Integrand size = 68, antiderivative size = 25 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {e^x \left (-\frac {1}{2}+2 x\right )^2}{\left (-2+\frac {e^2 x}{4}\right )^2} \end {dmath*}
Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {4 e^x (1-4 x)^2}{\left (-8+e^2 x\right )^2} \end {dmath*}
Integrate[(64*E^(-6 + x)*(-14 + 48*x + 32*x^2) + 16*E^(-4 + x)*(2 - 9*x + 8*x^2 - 16*x^3))/(2048/E^6 - (768*x)/E^4 + (96*x^2)/E^2 - 4*x^3),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.73 (sec) , antiderivative size = 192, normalized size of antiderivative = 7.68, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2007, 7292, 27, 27, 2629, 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.
| \(\displaystyle \int \frac {64 e^{x-6} \left (32 x^2+48 x-14\right )+16 e^{x-4} \left (-16 x^3+8 x^2-9 x+2\right )}{-4 x^3+\frac {96 x^2}{e^2}-\frac {768 x}{e^4}+\frac {2048}{e^6}} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {64 e^{x-6} \left (32 x^2+48 x-14\right )+16 e^{x-4} \left (-16 x^3+8 x^2-9 x+2\right )}{\left (\frac {8\ 2^{2/3}}{e^2}-2^{2/3} x\right )^3}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {16 e^{x-6} (1-4 x) \left (4 e^2 x^2-\left (32+e^2\right ) x-2 \left (28-e^2\right )\right )}{\left (\frac {8\ 2^{2/3}}{e^2}-2^{2/3} x\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 16 \int -\frac {e^x (1-4 x) \left (-4 e^2 x^2+\left (32+e^2\right ) x+2 \left (28-e^2\right )\right )}{4 \left (8-e^2 x\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 \int \frac {e^x (1-4 x) \left (-4 e^2 x^2+\left (32+e^2\right ) x+2 \left (28-e^2\right )\right )}{\left (8-e^2 x\right )^3}dx\) |
| \(\Big \downarrow \) 2629 |
| \(\displaystyle -4 \int \left (-16 e^{x-4}+\frac {8 e^{x-4} \left (-32+e^2\right )}{e^2 x-8}+\frac {e^{x-4} \left (-1024+320 e^2-9 e^4\right )}{\left (e^2 x-8\right )^2}+\frac {2 e^{x-2} \left (-32+e^2\right )^2}{\left (e^2 x-8\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \left (-e^{\frac {8}{e^2}-8} \left (1024-320 e^2+9 e^4\right ) \operatorname {ExpIntegralEi}\left (-\frac {8-e^2 x}{e^2}\right )+e^{\frac {8}{e^2}-8} \left (32-e^2\right )^2 \operatorname {ExpIntegralEi}\left (-\frac {8-e^2 x}{e^2}\right )-8 e^{\frac {8}{e^2}-6} \left (32-e^2\right ) \operatorname {ExpIntegralEi}\left (-\frac {8-e^2 x}{e^2}\right )-16 e^{x-4}-\frac {\left (1024-320 e^2+9 e^4\right ) e^{x-6}}{8-e^2 x}+\frac {\left (32-e^2\right )^2 e^{x-6}}{8-e^2 x}-\frac {\left (32-e^2\right )^2 e^{x-4}}{\left (8-e^2 x\right )^2}\right )\) |
Int[(64*E^(-6 + x)*(-14 + 48*x + 32*x^2) + 16*E^(-4 + x)*(2 - 9*x + 8*x^2 - 16*x^3))/(2048/E^6 - (768*x)/E^4 + (96*x^2)/E^2 - 4*x^3),x]
-4*(-16*E^(-4 + x) - (E^(-4 + x)*(32 - E^2)^2)/(8 - E^2*x)^2 + (E^(-6 + x) *(32 - E^2)^2)/(8 - E^2*x) - (E^(-6 + x)*(1024 - 320*E^2 + 9*E^4))/(8 - E^ 2*x) - 8*E^(-6 + 8/E^2)*(32 - E^2)*ExpIntegralEi[-((8 - E^2*x)/E^2)] + E^( -8 + 8/E^2)*(32 - E^2)^2*ExpIntegralEi[-((8 - E^2*x)/E^2)] - E^(-8 + 8/E^2 )*(1024 - 320*E^2 + 9*E^4)*ExpIntegralEi[-((8 - E^2*x)/E^2)])
