Integrand size = 64, antiderivative size = 28 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=\frac {1}{3} \left (-1+e^{\frac {\left (3+\frac {-3+x}{x}\right )^2}{4 e}}\right )^2 \]
Time = 0.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=\frac {1}{3} e^{\frac {(3-4 x)^2}{4 e x^2}} \left (-2+e^{\frac {(3-4 x)^2}{4 e x^2}}\right ) \]
Integrate[(E^((9 - 24*x + 16*x^2)/(4*E*x^2))*(3 - 4*x) + E^((9 - 24*x + 16 *x^2)/(2*E*x^2))*(-3 + 4*x))/(E*x^3),x]
Time = 0.42 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, 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.
\(\displaystyle \int \frac {e^{\frac {16 x^2-24 x+9}{4 e x^2}} (3-4 x)+e^{\frac {16 x^2-24 x+9}{2 e x^2}} (4 x-3)}{e x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^{\frac {16 x^2-24 x+9}{4 e x^2}} (3-4 x)-e^{\frac {16 x^2-24 x+9}{2 e x^2}} (3-4 x)}{x^3}dx}{e}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {\int \left (\frac {e^{\frac {(3-4 x)^2}{2 e x^2}} (4 x-3)}{x^3}-\frac {e^{\frac {(3-4 x)^2}{4 e x^2}} (4 x-3)}{x^3}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} e^{\frac {(3-4 x)^2}{2 e x^2}+1}-\frac {2}{3} e^{\frac {(3-4 x)^2}{4 e x^2}+1}}{e}\) |
Int[(E^((9 - 24*x + 16*x^2)/(4*E*x^2))*(3 - 4*x) + E^((9 - 24*x + 16*x^2)/ (2*E*x^2))*(-3 + 4*x))/(E*x^3),x]
3.10.18.3.1 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 = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
method | result | size |
risch | \(\frac {{\mathrm e}^{\frac {\left (-3+4 x \right )^{2} {\mathrm e}^{-1}}{2 x^{2}}}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (-3+4 x \right )^{2} {\mathrm e}^{-1}}{4 x^{2}}}}{3}\) | \(36\) |
parts | \(\frac {{\mathrm e}^{\frac {\left (16 x^{2}-24 x +9\right ) {\mathrm e}^{-1}}{2 x^{2}}}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (16 x^{2}-24 x +9\right ) {\mathrm e}^{-1}}{4 x^{2}}}}{3}\) | \(48\) |
norman | \(\frac {\frac {x^{2} {\mathrm e}^{\frac {\left (16 x^{2}-24 x +9\right ) {\mathrm e}^{-1}}{2 x^{2}}}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (16 x^{2}-24 x +9\right ) {\mathrm e}^{-1}}{4 x^{2}}} x^{2}}{3}}{x^{2}}\) | \(58\) |
default | \({\mathrm e}^{-1} \left (-\frac {4 i {\mathrm e}^{4 \,{\mathrm e}^{-1}} \sqrt {\pi }\, {\mathrm e}^{-4 \,{\mathrm e}^{-1}} {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} {\mathrm e}^{\frac {1}{2}}\right )}{3}+\frac {2 i {\mathrm e}^{8 \,{\mathrm e}^{-1}} \sqrt {\pi }\, {\mathrm e}^{-8 \,{\mathrm e}^{-1}} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i \sqrt {2}\, {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}}\right )}{3}-3 \,{\mathrm e}^{4 \,{\mathrm e}^{-1}} \left (\frac {2 \,{\mathrm e} \,{\mathrm e}^{\frac {9 \,{\mathrm e}^{-1}}{4 x^{2}}-\frac {6 \,{\mathrm e}^{-1}}{x}}}{9}-\frac {4 i \sqrt {\pi }\, {\mathrm e}^{-4 \,{\mathrm e}^{-1}} {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} {\mathrm e}^{\frac {1}{2}}\right )}{9}\right )+3 \,{\mathrm e}^{8 \,{\mathrm e}^{-1}} \left (\frac {{\mathrm e} \,{\mathrm e}^{\frac {9 \,{\mathrm e}^{-1}}{2 x^{2}}-\frac {12 \,{\mathrm e}^{-1}}{x}}}{9}-\frac {2 i \sqrt {\pi }\, {\mathrm e}^{-8 \,{\mathrm e}^{-1}} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i \sqrt {2}\, {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}}\right )}{9}\right )\right )\) | \(239\) |
