Integrand size = 109, antiderivative size = 32 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=-5-x-\left (4+\frac {2}{x}+\frac {x}{4}\right ) \left (-1+\left (e^x-x+\log (2)\right )^2\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(82\) vs. \(2(32)=64\).
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.56 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=-\frac {x^4+e^{2 x} \left (8+16 x+x^2\right )+8 \left (-1+\log ^2(2)\right )-x^3 (-16+\log (4))+e^x \left (-2 x^3+x^2 (-32+\log (4))+\log (65536)+x (-16+\log (4294967296))\right )+x^2 \left (11+\log ^2(2)-\log (4294967296)\right )}{4 x} \]
Integrate[(-8 - 11*x^2 - 32*x^3 - 3*x^4 + E^(2*x)*(8 - 16*x - 33*x^2 - 2*x ^3) + (32*x^2 + 4*x^3)*Log[2] + (8 - x^2)*Log[2]^2 + E^x*(48*x^2 + 36*x^3 + 2*x^4 + (16 - 16*x - 34*x^2 - 2*x^3)*Log[2]))/(4*x^2),x]
-1/4*(x^4 + E^(2*x)*(8 + 16*x + x^2) + 8*(-1 + Log[2]^2) - x^3*(-16 + Log[ 4]) + E^x*(-2*x^3 + x^2*(-32 + Log[4]) + Log[65536] + x*(-16 + Log[4294967 296])) + x^2*(11 + Log[2]^2 - Log[4294967296]))/x
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.56 (sec) , antiderivative size = 147, normalized size of antiderivative = 4.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {-3 x^4-32 x^3-11 x^2+\left (8-x^2\right ) \log ^2(2)+e^{2 x} \left (-2 x^3-33 x^2-16 x+8\right )+\left (4 x^3+32 x^2\right ) \log (2)+e^x \left (2 x^4+36 x^3+48 x^2+\left (-2 x^3-34 x^2-16 x+16\right ) \log (2)\right )-8}{4 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {3 x^4+32 x^3+11 x^2-e^{2 x} \left (-2 x^3-33 x^2-16 x+8\right )-2 e^x \left (x^4+18 x^3+24 x^2+\left (-x^3-17 x^2-8 x+8\right ) \log (2)\right )-\left (8-x^2\right ) \log ^2(2)-4 \left (x^3+8 x^2\right ) \log (2)+8}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {3 x^4+32 x^3+11 x^2-e^{2 x} \left (-2 x^3-33 x^2-16 x+8\right )-2 e^x \left (x^4+18 x^3+24 x^2+\left (-x^3-17 x^2-8 x+8\right ) \log (2)\right )-\left (8-x^2\right ) \log ^2(2)-4 \left (x^3+8 x^2\right ) \log (2)+8}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {e^{2 x} \left (2 x^3+33 x^2+16 x-8\right )}{x^2}+\frac {3 x^4+4 (8-\log (2)) x^3+\left (11-32 \log (2)+\log ^2(2)\right ) x^2+8 \left (1-\log ^2(2)\right )}{x^2}+\frac {2 e^x \left (-x^4-(18-\log (2)) x^3-(24-17 \log (2)) x^2+8 \log (2) x-\log (256)\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (2 \log (256) \operatorname {ExpIntegralEi}(x)-16 \log (2) \operatorname {ExpIntegralEi}(x)-x^3+2 e^x x^2-2 x^2 (8-\log (2))-4 e^x x-e^{2 x} x+4 e^x-16 e^{2 x}-\frac {8 e^{2 x}}{x}-x \left (11+\log ^2(2)-32 \log (2)\right )+\frac {8 \left (1-\log ^2(2)\right )}{x}+2 e^x x (18-\log (2))-2 e^x (18-\log (2))+2 e^x (24-17 \log (2))-\frac {2 e^x \log (256)}{x}\right )\) |
Int[(-8 - 11*x^2 - 32*x^3 - 3*x^4 + E^(2*x)*(8 - 16*x - 33*x^2 - 2*x^3) + (32*x^2 + 4*x^3)*Log[2] + (8 - x^2)*Log[2]^2 + E^x*(48*x^2 + 36*x^3 + 2*x^ 4 + (16 - 16*x - 34*x^2 - 2*x^3)*Log[2]))/(4*x^2),x]
(4*E^x - 16*E^(2*x) - (8*E^(2*x))/x - 4*E^x*x - E^(2*x)*x + 2*E^x*x^2 - x^ 3 + 2*E^x*(24 - 17*Log[2]) - 2*x^2*(8 - Log[2]) - 2*E^x*(18 - Log[2]) + 2* E^x*x*(18 - Log[2]) - 16*ExpIntegralEi[x]*Log[2] + (8*(1 - Log[2]^2))/x - x*(11 - 32*Log[2] + Log[2]^2) - (2*E^x*Log[256])/x + 2*ExpIntegralEi[x]*Lo g[256])/4
3.25.39.3.1 Defintions of rubi rules used
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]]
Leaf count of result is larger than twice the leaf count of optimal. \(100\) vs. \(2(29)=58\).
