Integrand size = 45, antiderivative size = 28 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \left (-4-e^{-5+\frac {9 (5-x)}{x^2}}+e^{2-2 x}+x\right ) \] Output:
3*exp(2-2*x)+3*x-3*exp(9*(5-x)/x^2)/exp(5)-12
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=-3 e^{-5+\frac {45}{x^2}-\frac {9}{x}}+3 e^{2-2 x}+3 x \] Input:
Integrate[(E^((45 - 9*x)/x^2)*(270 - 27*x) + 3*E^5*x^3 - 6*E^(7 - 2*x)*x^3 )/(E^5*x^3),x]
Output:
-3*E^(-5 + 45/x^2 - 9/x) + 3*E^(2 - 2*x) + 3*x
Time = 0.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {27, 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.
\(\displaystyle \int \frac {-6 e^{7-2 x} x^3+3 e^5 x^3+e^{\frac {45-9 x}{x^2}} (270-27 x)}{e^5 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 \left (-2 e^{7-2 x} x^3+e^5 x^3+9 e^{\frac {9 (5-x)}{x^2}} (10-x)\right )}{x^3}dx}{e^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \frac {-2 e^{7-2 x} x^3+e^5 x^3+9 e^{\frac {9 (5-x)}{x^2}} (10-x)}{x^3}dx}{e^5}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {3 \int \left (\frac {e^{-9/x} \left (e^{5+\frac {9}{x}} x^3-9 e^{\frac {45}{x^2}} x+90 e^{\frac {45}{x^2}}\right )}{x^3}-2 e^{7-2 x}\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \left (-e^{\frac {9 (5-x)}{x^2}}+e^5 x+e^{7-2 x}\right )}{e^5}\) |
Input:
Int[(E^((45 - 9*x)/x^2)*(270 - 27*x) + 3*E^5*x^3 - 6*E^(7 - 2*x)*x^3)/(E^5 *x^3),x]
Output:
(3*(E^(7 - 2*x) - E^((9*(5 - x))/x^2) + E^5*x))/E^5
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 = 1.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04
method | result | size |
parts | \(3 x +3 \,{\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-5} {\mathrm e}^{\frac {-9 x +45}{x^{2}}}\) | \(29\) |
risch | \(3 x +3 \,{\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-\frac {5 x^{2}+9 x -45}{x^{2}}}\) | \(31\) |
parallelrisch | \({\mathrm e}^{-5} \left (3 x \,{\mathrm e}^{5}+3 \,{\mathrm e}^{5} {\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-\frac {9 \left (-5+x \right )}{x^{2}}}\right )\) | \(33\) |
default | \({\mathrm e}^{-5} \left (-3 \,{\mathrm e}^{-\frac {9}{x}+\frac {45}{x^{2}}}+3 \,{\mathrm e}^{5} {\mathrm e}^{2} {\mathrm e}^{-2 x}+3 x \,{\mathrm e}^{5}\right )\) | \(36\) |
norman | \(\frac {3 x^{3}+3 x^{2} {\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-5} x^{2} {\mathrm e}^{\frac {-9 x +45}{x^{2}}}}{x^{2}}\) | \(41\) |
orering | \(\frac {\left (11+x \right ) \left (\left (-27 x +270\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-6 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+3 x^{3} {\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{3}}-\frac {\left (8 x^{9}-72 x^{8}-72 x^{7}+8028 x^{6}-1890 x^{5}-10260 x^{4}+106677 x^{3}-1742310 x^{2}+3572100 x -16767000\right ) \left (\frac {\left (-27 \,{\mathrm e}^{\frac {-9 x +45}{x^{2}}}+\left (-27 x +270\right ) \left (-\frac {9}{x^{2}}-\frac {2 \left (-9 x +45\right )}{x^{3}}\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-18 x^{2} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+12 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+9 x^{2} {\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{3}}-\frac {3 \left (\left (-27 x +270\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-6 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+3 x^{3} {\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{4}}\right )}{36 \left (4 x^{6}-84 x^{5}+534 x^{4}-3231 x^{3}+10530 x^{2}-24300 x +81000\right )}-\frac {\left (4 x^{6}-36 x^{5}+3672 x^{3}-4833 x^{2}+21060 x -186300\right ) x^{3} \left (\frac {\left (-54 \left (-\frac {9}{x^{2}}-\frac {2 \left (-9 x +45\right )}{x^{3}}\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}+\left (-27 x +270\right ) \left (\frac {36}{x^{3}}+\frac {-54 x +270}{x^{4}}\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}+\left (-27 x +270\right ) \left (-\frac {9}{x^{2}}-\frac {2 \left (-9 x +45\right )}{x^{3}}\right )^{2} {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-36 x \,{\mathrm e}^{5} {\mathrm e}^{2-2 x}+72 x^{2} {\mathrm e}^{5} {\mathrm e}^{2-2 x}-24 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+18 x \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{3}}-\frac {6 \left (-27 \,{\mathrm e}^{\frac {-9 x +45}{x^{2}}}+\left (-27 x +270\right ) \left (-\frac {9}{x^{2}}-\frac {2 \left (-9 x +45\right )}{x^{3}}\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-18 x^{2} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+12 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+9 x^{2} {\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{4}}+\frac {12 \left (\left (-27 x +270\right ) {\mathrm e}^{\frac {-9 x +45}{x^{2}}}-6 x^{3} {\mathrm e}^{5} {\mathrm e}^{2-2 x}+3 x^{3} {\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{x^{5}}\right )}{36 \left (4 x^{6}-84 x^{5}+534 x^{4}-3231 x^{3}+10530 x^{2}-24300 x +81000\right )}\) | \(603\) |
Input:
int(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/ x^3/exp(5),x,method=_RETURNVERBOSE)
Output:
3*x+3*exp(2-2*x)-3/exp(5)*exp((-9*x+45)/x^2)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} + e^{\left (-2 \, x + 7\right )} - e^{\left (-\frac {9 \, {\left (x - 5\right )}}{x^{2}}\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*ex p(5))/x^3/exp(5),x, algorithm="fricas")
Output:
3*(x*e^5 + e^(-2*x + 7) - e^(-9*(x - 5)/x^2))*e^(-5)
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 x - \frac {3 e^{\frac {45 - 9 x}{x^{2}}}}{e^{5}} + 3 e^{2 - 2 x} \] Input:
integrate(((-27*x+270)*exp((-9*x+45)/x**2)-6*x**3*exp(5)*exp(2-2*x)+3*x**3 *exp(5))/x**3/exp(5),x)
Output:
3*x - 3*exp(-5)*exp((45 - 9*x)/x**2) + 3*exp(2 - 2*x)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} - {\left (e^{\left (2 \, x + \frac {45}{x^{2}}\right )} - e^{\left (\frac {9}{x} + 7\right )}\right )} e^{\left (-2 \, x - \frac {9}{x}\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*ex p(5))/x^3/exp(5),x, algorithm="maxima")
Output:
3*(x*e^5 - (e^(2*x + 45/x^2) - e^(9/x + 7))*e^(-2*x - 9/x))*e^(-5)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} + e^{\left (-2 \, x + 7\right )} - e^{\left (-\frac {9}{x} + \frac {45}{x^{2}}\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*ex p(5))/x^3/exp(5),x, algorithm="giac")
Output:
3*(x*e^5 + e^(-2*x + 7) - e^(-9/x + 45/x^2))*e^(-5)
Time = 3.87 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3\,x+3\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2-3\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {9}{x}}\,{\mathrm {e}}^{\frac {45}{x^2}} \] Input:
int(-(exp(-5)*(exp(-(9*x - 45)/x^2)*(27*x - 270) - 3*x^3*exp(5) + 6*x^3*ex p(5)*exp(2 - 2*x)))/x^3,x)
Output:
3*x + 3*exp(-2*x)*exp(2) - 3*exp(-5)*exp(-9/x)*exp(45/x^2)
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.32 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=\frac {-3 e^{\frac {2 x^{3}+45}{x^{2}}}+3 e^{\frac {2 x^{2}+9}{x}} e^{5} x +3 e^{\frac {9}{x}} e^{7}}{e^{\frac {2 x^{2}+9}{x}} e^{5}} \] Input:
int(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/ x^3/exp(5),x)
Output:
(3*( - e**((2*x**3 + 45)/x**2) + e**((2*x**2 + 9)/x)*e**5*x + e**(9/x)*e** 7))/(e**((2*x**2 + 9)/x)*e**5)