Integrand size = 102, antiderivative size = 32 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=\frac {20}{5+x}-\left (-1-e^{5/x}+x+\frac {3 (5+x)}{x}\right )^2 \] Output:
20/(5+x)-(-1-exp(5/x)+3/x*(5+x)+x)^2
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=2 \left (-\frac {e^{10/x}}{2}-\frac {225}{2 x^2}-\frac {30}{x}-2 x-\frac {x^2}{2}+\frac {10}{5+x}+e^{5/x} \left (2+\frac {15}{x}+x\right )\right ) \] Input:
Integrate[(11250 + 6000*x + 1050*x^2 - 60*x^3 - 90*x^4 - 24*x^5 - 2*x^6 + E^(10/x)*(250*x + 100*x^2 + 10*x^3) + E^(5/x)*(-3750 - 2750*x - 900*x^2 - 100*x^3 + 10*x^4 + 2*x^5))/(25*x^3 + 10*x^4 + x^5),x]
Output:
2*(-1/2*E^(10/x) - 225/(2*x^2) - 30/x - 2*x - x^2/2 + 10/(5 + x) + E^(5/x) *(2 + 15/x + x))
Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(32)=64\).
Time = 1.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 2007, 7293, 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.
\(\displaystyle \int \frac {-2 x^6-24 x^5-90 x^4-60 x^3+1050 x^2+e^{10/x} \left (10 x^3+100 x^2+250 x\right )+e^{5/x} \left (2 x^5+10 x^4-100 x^3-900 x^2-2750 x-3750\right )+6000 x+11250}{x^5+10 x^4+25 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-2 x^6-24 x^5-90 x^4-60 x^3+1050 x^2+e^{10/x} \left (10 x^3+100 x^2+250 x\right )+e^{5/x} \left (2 x^5+10 x^4-100 x^3-900 x^2-2750 x-3750\right )+6000 x+11250}{x^3 \left (x^2+10 x+25\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-2 x^6-24 x^5-90 x^4-60 x^3+1050 x^2+e^{10/x} \left (10 x^3+100 x^2+250 x\right )+e^{5/x} \left (2 x^5+10 x^4-100 x^3-900 x^2-2750 x-3750\right )+6000 x+11250}{x^3 (x+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 x^3}{(x+5)^2}+\frac {11250}{(x+5)^2 x^3}-\frac {24 x^2}{(x+5)^2}+\frac {10 e^{10/x}}{x^2}+\frac {6000}{(x+5)^2 x^2}+\frac {2 e^{5/x} \left (x^3-5 x^2-25 x-75\right )}{x^3}-\frac {90 x}{(x+5)^2}-\frac {60}{(x+5)^2}+\frac {1050}{(x+5)^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^2-\frac {225}{x^2}+2 e^{5/x} x-4 x+4 e^{5/x}-e^{10/x}+\frac {20}{x+5}+\frac {30 e^{5/x}}{x}-\frac {60}{x}\) |
Input:
Int[(11250 + 6000*x + 1050*x^2 - 60*x^3 - 90*x^4 - 24*x^5 - 2*x^6 + E^(10/ x)*(250*x + 100*x^2 + 10*x^3) + E^(5/x)*(-3750 - 2750*x - 900*x^2 - 100*x^ 3 + 10*x^4 + 2*x^5))/(25*x^3 + 10*x^4 + x^5),x]
Output:
4*E^(5/x) - E^(10/x) - 225/x^2 - 60/x + (30*E^(5/x))/x - 4*x + 2*E^(5/x)*x - x^2 + 20/(5 + x)
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 1.69 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75
method | result | size |
risch | \(-x^{2}-4 x +\frac {-40 x^{2}-525 x -1125}{x^{2} \left (5+x \right )}-{\mathrm e}^{\frac {10}{x}}+\frac {2 \left (x^{2}+2 x +15\right ) {\mathrm e}^{\frac {5}{x}}}{x}\) | \(56\) |
parts | \(-{\mathrm e}^{\frac {10}{x}}+2 \,{\mathrm e}^{\frac {5}{x}} x +\frac {30 \,{\mathrm e}^{\frac {5}{x}}}{x}+4 \,{\mathrm e}^{\frac {5}{x}}+\frac {20}{5+x}-x^{2}-4 x -\frac {225}{x^{2}}-\frac {60}{x}\) | \(65\) |
derivativedivides | \(-\frac {4}{1+\frac {5}{x}}-x^{2}-4 x -\frac {60}{x}-\frac {225}{x^{2}}+2 \,{\mathrm e}^{\frac {5}{x}} x +4 \,{\mathrm e}^{\frac {5}{x}}-{\mathrm e}^{\frac {10}{x}}+\frac {30 \,{\mathrm e}^{\frac {5}{x}}}{x}\) | \(69\) |
