Integrand size = 139, antiderivative size = 30 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=e^x-x-\log \left (-2 x+e^4 x \left (x-\frac {1}{5} (3+x)^2\right )\right ) \] Output:
exp(x)-ln(exp(ln((x-1/5*(3+x)^2)*x)+4)-2*x)-x
Time = 2.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=e^x-x-\log (x)-\log \left (10+e^4 \left (9+x+x^2\right )\right ) \] Input:
Integrate[(18*x + 20*x^2 + 4*x^3 + 2*x^4 + E^x*(-18*x^2 - 2*x^3 - 2*x^4) + (E^4*(-9*x - x^2 - x^3)*(-9 - 11*x - 4*x^2 - x^3 + E^x*(9*x + x^2 + x^3)) )/5)/(-18*x^2 - 2*x^3 - 2*x^4 + (E^4*(-9*x - x^2 - x^3)*(9*x + x^2 + x^3)) /5),x]
Output:
E^x - x - Log[x] - Log[10 + E^4*(9 + x + x^2)]
Leaf count is larger than twice the leaf count of optimal. \(113\) vs. \(2(30)=60\).
Time = 2.00 (sec) , antiderivative size = 113, normalized size of antiderivative = 3.77, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2026, 2463, 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.
\(\displaystyle \int \frac {2 x^4+4 x^3+20 x^2+\frac {1}{5} e^4 \left (-x^3-x^2-9 x\right ) \left (-x^3-4 x^2+e^x \left (x^3+x^2+9 x\right )-11 x-9\right )+e^x \left (-2 x^4-2 x^3-18 x^2\right )+18 x}{-2 x^4-2 x^3-18 x^2+\frac {1}{5} e^4 \left (-x^3-x^2-9 x\right ) \left (x^3+x^2+9 x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2 x^4+4 x^3+20 x^2+\frac {1}{5} e^4 \left (-x^3-x^2-9 x\right ) \left (-x^3-4 x^2+e^x \left (x^3+x^2+9 x\right )-11 x-9\right )+e^x \left (-2 x^4-2 x^3-18 x^2\right )+18 x}{x^2 \left (-\frac {1}{5} e^4 x^4-\frac {2 e^4 x^3}{5}-\frac {1}{5} \left (10+19 e^4\right ) x^2-\frac {2}{5} \left (5+9 e^4\right ) x-\frac {9}{5} \left (10+9 e^4\right )\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^4 \left (2 x^4+4 x^3+20 x^2+\frac {1}{5} e^4 \left (-x^3-x^2-9 x\right ) \left (-x^3-4 x^2+e^x \left (x^3+x^2+9 x\right )-11 x-9\right )+e^x \left (-2 x^4-2 x^3-18 x^2\right )+18 x\right )}{2 x^2 \left (e^4 x^2+e^4 x+9 e^4+10\right )}-\frac {2 x^4+4 x^3+20 x^2+\frac {1}{5} e^4 \left (-x^3-x^2-9 x\right ) \left (-x^3-4 x^2+e^x \left (x^3+x^2+9 x\right )-11 x-9\right )+e^x \left (-2 x^4-2 x^3-18 x^2\right )+18 x}{2 x^2 \left (x^2+x+9\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log \left (e^4 x^2+e^4 x+9 e^4+10\right )-\frac {1}{10} \left (10+11 e^4\right ) x+\frac {11 e^4 x}{10}-\frac {9 e^{x+4}}{10}+\frac {1}{10} \left (10+9 e^4\right ) e^x-\frac {1}{10} \left (10+9 e^4\right ) \log (x)+\frac {81 e^8 \log (x)}{10 \left (10+9 e^4\right )}+\frac {9 e^4 \log (x)}{10+9 e^4}\) |
Input:
Int[(18*x + 20*x^2 + 4*x^3 + 2*x^4 + E^x*(-18*x^2 - 2*x^3 - 2*x^4) + (E^4* (-9*x - x^2 - x^3)*(-9 - 11*x - 4*x^2 - x^3 + E^x*(9*x + x^2 + x^3)))/5)/( -18*x^2 - 2*x^3 - 2*x^4 + (E^4*(-9*x - x^2 - x^3)*(9*x + x^2 + x^3))/5),x]
Output:
(-9*E^(4 + x))/10 + (E^x*(10 + 9*E^4))/10 + (11*E^4*x)/10 - ((10 + 11*E^4) *x)/10 + (9*E^4*Log[x])/(10 + 9*E^4) + (81*E^8*Log[x])/(10*(10 + 9*E^4)) - ((10 + 9*E^4)*Log[x])/10 - Log[10 + 9*E^4 + E^4*x + E^4*x^2]