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.16
method | result | size |
norman | \(\frac {\left (\frac {\ln \left (2\right )}{2}-4\right ) x^{3}+\left (-\frac {\ln \left (2\right )^{2}}{4}+8 \ln \left (2\right )-\frac {11}{4}\right ) x^{2}+\left (4-8 \ln \left (2\right )\right ) x \,{\mathrm e}^{x}+\left (8-\frac {\ln \left (2\right )}{2}\right ) x^{2} {\mathrm e}^{x}-\frac {x^{4}}{4}-2 \,{\mathrm e}^{2 x}-4 x \,{\mathrm e}^{2 x}+\frac {{\mathrm e}^{x} x^{3}}{2}-4 \,{\mathrm e}^{x} \ln \left (2\right )-\frac {{\mathrm e}^{2 x} x^{2}}{4}-2 \ln \left (2\right )^{2}+2}{x}\) | \(101\) |
risch | \(-\frac {x \ln \left (2\right )^{2}}{4}+\frac {x^{2} \ln \left (2\right )}{2}-\frac {x^{3}}{4}+8 x \ln \left (2\right )-4 x^{2}-\frac {11 x}{4}-\frac {2 \ln \left (2\right )^{2}}{x}+\frac {2}{x}-\frac {\left (x^{2}+16 x +8\right ) {\mathrm e}^{2 x}}{4 x}-\frac {\left (x^{2} \ln \left (2\right )-x^{3}+16 x \ln \left (2\right )-16 x^{2}+8 \ln \left (2\right )-8 x \right ) {\mathrm e}^{x}}{2 x}\) | \(101\) |
parts | \(-\frac {x \ln \left (2\right )^{2}}{4}+\frac {x^{2} \ln \left (2\right )}{2}-\frac {x^{3}}{4}+8 x \ln \left (2\right )-4 x^{2}-\frac {11 x}{4}-\frac {2 \ln \left (2\right )^{2}}{x}+\frac {2}{x}-\frac {x \ln \left (2\right ) {\mathrm e}^{x}}{2}-8 \,{\mathrm e}^{x} \ln \left (2\right )+8 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{x} x^{2}}{2}-\frac {4 \ln \left (2\right ) {\mathrm e}^{x}}{x}-4 \,{\mathrm e}^{2 x}-\frac {2 \,{\mathrm e}^{2 x}}{x}-\frac {x \,{\mathrm e}^{2 x}}{4}\) | \(108\) |
parallelrisch | \(-\frac {2 x^{2} \ln \left (2\right ) {\mathrm e}^{x}-2 \,{\mathrm e}^{x} x^{3}+x^{2} \ln \left (2\right )^{2}-2 x^{3} \ln \left (2\right )+x^{4}+32 x \ln \left (2\right ) {\mathrm e}^{x}-32 \,{\mathrm e}^{x} x^{2}-32 x^{2} \ln \left (2\right )+{\mathrm e}^{2 x} x^{2}+16 x^{3}+16 \,{\mathrm e}^{x} \ln \left (2\right )-16 \,{\mathrm e}^{x} x +8 \ln \left (2\right )^{2}+16 x \,{\mathrm e}^{2 x}+11 x^{2}+8 \,{\mathrm e}^{2 x}-8}{4 x}\) | \(111\) |
default | \(-4 x^{2}-\frac {11 x}{4}+\frac {2}{x}-\frac {x^{3}}{4}-4 \,{\mathrm e}^{2 x}-\frac {x \ln \left (2\right )^{2}}{4}-\frac {x \,{\mathrm e}^{2 x}}{4}+\frac {x^{2} \ln \left (2\right )}{2}+8 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{x} x^{2}}{2}-\frac {17 \,{\mathrm e}^{x} \ln \left (2\right )}{2}-\frac {2 \,{\mathrm e}^{2 x}}{x}-\frac {2 \ln \left (2\right )^{2}}{x}+4 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )-\frac {\ln \left (2\right ) \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )}{2}+4 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )+8 x \ln \left (2\right )\) | \(133\) |
