Integrand size = 42, antiderivative size = 21 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {1}{2} \left (7-x-\frac {8}{5} e^{-8 x} x\right )^2 \] Output:
1/2*(7-2/5*exp(ln(2)-4*x)^2*x-x)^2
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=-7 x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2-\frac {8}{5} e^{-8 x} \left (7 x-x^2\right ) \] Input:
Integrate[(-175 + 25*x + (4*(-70 + 580*x - 80*x^2))/E^(8*x) + (16*(4*x - 3 2*x^2))/E^(16*x))/25,x]
Output:
-7*x + x^2/2 + (32*x^2)/(25*E^(16*x)) - (8*(7*x - x^2))/(5*E^(8*x))
Leaf count is larger than twice the leaf count of optimal. \(43\) vs. \(2(21)=42\).
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{25} \left (4 e^{-8 x} \left (-80 x^2+580 x-70\right )+16 e^{-16 x} \left (4 x-32 x^2\right )+25 x-175\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \left (25 x+64 e^{-16 x} \left (x-8 x^2\right )-40 e^{-8 x} \left (8 x^2-58 x+7\right )-175\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (32 e^{-16 x} x^2+40 e^{-8 x} x^2+\frac {25 x^2}{2}-280 e^{-8 x} x-175 x\right )\) |
Input:
Int[(-175 + 25*x + (4*(-70 + 580*x - 80*x^2))/E^(8*x) + (16*(4*x - 32*x^2) )/E^(16*x))/25,x]
Output:
(-175*x - (280*x)/E^(8*x) + (25*x^2)/2 + (32*x^2)/E^(16*x) + (40*x^2)/E^(8 *x))/25
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62
method | result | size |
risch | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {4 \left (10 x^{2}-70 x \right ) {\mathrm e}^{-8 x}}{25}+\frac {x^{2}}{2}-7 x\) | \(34\) |
norman | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {8 \,{\mathrm e}^{-8 x} x^{2}}{5}-\frac {56 \,{\mathrm e}^{-8 x} x}{5}+\frac {x^{2}}{2}-7 x\) | \(50\) |
parallelrisch | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {8 \,{\mathrm e}^{-8 x} x^{2}}{5}-\frac {56 \,{\mathrm e}^{-8 x} x}{5}+\frac {x^{2}}{2}-7 x\) | \(50\) |
default | \(-7 x +\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \,{\mathrm e}^{-16 x} \ln \left (2\right )^{2}}{25}-\frac {4 \,{\mathrm e}^{-16 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \,{\mathrm e}^{-8 x} \ln \left (2\right )}{5}+\frac {{\mathrm e}^{-8 x} \ln \left (2\right )^{2}}{10}-\frac {{\mathrm e}^{-8 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{5}+\frac {x^{2}}{2}\) | \(146\) |
parts | \(-7 x +\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \,{\mathrm e}^{-16 x} \ln \left (2\right )^{2}}{25}-\frac {4 \,{\mathrm e}^{-16 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \,{\mathrm e}^{-8 x} \ln \left (2\right )}{5}+\frac {{\mathrm e}^{-8 x} \ln \left (2\right )^{2}}{10}-\frac {{\mathrm e}^{-8 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{5}+\frac {x^{2}}{2}\) | \(146\) |
derivativedivides | \(\frac {7 \ln \left (2\right )}{4}-7 x +\frac {\left (\ln \left (2\right )-4 x \right )^{2}}{32}+\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \,{\mathrm e}^{-16 x} \ln \left (2\right )^{2}}{25}-\frac {4 \,{\mathrm e}^{-16 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \,{\mathrm e}^{-8 x} \ln \left (2\right )}{5}+\frac {{\mathrm e}^{-8 x} \ln \left (2\right )^{2}}{10}-\frac {{\mathrm e}^{-8 x} \ln \left (2\right ) \left (\ln \left (2\right )-4 x \right )}{5}-\frac {\left (\ln \left (2\right )-4 x \right ) \ln \left (2\right )}{16}\) | \(165\) |