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])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 5.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
method | result | size |
default | \(-x -\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )-\ln \left (x \right )+{\mathrm e}^{x}\) | \(30\) |
parts | \(-x -\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )-\ln \left (x \right )+{\mathrm e}^{x}\) | \(30\) |
norman | \(\frac {{\mathrm e}^{x} x -x^{2}}{x}-\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )\) | \(39\) |
parallelrisch | \(\frac {2 x^{6} {\mathrm e}^{x}+38 \,{\mathrm e}^{x} x^{4}+162 \,{\mathrm e}^{x} x^{2}+36 \,{\mathrm e}^{x} x^{3}+4 x^{5} {\mathrm e}^{x}-2 x^{7}+15 x^{6}+1539 x^{2}+180 x^{3}+325 x^{4}-2 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{6}-4 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{5}-38 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{4}-36 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{3}-162 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{2}}{2 x^{2} \left (x^{2}+x +9\right )^{2}}\) | \(215\) |
risch | \(-x -\ln \left (x^{3}+x^{2}+9 x \right )+{\mathrm e}^{x}-\ln \left (5\right )+4+\ln \left (x \right )+\ln \left (x^{2}+x +9\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \left (\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )-1\right )-\ln \left (2 x -\frac {x \left (x^{2}+x +9\right ) {\mathrm e}^{4} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{3}}{2}} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \operatorname {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}}}{5}\right )\) | \(240\) |
Input:
int((((x^3+x^2+9*x)*exp(x)-x^3-4*x^2-11*x-9)*exp(ln(-1/5*x^3-1/5*x^2-9/5*x )+4)+(-2*x^4-2*x^3-18*x^2)*exp(x)+2*x^4+4*x^3+20*x^2+18*x)/((x^3+x^2+9*x)* exp(ln(-1/5*x^3-1/5*x^2-9/5*x)+4)-2*x^4-2*x^3-18*x^2),x,method=_RETURNVERB OSE)
Output:
-x-ln(x^2*exp(4)+x*exp(4)+9*exp(4)+10)-ln(x)+exp(x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=-x + e^{x} - \log \left ({\left (x^{3} + x^{2} + 9 \, x\right )} e^{4} + 10 \, x\right ) \] Input:
integrate((((x^3+x^2+9*x)*exp(x)-x^3-4*x^2-11*x-9)*exp(log(-1/5*x^3-1/5*x^ 2-9/5*x)+4)+(-2*x^4-2*x^3-18*x^2)*exp(x)+2*x^4+4*x^3+20*x^2+18*x)/((x^3+x^ 2+9*x)*exp(log(-1/5*x^3-1/5*x^2-9/5*x)+4)-2*x^4-2*x^3-18*x^2),x, algorithm ="fricas")
Output:
-x + e^x - log((x^3 + x^2 + 9*x)*e^4 + 10*x)
Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=- x + e^{x} - \log {\left (x^{3} e^{4} + x^{2} e^{4} + x \left (10 + 9 e^{4}\right ) \right )} \] Input:
integrate((((x**3+x**2+9*x)*exp(x)-x**3-4*x**2-11*x-9)*exp(ln(-1/5*x**3-1/ 5*x**2-9/5*x)+4)+(-2*x**4-2*x**3-18*x**2)*exp(x)+2*x**4+4*x**3+20*x**2+18* x)/((x**3+x**2+9*x)*exp(ln(-1/5*x**3-1/5*x**2-9/5*x)+4)-2*x**4-2*x**3-18*x **2),x)
Output:
-x + exp(x) - log(x**3*exp(4) + x**2*exp(4) + x*(10 + 9*exp(4)))
Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (29) = 58\).
