Integrand size = 39, antiderivative size = 28 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=4 e^{\frac {(3+2 x)^2}{x^2}} \left (5-e^5\right )+\frac {50}{x} \] Output:
4*exp((3+2*x)^2/x^2)*(5-exp(5))+50/x
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=-4 e^{4+\frac {9}{x^2}+\frac {12}{x}} \left (-5+e^5\right )+\frac {50}{x} \] Input:
Integrate[(-50*x + E^((9 + 12*x + 4*x^2)/x^2)*(-360 - 240*x + E^5*(72 + 48 *x)))/x^3,x]
Output:
-4*E^(4 + 9/x^2 + 12/x)*(-5 + E^5) + 50/x
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, 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.
\(\displaystyle \int \frac {e^{\frac {4 x^2+12 x+9}{x^2}} \left (-240 x+e^5 (48 x+72)-360\right )-50 x}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {24 \left (e^5-5\right ) e^{\frac {(2 x+3)^2}{x^2}} (2 x+3)}{x^3}-\frac {50}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (5-e^5\right ) e^{\frac {(2 x+3)^2}{x^2}}+\frac {50}{x}\) |
Input:
Int[(-50*x + E^((9 + 12*x + 4*x^2)/x^2)*(-360 - 240*x + E^5*(72 + 48*x)))/ x^3,x]
Output:
4*E^((3 + 2*x)^2/x^2)*(5 - E^5) + 50/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 = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
norman | \(\frac {\left (-4 \,{\mathrm e}^{5}+20\right ) x^{2} {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}+50 x}{x^{2}}\) | \(34\) |
risch | \(\frac {50}{x}-4 \,{\mathrm e}^{\frac {\left (3+2 x \right )^{2}}{x^{2}}} {\mathrm e}^{5}+20 \,{\mathrm e}^{\frac {\left (3+2 x \right )^{2}}{x^{2}}}\) | \(37\) |
parallelrisch | \(-\frac {4 \,{\mathrm e}^{5} {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}} x -20 \,{\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}} x -50}{x}\) | \(46\) |
derivativedivides | \(\frac {50}{x}-40 i {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )+360 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )+8 i {\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )-72 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )\) | \(125\) |
default | \(\frac {50}{x}-40 i {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )+360 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )+8 i {\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )-72 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )\) | \(125\) |
parts | \(\frac {50}{x}-40 i {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )+360 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )+8 i {\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )-72 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}}}{18}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-4} \operatorname {erf}\left (\frac {3 i}{x}+2 i\right )}{9}\right )\) | \(125\) |
orering | \(-\frac {\left (9 x^{3}+52 x^{2}+102 x +72\right ) \left (\left (\left (48 x +72\right ) {\mathrm e}^{5}-240 x -360\right ) {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}-50 x \right )}{12 x^{2} \left (3 x^{2}+8 x +6\right )}-\frac {\left (4+3 x \right ) x^{4} \left (\frac {\left (48 \,{\mathrm e}^{5}-240\right ) {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}+\left (\left (48 x +72\right ) {\mathrm e}^{5}-240 x -360\right ) \left (\frac {8 x +12}{x^{2}}-\frac {2 \left (4 x^{2}+12 x +9\right )}{x^{3}}\right ) {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}-50}{x^{3}}-\frac {3 \left (\left (\left (48 x +72\right ) {\mathrm e}^{5}-240 x -360\right ) {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}-50 x \right )}{x^{4}}\right )}{12 \left (3 x^{2}+8 x +6\right )}\) | \(210\) |
Input:
int((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x,meth od=_RETURNVERBOSE)
Output:
((-4*exp(5)+20)*x^2*exp((4*x^2+12*x+9)/x^2)+50*x)/x^2
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=-\frac {2 \, {\left (2 \, {\left (x e^{5} - 5 \, x\right )} e^{\left (\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )} - 25\right )}}{x} \] Input:
integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3, x, algorithm="fricas")
Output:
-2*(2*(x*e^5 - 5*x)*e^((4*x^2 + 12*x + 9)/x^2) - 25)/x
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=\left (20 - 4 e^{5}\right ) e^{\frac {4 x^{2} + 12 x + 9}{x^{2}}} + \frac {50}{x} \] Input:
integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x**2+12*x+9)/x**2)-50*x)/x* *3,x)
Output:
(20 - 4*exp(5))*exp((4*x**2 + 12*x + 9)/x**2) + 50/x
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=-4 \, {\left (e^{9} - 5 \, e^{4}\right )} e^{\left (\frac {12}{x} + \frac {9}{x^{2}}\right )} + \frac {50}{x} \] Input:
integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3, x, algorithm="maxima")
Output:
-4*(e^9 - 5*e^4)*e^(12/x + 9/x^2) + 50/x
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (24) = 48\).
Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.29 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=\frac {2 \, {\left (12 \, x e^{\left (\frac {3 \, {\left (3 \, x^{2} + 4 \, x + 3\right )}}{x^{2}}\right )} + 25 \, e^{\left (\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )} + 18 \, e^{\left (\frac {3 \, {\left (3 \, x^{2} + 4 \, x + 3\right )}}{x^{2}}\right )}\right )} e^{\left (-\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )}}{x} + 20 \, e^{\left (\frac {12}{x} + \frac {9}{x^{2}} + 4\right )} \] Input:
integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3, x, algorithm="giac")
Output:
2*(12*x*e^(3*(3*x^2 + 4*x + 3)/x^2) + 25*e^((4*x^2 + 12*x + 9)/x^2) + 18*e ^(3*(3*x^2 + 4*x + 3)/x^2))*e^(-(4*x^2 + 12*x + 9)/x^2)/x + 20*e^(12/x + 9 /x^2 + 4)
Time = 3.57 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=\frac {50}{x}-{\mathrm {e}}^{\frac {12}{x}+\frac {9}{x^2}+4}\,\left (4\,{\mathrm {e}}^5-20\right ) \] Input:
int(-(50*x + exp((12*x + 4*x^2 + 9)/x^2)*(240*x - exp(5)*(48*x + 72) + 360 ))/x^3,x)
Output:
50/x - exp(12/x + 9/x^2 + 4)*(4*exp(5) - 20)
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} \left (-360-240 x+e^5 (72+48 x)\right )}{x^3} \, dx=\frac {-4 e^{\frac {12 x +9}{x^{2}}} e^{9} x +20 e^{\frac {12 x +9}{x^{2}}} e^{4} x +50}{x} \] Input:
int((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x)
Output:
(2*( - 2*e**((12*x + 9)/x**2)*e**9*x + 10*e**((12*x + 9)/x**2)*e**4*x + 25 ))/x