Integrand size = 76, antiderivative size = 39 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=x+x \left (-x+e^{x^2} \left (-4+\left (-\frac {5}{3 x}+\frac {-e^x+4 x}{x}\right )^2\right )\right ) \]
Time = 1.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=\frac {9 e^{x (2+x)}+e^{x+x^2} (30-72 x)-9 (-1+x) x^2+e^{x^2} \left (25-120 x+108 x^2\right )}{9 x} \]
Integrate[(9*x^2 - 18*x^3 + E^x^2*(-25 + 158*x^2 - 240*x^3 + 216*x^4 + E^( 2*x)*(-9 + 18*x + 18*x^2) + E^x*(-30 + 30*x - 12*x^2 - 144*x^3)))/(9*x^2), x]
(9*E^(x*(2 + x)) + E^(x + x^2)*(30 - 72*x) - 9*(-1 + x)*x^2 + E^x^2*(25 - 120*x + 108*x^2))/(9*x)
Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(39)=78\).
Time = 6.97 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {27, 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.
\(\displaystyle \int \frac {-18 x^3+9 x^2+e^{x^2} \left (216 x^4-240 x^3+158 x^2+e^{2 x} \left (18 x^2+18 x-9\right )+e^x \left (-144 x^3-12 x^2+30 x-30\right )-25\right )}{9 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {-18 x^3+9 x^2-e^{x^2} \left (-216 x^4+240 x^3-158 x^2+9 e^{2 x} \left (-2 x^2-2 x+1\right )+6 e^x \left (24 x^3+2 x^2-5 x+5\right )+25\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{9} \int \left (\frac {e^{x^2} \left (216 x^4-144 e^x x^3-240 x^3-12 e^x x^2+18 e^{2 x} x^2+158 x^2+30 e^x x+18 e^{2 x} x-30 e^x-9 e^{2 x}-25\right )}{x^2}-9 (2 x-1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (-120 e^{x^2}+108 e^{x^2} x+\frac {9 e^{x^2+2 x} \left (x^2+x\right )}{x^2 (x+1)}+\frac {25 e^{x^2}}{x}+\frac {6 e^{x^2+x} \left (-24 x^3-2 x^2+5 x\right )}{x^2 (2 x+1)}-\frac {9}{4} (1-2 x)^2\right )\) |
Int[(9*x^2 - 18*x^3 + E^x^2*(-25 + 158*x^2 - 240*x^3 + 216*x^4 + E^(2*x)*( -9 + 18*x + 18*x^2) + E^x*(-30 + 30*x - 12*x^2 - 144*x^3)))/(9*x^2),x]
(-120*E^x^2 - (9*(1 - 2*x)^2)/4 + (25*E^x^2)/x + 108*E^x^2*x + (9*E^(2*x + x^2)*(x + x^2))/(x^2*(1 + x)) + (6*E^(x + x^2)*(5*x - 2*x^2 - 24*x^3))/(x ^2*(1 + 2*x)))/9
3.13.83.3.1 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_)*((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 = 0.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-x^{2}+x +\frac {\left (108 x^{2}-72 \,{\mathrm e}^{x} x +9 \,{\mathrm e}^{2 x}-120 x +30 \,{\mathrm e}^{x}+25\right ) {\mathrm e}^{x^{2}}}{9 x}\) | \(42\) |
norman | \(\frac {x^{2}+{\mathrm e}^{x^{2}} {\mathrm e}^{2 x}-x^{3}+12 x^{2} {\mathrm e}^{x^{2}}+\frac {10 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{3}-\frac {40 \,{\mathrm e}^{x^{2}} x}{3}-8 x \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}+\frac {25 \,{\mathrm e}^{x^{2}}}{9}}{x}\) | \(62\) |
parts | \(x +\frac {{\mathrm e}^{2 x} {\mathrm e}^{x^{2}}}{x}-x^{2}+12 \,{\mathrm e}^{x^{2}} x +\frac {25 \,{\mathrm e}^{x^{2}}}{9 x}-\frac {40 \,{\mathrm e}^{x^{2}}}{3}+\frac {\frac {10 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{3}-8 x \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{x}\) | \(64\) |