derivativedivides | \(-{\mathrm e}^{-1} \left (\frac {4 i {\mathrm e}^{4 \,{\mathrm e}^{-1}} \sqrt {\pi }\, {\mathrm e}^{-4 \,{\mathrm e}^{-1}} {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} {\mathrm e}^{\frac {1}{2}}\right )}{3}-\frac {2 i {\mathrm e}^{8 \,{\mathrm e}^{-1}} \sqrt {\pi }\, {\mathrm e}^{-8 \,{\mathrm e}^{-1}} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i \sqrt {2}\, {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}}\right )}{3}+3 \,{\mathrm e}^{4 \,{\mathrm e}^{-1}} \left (\frac {2 \,{\mathrm e} \,{\mathrm e}^{\frac {9 \,{\mathrm e}^{-1}}{4 x^{2}}-\frac {6 \,{\mathrm e}^{-1}}{x}}}{9}-\frac {4 i \sqrt {\pi }\, {\mathrm e}^{-4 \,{\mathrm e}^{-1}} {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} {\mathrm e}^{\frac {1}{2}}\right )}{9}\right )-3 \,{\mathrm e}^{8 \,{\mathrm e}^{-1}} \left (\frac {{\mathrm e} \,{\mathrm e}^{\frac {9 \,{\mathrm e}^{-1}}{2 x^{2}}-\frac {12 \,{\mathrm e}^{-1}}{x}}}{9}-\frac {2 i \sqrt {\pi }\, {\mathrm e}^{-8 \,{\mathrm e}^{-1}} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}} \operatorname {erf}\left (\frac {3 i \sqrt {2}\, {\mathrm e}^{-\frac {1}{2}}}{2 x}-2 i {\mathrm e}^{-1} \sqrt {2}\, {\mathrm e}^{\frac {1}{2}}\right )}{9}\right )\right )\) | \(240\) |
int(((-3+4*x)*exp(1/4*(16*x^2-24*x+9)/x^2/exp(1))^2+(3-4*x)*exp(1/4*(16*x^ 2-24*x+9)/x^2/exp(1)))/x^3/exp(1),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=\frac {1}{3} \, e^{\left (\frac {{\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (-1\right )}}{2 \, x^{2}}\right )} - \frac {2}{3} \, e^{\left (\frac {{\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (-1\right )}}{4 \, x^{2}}\right )} \]
integrate(((-3+4*x)*exp(1/4*(16*x^2-24*x+9)/x^2/exp(1))^2+(3-4*x)*exp(1/4* (16*x^2-24*x+9)/x^2/exp(1)))/x^3/exp(1),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=\frac {e^{\frac {2 \cdot \left (4 x^{2} - 6 x + \frac {9}{4}\right )}{e x^{2}}}}{3} - \frac {2 e^{\frac {4 x^{2} - 6 x + \frac {9}{4}}{e x^{2}}}}{3} \]
integrate(((-3+4*x)*exp(1/4*(16*x**2-24*x+9)/x**2/exp(1))**2+(3-4*x)*exp(1 /4*(16*x**2-24*x+9)/x**2/exp(1)))/x**3/exp(1),x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=-\frac {1}{3} \, {\left (2 \, e^{\left (\frac {6 \, e^{\left (-1\right )}}{x} + \frac {9 \, e^{\left (-1\right )}}{4 \, x^{2}} + 4 \, e^{\left (-1\right )} + 1\right )} - e^{\left (\frac {9 \, e^{\left (-1\right )}}{2 \, x^{2}} + 8 \, e^{\left (-1\right )} + 1\right )}\right )} e^{\left (-\frac {12 \, e^{\left (-1\right )}}{x} - 1\right )} \]
integrate(((-3+4*x)*exp(1/4*(16*x^2-24*x+9)/x^2/exp(1))^2+(3-4*x)*exp(1/4* (16*x^2-24*x+9)/x^2/exp(1)))/x^3/exp(1),x, algorithm=\
-1/3*(2*e^(6*e^(-1)/x + 9/4*e^(-1)/x^2 + 4*e^(-1) + 1) - e^(9/2*e^(-1)/x^2 + 8*e^(-1) + 1))*e^(-12*e^(-1)/x - 1)
\[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx=\int { \frac {{\left ({\left (4 \, x - 3\right )} e^{\left (\frac {{\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (-1\right )}}{2 \, x^{2}}\right )} - {\left (4 \, x - 3\right )} e^{\left (\frac {{\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (-1\right )}}{4 \, x^{2}}\right )}\right )} e^{\left (-1\right )}}{x^{3}} \,d x } \]
integrate(((-3+4*x)*exp(1/4*(16*x^2-24*x+9)/x^2/exp(1))^2+(3-4*x)*exp(1/4* (16*x^2-24*x+9)/x^2/exp(1)))/x^3/exp(1),x, algorithm=\
integrate(((4*x - 3)*e^(1/2*(16*x^2 - 24*x + 9)*e^(-1)/x^2) - (4*x - 3)*e^ (1/4*(16*x^2 - 24*x + 9)*e^(-1)/x^2))*e^(-1)/x^3, x)
Time = 12.88 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {e^{\frac {9-24 x+16 x^2}{4 e x^2}} (3-4 x)+e^{\frac {9-24 x+16 x^2}{2 e x^2}} (-3+4 x)}{e x^3} \, dx={\mathrm {e}}^{4\,{\mathrm {e}}^{-1}-\frac {6\,{\mathrm {e}}^{-1}}{x}+\frac {9\,{\mathrm {e}}^{-1}}{4\,x^2}}\,\left (\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^{-1}-\frac {6\,{\mathrm {e}}^{-1}}{x}+\frac {9\,{\mathrm {e}}^{-1}}{4\,x^2}}}{3}-\frac {2}{3}\right ) \]
int(-(exp(-1)*(exp((exp(-1)*(4*x^2 - 6*x + 9/4))/x^2)*(4*x - 3) - exp((2*e xp(-1)*(4*x^2 - 6*x + 9/4))/x^2)*(4*x - 3)))/x^3,x)