default | \(-\frac {4}{1+\frac {5}{x}}-x^{2}-4 x -\frac {60}{x}-\frac {225}{x^{2}}+2 \,{\mathrm e}^{\frac {5}{x}} x +4 \,{\mathrm e}^{\frac {5}{x}}-{\mathrm e}^{\frac {10}{x}}+\frac {30 \,{\mathrm e}^{\frac {5}{x}}}{x}\) | \(69\) |
parallelrisch | \(-\frac {x^{5}-2 \,{\mathrm e}^{\frac {5}{x}} x^{4}+x^{3} {\mathrm e}^{\frac {10}{x}}+9 x^{4}-14 \,{\mathrm e}^{\frac {5}{x}} x^{3}+1125+5 \,{\mathrm e}^{\frac {10}{x}} x^{2}-50 \,{\mathrm e}^{\frac {5}{x}} x^{2}-60 x^{2}-150 \,{\mathrm e}^{\frac {5}{x}} x +525 x}{x^{2} \left (5+x \right )}\) | \(96\) |
norman | \(\frac {-1125+60 x^{2}-525 x -9 x^{4}-x^{5}+150 \,{\mathrm e}^{\frac {5}{x}} x +50 \,{\mathrm e}^{\frac {5}{x}} x^{2}+14 \,{\mathrm e}^{\frac {5}{x}} x^{3}+2 \,{\mathrm e}^{\frac {5}{x}} x^{4}-5 \,{\mathrm e}^{\frac {10}{x}} x^{2}-x^{3} {\mathrm e}^{\frac {10}{x}}}{x^{2} \left (5+x \right )}\) | \(98\) |
orering | \(\text {Expression too large to display}\) | \(1454\) |
Input:
int(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750* x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4-60*x^3+1050*x^2+6000*x+11250)/(x^5+10 *x^4+25*x^3),x,method=_RETURNVERBOSE)
Output:
-x^2-4*x+(-40*x^2-525*x-1125)/x^2/(5+x)-exp(10/x)+2*(x^2+2*x+15)/x*exp(5/x )
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (31) = 62\).
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.41 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=-\frac {x^{5} + 9 \, x^{4} + 20 \, x^{3} + 40 \, x^{2} + {\left (x^{3} + 5 \, x^{2}\right )} e^{\frac {10}{x}} - 2 \, {\left (x^{4} + 7 \, x^{3} + 25 \, x^{2} + 75 \, x\right )} e^{\frac {5}{x}} + 525 \, x + 1125}{x^{3} + 5 \, x^{2}} \] Input:
integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2 -2750*x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4-60*x^3+1050*x^2+6000*x+11250)/( x^5+10*x^4+25*x^3),x, algorithm="fricas")
Output:
-(x^5 + 9*x^4 + 20*x^3 + 40*x^2 + (x^3 + 5*x^2)*e^(10/x) - 2*(x^4 + 7*x^3 + 25*x^2 + 75*x)*e^(5/x) + 525*x + 1125)/(x^3 + 5*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=- x^{2} - 4 x - \frac {40 x^{2} + 525 x + 1125}{x^{3} + 5 x^{2}} + \frac {- x e^{\frac {10}{x}} + \left (2 x^{2} + 4 x + 30\right ) e^{\frac {5}{x}}}{x} \] Input:
integrate(((10*x**3+100*x**2+250*x)*exp(5/x)**2+(2*x**5+10*x**4-100*x**3-9 00*x**2-2750*x-3750)*exp(5/x)-2*x**6-24*x**5-90*x**4-60*x**3+1050*x**2+600 0*x+11250)/(x**5+10*x**4+25*x**3),x)
Output:
-x**2 - 4*x - (40*x**2 + 525*x + 1125)/(x**3 + 5*x**2) + (-x*exp(10/x) + ( 2*x**2 + 4*x + 30)*exp(5/x))/x
\[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=\int { -\frac {2 \, {\left (x^{6} + 12 \, x^{5} + 45 \, x^{4} + 30 \, x^{3} - 525 \, x^{2} - 5 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{\frac {10}{x}} - {\left (x^{5} + 5 \, x^{4} - 50 \, x^{3} - 450 \, x^{2} - 1375 \, x - 1875\right )} e^{\frac {5}{x}} - 3000 \, x - 5625\right )}}{x^{5} + 10 \, x^{4} + 25 \, x^{3}} \,d x } \] Input:
integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2 -2750*x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4-60*x^3+1050*x^2+6000*x+11250)/( x^5+10*x^4+25*x^3),x, algorithm="maxima")
Output:
-x^2 + 2*x*e^(5/x) - 4*x + 45*(6*x^2 + 15*x - 25)/(x^3 + 5*x^2) - 240*(2*x + 5)/(x^2 + 5*x) + 170/(x + 5) - e^(10/x) - 2*integrate(25*(x + 3)*e^(5/x )/x^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (31) = 62\).