parallelrisch | \(-\frac {72 x \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}-108 x^{2} {\mathrm e}^{x^{2}}+9 x^{3}-30 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}-9 \,{\mathrm e}^{x^{2}} {\mathrm e}^{2 x}+120 \,{\mathrm e}^{x^{2}} x -9 x^{2}-25 \,{\mathrm e}^{x^{2}}}{9 x}\) | \(66\) |
int(1/9*(((18*x^2+18*x-9)*exp(x)^2+(-144*x^3-12*x^2+30*x-30)*exp(x)+216*x^ 4-240*x^3+158*x^2-25)*exp(x^2)-18*x^3+9*x^2)/x^2,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=-\frac {9 \, x^{3} - 9 \, x^{2} - {\left (108 \, x^{2} - 6 \, {\left (12 \, x - 5\right )} e^{x} - 120 \, x + 9 \, e^{\left (2 \, x\right )} + 25\right )} e^{\left (x^{2}\right )}}{9 \, x} \]
integrate(1/9*(((18*x^2+18*x-9)*exp(x)^2+(-144*x^3-12*x^2+30*x-30)*exp(x)+ 216*x^4-240*x^3+158*x^2-25)*exp(x^2)-18*x^3+9*x^2)/x^2,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=- x^{2} + x + \frac {\left (108 x^{2} - 72 x e^{x} - 120 x + 9 e^{2 x} + 30 e^{x} + 25\right ) e^{x^{2}}}{9 x} \]
integrate(1/9*(((18*x**2+18*x-9)*exp(x)**2+(-144*x**3-12*x**2+30*x-30)*exp (x)+216*x**4-240*x**3+158*x**2-25)*exp(x**2)-18*x**3+9*x**2)/x**2,x)
\[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=\int { -\frac {18 \, x^{3} - 9 \, x^{2} - {\left (216 \, x^{4} - 240 \, x^{3} + 158 \, x^{2} + 9 \, {\left (2 \, x^{2} + 2 \, x - 1\right )} e^{\left (2 \, x\right )} - 6 \, {\left (24 \, x^{3} + 2 \, x^{2} - 5 \, x + 5\right )} e^{x} - 25\right )} e^{\left (x^{2}\right )}}{9 \, x^{2}} \,d x } \]
integrate(1/9*(((18*x^2+18*x-9)*exp(x)^2+(-144*x^3-12*x^2+30*x-30)*exp(x)+ 216*x^4-240*x^3+158*x^2-25)*exp(x^2)-18*x^3+9*x^2)/x^2,x, algorithm=\
2/3*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4) - x^2 + 12*x*e^(x^2) - 25/9*I*sqr t(pi)*erf(I*x) + x + 25/18*sqrt(-x^2)*gamma(-1/2, -x^2)/x + e^(x^2 + 2*x)/ x - 40/3*e^(x^2) - 1/9*integrate(6*(24*x^3 - 5*x + 5)*e^(x^2 + x)/x^2, x)
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.67 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=-\frac {9 \, x^{3} - 108 \, x^{2} e^{\left (x^{2}\right )} - 9 \, x^{2} + 72 \, x e^{\left (x^{2} + x\right )} + 120 \, x e^{\left (x^{2}\right )} - 9 \, e^{\left (x^{2} + 2 \, x\right )} - 30 \, e^{\left (x^{2} + x\right )} - 25 \, e^{\left (x^{2}\right )}}{9 \, x} \]
integrate(1/9*(((18*x^2+18*x-9)*exp(x)^2+(-144*x^3-12*x^2+30*x-30)*exp(x)+ 216*x^4-240*x^3+158*x^2-25)*exp(x^2)-18*x^3+9*x^2)/x^2,x, algorithm=\
-1/9*(9*x^3 - 108*x^2*e^(x^2) - 9*x^2 + 72*x*e^(x^2 + x) + 120*x*e^(x^2) - 9*e^(x^2 + 2*x) - 30*e^(x^2 + x) - 25*e^(x^2))/x
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int \frac {9 x^2-18 x^3+e^{x^2} \left (-25+158 x^2-240 x^3+216 x^4+e^{2 x} \left (-9+18 x+18 x^2\right )+e^x \left (-30+30 x-12 x^2-144 x^3\right )\right )}{9 x^2} \, dx=\frac {\frac {10\,{\mathrm {e}}^{x^2+x}}{3}+\frac {25\,{\mathrm {e}}^{x^2}}{9}+{\mathrm {e}}^{x^2+2\,x}}{x}-\frac {40\,{\mathrm {e}}^{x^2}}{3}-8\,{\mathrm {e}}^{x^2+x}+x\,\left (12\,{\mathrm {e}}^{x^2}+1\right )-x^2 \]
int(((exp(x^2)*(exp(2*x)*(18*x + 18*x^2 - 9) + 158*x^2 - 240*x^3 + 216*x^4 - exp(x)*(12*x^2 - 30*x + 144*x^3 + 30) - 25))/9 + x^2 - 2*x^3)/x^2,x)