log(x)**2) + 6*e**(log(x)**2 + x) + 6*e**(log(x)**2)*x + 54*e* *(log(x)**2) + e**(2*x) + 2*e**x*x + 18*e**x + 18*x + 81))/x**2