3.3.17.3.1 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_.)*(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[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 0.38 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72
| method | result | size |
| norman | \(\frac {\left (4 \,{\mathrm e}^{4} {\mathrm e}^{x}-32 x \,{\mathrm e}^{4} {\mathrm e}^{x}+64 x^{2} {\mathrm e}^{4} {\mathrm e}^{x}\right ) {\mathrm e}^{-4}}{\left ({\mathrm e}^{2} x -8\right )^{2}}\) | \(43\) |
| gosper | \(\frac {4 \left (-1+4 x \right )^{2} {\mathrm e}^{x} {\mathrm e}^{-4}}{64 \,{\mathrm e}^{-4}-4 x \,{\mathrm e}^{2 \ln \left (2\right )-2}+x^{2}}\) | \(48\) |
| parallelrisch | \(\frac {16 \,{\mathrm e}^{-4} {\mathrm e}^{x}-128 \,{\mathrm e}^{-4} x \,{\mathrm e}^{x}+256 \,{\mathrm e}^{-4} x^{2} {\mathrm e}^{x}}{256 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{2 \ln \left (2\right )-2}+4 x^{2}}\) | \(73\) |
| default | \(8 \,{\mathrm e}^{-4} \left (-\frac {{\mathrm e}^{x} \left (-x +8 \,{\mathrm e}^{-2}-1\right )}{2 \left (64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}\right )}+\frac {{\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )}{2}\right )-224 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{x} \left (-x +8 \,{\mathrm e}^{-2}-1\right )}{2 \left (64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}\right )}+\frac {{\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )}{2}\right )-36 \,{\mathrm e}^{-4} \left (-\frac {{\mathrm e}^{x} \left (32 \,{\mathrm e}^{-4}-4 x \,{\mathrm e}^{-2}+4 \,{\mathrm e}^{-2}-x \right )}{64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}}-\left (-4 \,{\mathrm e}^{-2}-1\right ) {\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )\right )+32 \,{\mathrm e}^{-4} \left (-\frac {16 \,{\mathrm e}^{x} {\mathrm e}^{-2} \left (16 \,{\mathrm e}^{-4}-2 x \,{\mathrm e}^{-2}+6 \,{\mathrm e}^{-2}-x \right )}{64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}}-\left (-32 \,{\mathrm e}^{-4}-16 \,{\mathrm e}^{-2}-1\right ) {\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )\right )-64 \,{\mathrm e}^{-4} \left (-{\mathrm e}^{x}-\frac {64 \,{\mathrm e}^{x} {\mathrm e}^{-4} \left (32 \,{\mathrm e}^{-4}-4 x \,{\mathrm e}^{-2}+20 \,{\mathrm e}^{-2}-3 x \right )}{64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}}-\left (-256 \,{\mathrm e}^{-6}-192 \,{\mathrm e}^{-4}-24 \,{\mathrm e}^{-2}\right ) {\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )\right )+768 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{x} \left (32 \,{\mathrm e}^{-4}-4 x \,{\mathrm e}^{-2}+4 \,{\mathrm e}^{-2}-x \right )}{64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}}-\left (-4 \,{\mathrm e}^{-2}-1\right ) {\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )\right )+512 \,{\mathrm e}^{-6} \left (-\frac {16 \,{\mathrm e}^{x} {\mathrm e}^{-2} \left (16 \,{\mathrm e}^{-4}-2 x \,{\mathrm e}^{-2}+6 \,{\mathrm e}^{-2}-x \right )}{64 \,{\mathrm e}^{-4}-16 x \,{\mathrm e}^{-2}+x^{2}}-\left (-32 \,{\mathrm e}^{-4}-16 \,{\mathrm e}^{-2}-1\right ) {\mathrm e}^{8 \,{\mathrm e}^{-2}} \operatorname {Ei}_{1}\left (8 \,{\mathrm e}^{-2}-x \right )\right )\) | \(487\) |