orering | \(\frac {\left (32768 x^{6}-712704 x^{5}+406016 x^{4}+23439360 x^{3}-14637728 x^{2}+3095176 x -205275\right ) \left (\frac {16 \left (-32 x^{2}+4 x \right ) {\mathrm e}^{-16 x}}{25}+\frac {4 \left (-80 x^{2}+580 x -70\right ) {\mathrm e}^{-8 x}}{25}+x -7\right )}{65536 x^{5}-954368 x^{4}+3762176 x^{3}-2114560 x^{2}+431872 x -28224}+\frac {3 \left (2048 x^{6}-43520 x^{5}+8960 x^{4}+1482208 x^{3}-735896 x^{2}+113505 x -5635\right ) \left (\frac {16 \left (-64 x +4\right ) {\mathrm e}^{-16 x}}{25}-\frac {256 \left (-32 x^{2}+4 x \right ) {\mathrm e}^{-16 x}}{25}+\frac {4 \left (-160 x +580\right ) {\mathrm e}^{-8 x}}{25}-\frac {32 \left (-80 x^{2}+580 x -70\right ) {\mathrm e}^{-8 x}}{25}+1\right )}{64 \left (1024 x^{5}-14912 x^{4}+58784 x^{3}-33040 x^{2}+6748 x -441\right )}+\frac {\left (4096 x^{6}-86016 x^{5}+2964416 x^{3}-1101240 x^{2}+135240 x -5635\right ) \left (-\frac {1024 \,{\mathrm e}^{-16 x}}{25}-\frac {512 \left (-64 x +4\right ) {\mathrm e}^{-16 x}}{25}+\frac {4096 \left (-32 x^{2}+4 x \right ) {\mathrm e}^{-16 x}}{25}-\frac {128 \,{\mathrm e}^{-8 x}}{5}-\frac {64 \left (-160 x +580\right ) {\mathrm e}^{-8 x}}{25}+\frac {256 \left (-80 x^{2}+580 x -70\right ) {\mathrm e}^{-8 x}}{25}\right )}{1024 \left (8 x^{2}-56 x +7\right ) \left (128 x^{3}-968 x^{2}+460 x -63\right )}\) | \(391\) |
Input:
int(1/25*(-32*x^2+4*x)*exp(ln(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(ln(2)- 4*x)^2+x-7,x,method=_RETURNVERBOSE)
Output:
32/25*exp(-16*x)*x^2+4/25*(10*x^2-70*x)*exp(-8*x)+1/2*x^2-7*x
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \left (2\right )\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \left (2\right )\right )} - 7 \, x \] Input:
integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp (log(2)-4*x)^2+x-7,x, algorithm="fricas")
Output:
2/25*x^2*e^(-16*x + 4*log(2)) + 1/2*x^2 + 2/5*(x^2 - 7*x)*e^(-8*x + 2*log( 2)) - 7*x
Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {x^{2}}{2} + \frac {32 x^{2} e^{- 16 x}}{25} - 7 x + \frac {\left (200 x^{2} - 1400 x\right ) e^{- 8 x}}{125} \] Input:
integrate(1/25*(-32*x**2+4*x)*exp(ln(2)-4*x)**4+1/25*(-80*x**2+580*x-70)*e xp(ln(2)-4*x)**2+x-7,x)
Output:
x**2/2 + 32*x**2*exp(-16*x)/25 - 7*x + (200*x**2 - 1400*x)*exp(-8*x)/125
Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {32}{25} \, x^{2} e^{\left (-16 \, x\right )} + \frac {1}{2} \, x^{2} + \frac {8}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x\right )} - 7 \, x \] Input:
integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp (log(2)-4*x)^2+x-7,x, algorithm="maxima")
Output:
32/25*x^2*e^(-16*x) + 1/2*x^2 + 8/5*(x^2 - 7*x)*e^(-8*x) - 7*x
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \left (2\right )\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \left (2\right )\right )} - 7 \, x \] Input:
integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp (log(2)-4*x)^2+x-7,x, algorithm="giac")
Output:
2/25*x^2*e^(-16*x + 4*log(2)) + 1/2*x^2 + 2/5*(x^2 - 7*x)*e^(-8*x + 2*log( 2)) - 7*x
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{-16\,x}\,\left (5\,{\mathrm {e}}^{8\,x}+8\right )\,\left (8\,x-70\,{\mathrm {e}}^{8\,x}+5\,x\,{\mathrm {e}}^{8\,x}\right )}{50} \] Input:
int(x + (exp(4*log(2) - 16*x)*(4*x - 32*x^2))/25 - (exp(2*log(2) - 8*x)*(8 0*x^2 - 580*x + 70))/25 - 7,x)
Output:
(x*exp(-16*x)*(5*exp(8*x) + 8)*(8*x - 70*exp(8*x) + 5*x*exp(8*x)))/50
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.10 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {x \left (25 e^{16 x} x -350 e^{16 x}+80 e^{8 x} x -560 e^{8 x}+64 x \right )}{50 e^{16 x}} \] Input:
int(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(log(2 )-4*x)^2+x-7,x)
Output:
(x*(25*e**(16*x)*x - 350*e**(16*x) + 80*e**(8*x)*x - 560*e**(8*x) + 64*x)) /(50*e**(16*x))