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.28 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=\frac {\frac {2 \, e^{\frac {5}{x}}}{x} - \frac {9}{x} - \frac {e^{\frac {10}{x}}}{x^{2}} + \frac {14 \, e^{\frac {5}{x}}}{x^{2}} - \frac {24}{x^{2}} - \frac {5 \, e^{\frac {10}{x}}}{x^{3}} + \frac {50 \, e^{\frac {5}{x}}}{x^{3}} - \frac {60}{x^{3}} + \frac {150 \, e^{\frac {5}{x}}}{x^{4}} - \frac {525}{x^{4}} - \frac {1125}{x^{5}} - 1}{\frac {1}{x^{2}} + \frac {5}{x^{3}}} \] Input:
integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2 -2750*x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4-60*x^3+1050*x^2+6000*x+11250)/( x^5+10*x^4+25*x^3),x, algorithm="giac")
Output:
(2*e^(5/x)/x - 9/x - e^(10/x)/x^2 + 14*e^(5/x)/x^2 - 24/x^2 - 5*e^(10/x)/x ^3 + 50*e^(5/x)/x^3 - 60/x^3 + 150*e^(5/x)/x^4 - 525/x^4 - 1125/x^5 - 1)/( 1/x^2 + 5/x^3)
Time = 2.60 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=4\,{\mathrm {e}}^{5/x}-{\mathrm {e}}^{10/x}+x\,\left (2\,{\mathrm {e}}^{5/x}-4\right )-x^2+\frac {x\,\left (150\,{\mathrm {e}}^{5/x}-525\right )+x^2\,\left (30\,{\mathrm {e}}^{5/x}-40\right )-1125}{x^2\,\left (x+5\right )} \] Input:
int(-(exp(5/x)*(2750*x + 900*x^2 + 100*x^3 - 10*x^4 - 2*x^5 + 3750) - 6000 *x - exp(10/x)*(250*x + 100*x^2 + 10*x^3) - 1050*x^2 + 60*x^3 + 90*x^4 + 2 4*x^5 + 2*x^6 - 11250)/(25*x^3 + 10*x^4 + x^5),x)
Output:
4*exp(5/x) - exp(10/x) + x*(2*exp(5/x) - 4) - x^2 + (x*(150*exp(5/x) - 525 ) + x^2*(30*exp(5/x) - 40) - 1125)/(x^2*(x + 5))
Time = 0.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.09 \[ \int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{25 x^3+10 x^4+x^5} \, dx=\frac {-e^{\frac {10}{x}} x^{3}-5 e^{\frac {10}{x}} x^{2}+2 e^{\frac {5}{x}} x^{4}+14 e^{\frac {5}{x}} x^{3}+50 e^{\frac {5}{x}} x^{2}+150 e^{\frac {5}{x}} x -x^{5}-9 x^{4}-12 x^{3}-525 x -1125}{x^{2} \left (x +5\right )} \] Input:
int(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750* x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4-60*x^3+1050*x^2+6000*x+11250)/(x^5+10 *x^4+25*x^3),x)
Output:
( - e**(10/x)*x**3 - 5*e**(10/x)*x**2 + 2*e**(5/x)*x**4 + 14*e**(5/x)*x**3 + 50*e**(5/x)*x**2 + 150*e**(5/x)*x - x**5 - 9*x**4 - 12*x**3 - 525*x - 1 125)/(x**2*(x + 5))