int(((32*x^2+48*x-14)*exp(x)*exp(2*ln(2)-2)^3+(-16*x^3+8*x^2-9*x+2)*exp(x) *exp(2*ln(2)-2)^2)/(32*exp(2*ln(2)-2)^3-48*x*exp(2*ln(2)-2)^2+24*x^2*exp(2 *ln(2)-2)-4*x^3),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {{\left (16 \, x^{2} - 8 \, x + 1\right )} e^{\left (x + 6 \, \log \left (2\right ) - 6\right )}}{4 \, {\left (x^{2} e^{\left (2 \, \log \left (2\right ) - 2\right )} - 4 \, x e^{\left (4 \, \log \left (2\right ) - 4\right )} + 4 \, e^{\left (6 \, \log \left (2\right ) - 6\right )}\right )}} \end {dmath*}
integrate(((32*x^2+48*x-14)*exp(x)*exp(2*log(2)-2)^3+(-16*x^3+8*x^2-9*x+2) *exp(x)*exp(2*log(2)-2)^2)/(32*exp(2*log(2)-2)^3-48*x*exp(2*log(2)-2)^2+24 *x^2*exp(2*log(2)-2)-4*x^3),x, algorithm=\
1/4*(16*x^2 - 8*x + 1)*e^(x + 6*log(2) - 6)/(x^2*e^(2*log(2) - 2) - 4*x*e^ (4*log(2) - 4) + 4*e^(6*log(2) - 6))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {\left (64 x^{2} - 32 x + 4\right ) e^{x}}{x^{2} e^{4} - 16 x e^{2} + 64} \end {dmath*}
integrate(((32*x**2+48*x-14)*exp(x)*exp(2*ln(2)-2)**3+(-16*x**3+8*x**2-9*x +2)*exp(x)*exp(2*ln(2)-2)**2)/(32*exp(2*ln(2)-2)**3-48*x*exp(2*ln(2)-2)**2 +24*x**2*exp(2*ln(2)-2)-4*x**3),x)
\begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\int { -\frac {2 \, {\left (16 \, x^{2} + 24 \, x - 7\right )} e^{\left (x + 6 \, \log \left (2\right ) - 6\right )} - {\left (16 \, x^{3} - 8 \, x^{2} + 9 \, x - 2\right )} e^{\left (x + 4 \, \log \left (2\right ) - 4\right )}}{4 \, {\left (x^{3} - 6 \, x^{2} e^{\left (2 \, \log \left (2\right ) - 2\right )} + 12 \, x e^{\left (4 \, \log \left (2\right ) - 4\right )} - 8 \, e^{\left (6 \, \log \left (2\right ) - 6\right )}\right )}} \,d x } \end {dmath*}
integrate(((32*x^2+48*x-14)*exp(x)*exp(2*log(2)-2)^3+(-16*x^3+8*x^2-9*x+2) *exp(x)*exp(2*log(2)-2)^2)/(32*exp(2*log(2)-2)^3-48*x*exp(2*log(2)-2)^2+24 *x^2*exp(2*log(2)-2)-4*x^3),x, algorithm=\
4*(16*x^3*e^2 - 8*x^2*(e^2 + 16) + x*(e^2 + 64))*e^x/(x^3*e^6 - 24*x^2*e^4 + 192*x*e^2 - 512) - 224*e^(8*e^(-2) - 2)*exp_integral_e(3, -(x*e^2 - 8)* e^(-2))/(x*e^2 - 8)^2 - 1/4*integrate(128*(8*x*e^2 - 3*e^2 - 64)*e^x/(x^4* e^8 - 32*x^3*e^6 + 384*x^2*e^4 - 2048*x*e^2 + 4096), x)
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {4 \, {\left (16 \, x^{2} e^{x} - 8 \, x e^{x} + e^{x}\right )}}{x^{2} e^{4} - 16 \, x e^{2} + 64} \end {dmath*}
integrate(((32*x^2+48*x-14)*exp(x)*exp(2*log(2)-2)^3+(-16*x^3+8*x^2-9*x+2) *exp(x)*exp(2*log(2)-2)^2)/(32*exp(2*log(2)-2)^3-48*x*exp(2*log(2)-2)^2+24 *x^2*exp(2*log(2)-2)-4*x^3),x, algorithm=\
Time = 0.53 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {64 e^{-6+x} \left (-14+48 x+32 x^2\right )+16 e^{-4+x} \left (2-9 x+8 x^2-16 x^3\right )}{\frac {2048}{e^6}-\frac {768 x}{e^4}+\frac {96 x^2}{e^2}-4 x^3} \, dx=\frac {4\,{\mathrm {e}}^x\,{\left (4\,x-1\right )}^2}{{\left (x\,{\mathrm {e}}^2-8\right )}^2} \end {dmath*}