int(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*ln(2)+2* x^4+36*x^3+48*x^2)*exp(x)+(-x^2+8)*ln(2)^2+(4*x^3+32*x^2)*ln(2)-3*x^4-32*x ^3-11*x^2-8)/x^2,x,method=_RETURNVERBOSE)
((1/2*ln(2)-4)*x^3+(-1/4*ln(2)^2+8*ln(2)-11/4)*x^2+(4-8*ln(2))*x*exp(x)+(8 -1/2*ln(2))*x^2*exp(x)-1/4*x^4-2*exp(x)^2-4*x*exp(x)^2+1/2*exp(x)*x^3-4*ex p(x)*ln(2)-1/4*exp(x)^2*x^2-2*ln(2)^2+2)/x
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (29) = 58\).
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.62 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=-\frac {x^{4} + 16 \, x^{3} + {\left (x^{2} + 8\right )} \log \left (2\right )^{2} + 11 \, x^{2} + {\left (x^{2} + 16 \, x + 8\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{3} + 16 \, x^{2} - {\left (x^{2} + 16 \, x + 8\right )} \log \left (2\right ) + 8 \, x\right )} e^{x} - 2 \, {\left (x^{3} + 16 \, x^{2}\right )} \log \left (2\right ) - 8}{4 \, x} \]
integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*lo g(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3 *x^4-32*x^3-11*x^2-8)/x^2,x, algorithm=\
-1/4*(x^4 + 16*x^3 + (x^2 + 8)*log(2)^2 + 11*x^2 + (x^2 + 16*x + 8)*e^(2*x ) - 2*(x^3 + 16*x^2 - (x^2 + 16*x + 8)*log(2) + 8*x)*e^x - 2*(x^3 + 16*x^2 )*log(2) - 8)/x
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.44 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=- \frac {x^{3}}{4} - \frac {x^{2} \cdot \left (16 - 2 \log {\left (2 \right )}\right )}{4} - \frac {x \left (- 32 \log {\left (2 \right )} + \log {\left (2 \right )}^{2} + 11\right )}{4} - \frac {-8 + 8 \log {\left (2 \right )}^{2}}{4 x} + \frac {\left (- 2 x^{3} - 32 x^{2} - 16 x\right ) e^{2 x} + \left (4 x^{4} - 4 x^{3} \log {\left (2 \right )} + 64 x^{3} - 64 x^{2} \log {\left (2 \right )} + 32 x^{2} - 32 x \log {\left (2 \right )}\right ) e^{x}}{8 x^{2}} \]
integrate(1/4*((-2*x**3-33*x**2-16*x+8)*exp(x)**2+((-2*x**3-34*x**2-16*x+1 6)*ln(2)+2*x**4+36*x**3+48*x**2)*exp(x)+(-x**2+8)*ln(2)**2+(4*x**3+32*x**2 )*ln(2)-3*x**4-32*x**3-11*x**2-8)/x**2,x)
-x**3/4 - x**2*(16 - 2*log(2))/4 - x*(-32*log(2) + log(2)**2 + 11)/4 - (-8 + 8*log(2)**2)/(4*x) + ((-2*x**3 - 32*x**2 - 16*x)*exp(2*x) + (4*x**4 - 4 *x**3*log(2) + 64*x**3 - 64*x**2*log(2) + 32*x**2 - 32*x*log(2))*exp(x))/( 8*x**2)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.06 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=-\frac {1}{4} \, x^{3} + \frac {1}{2} \, x^{2} \log \left (2\right ) - \frac {1}{2} \, {\left (x - 1\right )} e^{x} \log \left (2\right ) - \frac {1}{4} \, x \log \left (2\right )^{2} - 4 \, x^{2} - \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 9 \, {\left (x - 1\right )} e^{x} + 8 \, x \log \left (2\right ) - 4 \, {\rm Ei}\left (x\right ) \log \left (2\right ) - \frac {17}{2} \, e^{x} \log \left (2\right ) + 4 \, \Gamma \left (-1, -x\right ) \log \left (2\right ) - \frac {11}{4} \, x - \frac {2 \, \log \left (2\right )^{2}}{x} + \frac {2}{x} - 4 \, {\rm Ei}\left (2 \, x\right ) - \frac {33}{8} \, e^{\left (2 \, x\right )} + 12 \, e^{x} + 4 \, \Gamma \left (-1, -2 \, x\right ) \]
integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*lo g(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3 *x^4-32*x^3-11*x^2-8)/x^2,x, algorithm=\
-1/4*x^3 + 1/2*x^2*log(2) - 1/2*(x - 1)*e^x*log(2) - 1/4*x*log(2)^2 - 4*x^ 2 - 1/8*(2*x - 1)*e^(2*x) + 1/2*(x^2 - 2*x + 2)*e^x + 9*(x - 1)*e^x + 8*x* log(2) - 4*Ei(x)*log(2) - 17/2*e^x*log(2) + 4*gamma(-1, -x)*log(2) - 11/4* x - 2*log(2)^2/x + 2/x - 4*Ei(2*x) - 33/8*e^(2*x) + 12*e^x + 4*gamma(-1, - 2*x)
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (29) = 58\).
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.44 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=-\frac {x^{4} - 2 \, x^{3} e^{x} - 2 \, x^{3} \log \left (2\right ) + 2 \, x^{2} e^{x} \log \left (2\right ) + x^{2} \log \left (2\right )^{2} + 16 \, x^{3} + x^{2} e^{\left (2 \, x\right )} - 32 \, x^{2} e^{x} - 32 \, x^{2} \log \left (2\right ) + 32 \, x e^{x} \log \left (2\right ) + 11 \, x^{2} + 16 \, x e^{\left (2 \, x\right )} - 16 \, x e^{x} + 16 \, e^{x} \log \left (2\right ) + 8 \, \log \left (2\right )^{2} + 8 \, e^{\left (2 \, x\right )} - 8}{4 \, x} \]
integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*lo g(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3 *x^4-32*x^3-11*x^2-8)/x^2,x, algorithm=\
-1/4*(x^4 - 2*x^3*e^x - 2*x^3*log(2) + 2*x^2*e^x*log(2) + x^2*log(2)^2 + 1 6*x^3 + x^2*e^(2*x) - 32*x^2*e^x - 32*x^2*log(2) + 32*x*e^x*log(2) + 11*x^ 2 + 16*x*e^(2*x) - 16*x*e^x + 16*e^x*log(2) + 8*log(2)^2 + 8*e^(2*x) - 8)/ x
Time = 13.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.81 \[ \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{4 x^2} \, dx=x^2\,\left (\frac {\ln \left (4\right )}{4}+\frac {{\mathrm {e}}^x}{2}-4\right )-\frac {2\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^x\,\ln \left (2\right )+2\,{\ln \left (2\right )}^2-2}{x}-{\mathrm {e}}^x\,\left (8\,\ln \left (2\right )-4\right )-4\,{\mathrm {e}}^{2\,x}-x\,\left (\frac {{\mathrm {e}}^{2\,x}}{4}-8\,\ln \left (2\right )+\frac {{\mathrm {e}}^x\,\left (\ln \left (4\right )-32\right )}{4}+\frac {{\ln \left (2\right )}^2}{4}+\frac {11}{4}\right )-\frac {x^3}{4} \]
int(-((exp(2*x)*(16*x + 33*x^2 + 2*x^3 - 8))/4 + (log(2)^2*(x^2 - 8))/4 - (log(2)*(32*x^2 + 4*x^3))/4 - (exp(x)*(48*x^2 - log(2)*(16*x + 34*x^2 + 2* x^3 - 16) + 36*x^3 + 2*x^4))/4 + (11*x^2)/4 + 8*x^3 + (3*x^4)/4 + 2)/x^2,x )