Time = 0.18 (sec) , antiderivative size = 703, normalized size of antiderivative = 23.43 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate((((x^3+x^2+9*x)*exp(x)-x^3-4*x^2-11*x-9)*exp(log(-1/5*x^3-1/5*x^ 2-9/5*x)+4)+(-2*x^4-2*x^3-18*x^2)*exp(x)+2*x^4+4*x^3+20*x^2+18*x)/((x^3+x^ 2+9*x)*exp(log(-1/5*x^3-1/5*x^2-9/5*x)+4)-2*x^4-2*x^3-18*x^2),x, algorithm ="maxima")
Output:
-(17*e^4 + 20)*arctan((2*x*e^4 + e^4)*e^(-2)/sqrt(35*e^4 + 40))*e^(-2)/sqr t(35*e^4 + 40) + 1/700*(35*(17*e^4 + 20)*e^(-4)*log(x^2*e^4 + x*e^4 + 9*e^ 4 + 10) + 70*(127*e^8 + 320*e^4 + 200)*arctan((2*x*e^4 + e^4)*e^(-2)/sqrt( 35*e^4 + 40))*e^(-6)/sqrt(35*e^4 + 40) - 700*x*e^(-4) - 254*sqrt(35)*arcta n(1/35*sqrt(35)*(2*x + 1)) - 595*log(x^2 + x + 9))*e^4 - 1/70*(35*(4*e^4 + 5)*e^(-4)*log(x^2*e^4 + x*e^4 + 9*e^4 + 10) - 70*(13*e^4 + 15)*arctan((2* x*e^4 + e^4)*e^(-2)/sqrt(35*e^4 + 40))*e^(-2)/sqrt(35*e^4 + 40) + 26*sqrt( 35)*arctan(1/35*sqrt(35)*(2*x + 1)) - 140*log(x^2 + x + 9))*e^4 - 6/175*(7 0*(17*e^4 + 20)*arctan((2*x*e^4 + e^4)*e^(-2)/sqrt(35*e^4 + 40))*e^(-2)/sq rt(35*e^4 + 40) - 34*sqrt(35)*arctan(1/35*sqrt(35)*(2*x + 1)) + 35*log(x^2 *e^4 + x*e^4 + 9*e^4 + 10) - 35*log(x^2 + x + 9))*e^4 + 2/25*(2*sqrt(35)*a rctan(1/35*sqrt(35)*(2*x + 1)) - 70*arctan((2*x*e^4 + e^4)*e^(-2)/sqrt(35* e^4 + 40))*e^2/sqrt(35*e^4 + 40) + 35*log(x^2*e^4 + x*e^4 + 9*e^4 + 10) - 35*log(x^2 + x + 9))*e^4 + 9/700*(2*sqrt(35)*arctan(1/35*sqrt(35)*(2*x + 1 )) - 315*e^4*log(x^2*e^4 + x*e^4 + 9*e^4 + 10)/(9*e^4 + 10) - 630*arctan(( 2*x*e^4 + e^4)*e^(-2)/sqrt(35*e^4 + 40))*e^6/(sqrt(35*e^4 + 40)*(9*e^4 + 1 0)) - 700*log(x)/(9*e^4 + 10) + 35*log(x^2 + x + 9))*e^4 - 108/175*(sqrt(3 5)*arctan(1/35*sqrt(35)*(2*x + 1)) - 35*arctan((2*x*e^4 + e^4)*e^(-2)/sqrt (35*e^4 + 40))*e^2/sqrt(35*e^4 + 40))*e^4 + 18*arctan((2*x*e^4 + e^4)*e^(- 2)/sqrt(35*e^4 + 40))*e^2/sqrt(35*e^4 + 40) - 9/2*e^4*log(x^2*e^4 + x*e...
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=-{\left (x e^{4} + e^{4} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) + e^{4} \log \left (x\right ) - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )} \] Input:
integrate((((x^3+x^2+9*x)*exp(x)-x^3-4*x^2-11*x-9)*exp(log(-1/5*x^3-1/5*x^ 2-9/5*x)+4)+(-2*x^4-2*x^3-18*x^2)*exp(x)+2*x^4+4*x^3+20*x^2+18*x)/((x^3+x^ 2+9*x)*exp(log(-1/5*x^3-1/5*x^2-9/5*x)+4)-2*x^4-2*x^3-18*x^2),x, algorithm ="giac")
Output:
-(x*e^4 + e^4*log(x^2*e^4 + x*e^4 + 9*e^4 + 10) + e^4*log(x) - e^(x + 4))* e^(-4)
Time = 0.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx={\mathrm {e}}^x-\ln \left (9\,x+10\,x\,{\mathrm {e}}^{-4}+x^2+x^3\right )-x \] Input:
int(-(18*x - exp(x)*(18*x^2 + 2*x^3 + 2*x^4) - exp(log(- (9*x)/5 - x^2/5 - x^3/5) + 4)*(11*x - exp(x)*(9*x + x^2 + x^3) + 4*x^2 + x^3 + 9) + 20*x^2 + 4*x^3 + 2*x^4)/(18*x^2 - exp(log(- (9*x)/5 - x^2/5 - x^3/5) + 4)*(9*x + x^2 + x^3) + 2*x^3 + 2*x^4),x)
Output:
exp(x) - log(9*x + 10*x*exp(-4) + x^2 + x^3) - x
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=e^{x}-\mathrm {log}\left (e^{4} x^{2}+e^{4} x +9 e^{4}+10\right )-\mathrm {log}\left (x \right )-x \] Input:
int((((x^3+x^2+9*x)*exp(x)-x^3-4*x^2-11*x-9)*exp(log(-1/5*x^3-1/5*x^2-9/5* x)+4)+(-2*x^4-2*x^3-18*x^2)*exp(x)+2*x^4+4*x^3+20*x^2+18*x)/((x^3+x^2+9*x) *exp(log(-1/5*x^3-1/5*x^2-9/5*x)+4)-2*x^4-2*x^3-18*x^2),x)
Output:
e**x - log(e**4*x**2 + e**4*x + 9*e**4 + 10) - log(x) - x