Integrand size = 62, antiderivative size = 27 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=-3+e^x \left (2+x \left (\frac {e^4}{x (2+x)^2}-\log (7)\right )\right ) \] Output:
exp(x)*((exp(2)^2/(2+x)^2/x-ln(7))*x+2)-3
Time = 5.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=-\frac {e^x \left (-8-e^4+x^3 \log (7)+x (-8+\log (2401))+x^2 (-2+\log (2401))\right )}{(2+x)^2} \] Input:
Integrate[(E^x*(16 + 24*x + E^4*x + 12*x^2 + 2*x^3 + (-8 - 20*x - 18*x^2 - 7*x^3 - x^4)*Log[7]))/(8 + 12*x + 6*x^2 + x^3),x]
Output:
-((E^x*(-8 - E^4 + x^3*Log[7] + x*(-8 + Log[2401]) + x^2*(-2 + Log[2401])) )/(2 + x)^2)
Time = 0.64 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6, 2007, 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.
\(\displaystyle \int \frac {e^x \left (2 x^3+12 x^2+\left (-x^4-7 x^3-18 x^2-20 x-8\right ) \log (7)+e^4 x+24 x+16\right )}{x^3+6 x^2+12 x+8} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x \left (2 x^3+12 x^2+\left (-x^4-7 x^3-18 x^2-20 x-8\right ) \log (7)+\left (24+e^4\right ) x+16\right )}{x^3+6 x^2+12 x+8}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^x \left (2 x^3+12 x^2+\left (-x^4-7 x^3-18 x^2-20 x-8\right ) \log (7)+\left (24+e^4\right ) x+16\right )}{(x+2)^3}dx\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \int \left (\frac {e^{x+4}}{(x+2)^2}-\frac {2 e^{x+4}}{(x+2)^3}-e^x (x+2) \log (7)+e^x (2+\log (7))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{x+4}}{(x+2)^2}-e^x (x+2) \log (7)+e^x (2+\log (7))+e^x \log (7)\) |
Input:
Int[(E^x*(16 + 24*x + E^4*x + 12*x^2 + 2*x^3 + (-8 - 20*x - 18*x^2 - 7*x^3 - x^4)*Log[7]))/(8 + 12*x + 6*x^2 + x^3),x]
Output:
E^(4 + x)/(2 + x)^2 + E^x*Log[7] - E^x*(2 + x)*Log[7] + E^x*(2 + Log[7])
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, 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 = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56
method | result | size |
risch | \(-\frac {\left (\ln \left (7\right ) x^{3}+4 x^{2} \ln \left (7\right )+4 \ln \left (7\right ) x -{\mathrm e}^{4}-2 x^{2}-8 x -8\right ) {\mathrm e}^{x}}{\left (2+x \right )^{2}}\) | \(42\) |
norman | \(\frac {\left ({\mathrm e}^{4}+8\right ) {\mathrm e}^{x}+\left (-4 \ln \left (7\right )+2\right ) x^{2} {\mathrm e}^{x}+\left (-4 \ln \left (7\right )+8\right ) x \,{\mathrm e}^{x}-\ln \left (7\right ) x^{3} {\mathrm e}^{x}}{\left (2+x \right )^{2}}\) | \(48\) |
gosper | \(-\frac {\left (\ln \left (7\right ) x^{3}+4 x^{2} \ln \left (7\right )+4 \ln \left (7\right ) x -{\mathrm e}^{4}-2 x^{2}-8 x -8\right ) {\mathrm e}^{x}}{x^{2}+4 x +4}\) | \(49\) |
parallelrisch | \(-\frac {\ln \left (7\right ) x^{3} {\mathrm e}^{x}+4 \ln \left (7\right ) x^{2} {\mathrm e}^{x}+4 \ln \left (7\right ) x \,{\mathrm e}^{x}-{\mathrm e}^{4} {\mathrm e}^{x}-2 \,{\mathrm e}^{x} x^{2}-8 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}}{x^{2}+4 x +4}\) | \(62\) |
default | \(\frac {{\mathrm e}^{4} {\mathrm e}^{x}}{\left (2+x \right )^{2}}+2 \,{\mathrm e}^{x}-8 \ln \left (7\right ) \left (-\frac {{\mathrm e}^{x}}{2 \left (2+x \right )^{2}}-\frac {{\mathrm e}^{x}}{2 \left (2+x \right )}-\frac {{\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-x \right )}{2}\right )-18 \ln \left (7\right ) \left (-\frac {2 \,{\mathrm e}^{x}}{\left (2+x \right )^{2}}+\frac {2 \,{\mathrm e}^{x}}{2+x}+{\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-x \right )\right )-\frac {20 \ln \left (7\right ) {\mathrm e}^{x}}{\left (2+x \right )^{2}}-7 \ln \left (7\right ) \left ({\mathrm e}^{x}+\frac {4 \,{\mathrm e}^{x}}{\left (2+x \right )^{2}}-\frac {8 \,{\mathrm e}^{x}}{2+x}-2 \,{\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-x \right )\right )-\ln \left (7\right ) \left ({\mathrm e}^{x} x -7 \,{\mathrm e}^{x}-\frac {8 \,{\mathrm e}^{x}}{\left (2+x \right )^{2}}+\frac {24 \,{\mathrm e}^{x}}{2+x}\right )\) | \(163\) |
Input:
int(((-x^4-7*x^3-18*x^2-20*x-8)*ln(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*exp (x)/(x^3+6*x^2+12*x+8),x,method=_RETURNVERBOSE)
Output:
-(ln(7)*x^3+4*x^2*ln(7)+4*ln(7)*x-exp(4)-2*x^2-8*x-8)/(2+x)^2*exp(x)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {{\left (2 \, x^{2} - {\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \log \left (7\right ) + 8 \, x + e^{4} + 8\right )} e^{x}}{x^{2} + 4 \, x + 4} \] Input:
integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+ 16)*exp(x)/(x^3+6*x^2+12*x+8),x, algorithm="fricas")
Output:
(2*x^2 - (x^3 + 4*x^2 + 4*x)*log(7) + 8*x + e^4 + 8)*e^x/(x^2 + 4*x + 4)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).
Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {\left (- x^{3} \log {\left (7 \right )} - 4 x^{2} \log {\left (7 \right )} + 2 x^{2} - 4 x \log {\left (7 \right )} + 8 x + 8 + e^{4}\right ) e^{x}}{x^{2} + 4 x + 4} \] Input:
integrate(((-x**4-7*x**3-18*x**2-20*x-8)*ln(7)+x*exp(2)**2+2*x**3+12*x**2+ 24*x+16)*exp(x)/(x**3+6*x**2+12*x+8),x)
Output:
(-x**3*log(7) - 4*x**2*log(7) + 2*x**2 - 4*x*log(7) + 8*x + 8 + exp(4))*ex p(x)/(x**2 + 4*x + 4)
\[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=\int { \frac {{\left (2 \, x^{3} + 12 \, x^{2} + x e^{4} - {\left (x^{4} + 7 \, x^{3} + 18 \, x^{2} + 20 \, x + 8\right )} \log \left (7\right ) + 24 \, x + 16\right )} e^{x}}{x^{3} + 6 \, x^{2} + 12 \, x + 8} \,d x } \] Input:
integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+ 16)*exp(x)/(x^3+6*x^2+12*x+8),x, algorithm="maxima")
Output:
-(x^4*log(7) + 2*x^3*(3*log(7) - 1) + 12*x^2*(log(7) - 1) - x*(e^4 + 12*lo g(7)))*e^x/(x^3 + 6*x^2 + 12*x + 8) - 20*e^x*log(7)/(x^2 + 4*x + 4) + 8*e^ (-2)*exp_integral_e(3, -x - 2)*log(7)/(x + 2)^2 + 24*e^x/(x^2 + 4*x + 4) - 16*e^(-2)*exp_integral_e(3, -x - 2)/(x + 2)^2 + integrate(2*(x*(e^4 + 24* log(7) - 24) - e^4 - 12*log(7))*e^x/(x^4 + 8*x^3 + 24*x^2 + 32*x + 16), x)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=-\frac {x^{3} e^{x} \log \left (7\right ) + 4 \, x^{2} e^{x} \log \left (7\right ) - 2 \, x^{2} e^{x} + 4 \, x e^{x} \log \left (7\right ) - 8 \, x e^{x} - e^{\left (x + 4\right )} - 8 \, e^{x}}{x^{2} + 4 \, x + 4} \] Input:
integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+ 16)*exp(x)/(x^3+6*x^2+12*x+8),x, algorithm="giac")
Output:
-(x^3*e^x*log(7) + 4*x^2*e^x*log(7) - 2*x^2*e^x + 4*x*e^x*log(7) - 8*x*e^x - e^(x + 4) - 8*e^x)/(x^2 + 4*x + 4)
Time = 2.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {{\mathrm {e}}^{x+4}}{{\left (x+2\right )}^2}-{\mathrm {e}}^x\,\left (x\,\ln \left (7\right )-2\right ) \] Input:
int((exp(x)*(24*x + x*exp(4) - log(7)*(20*x + 18*x^2 + 7*x^3 + x^4 + 8) + 12*x^2 + 2*x^3 + 16))/(12*x + 6*x^2 + x^3 + 8),x)
Output:
exp(x + 4)/(x + 2)^2 - exp(x)*(x*log(7) - 2)
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {e^x \left (16+24 x+e^4 x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {e^{x} \left (-\mathrm {log}\left (7\right ) x^{3}-4 x^{2} \mathrm {log}\left (7\right )-4 \,\mathrm {log}\left (7\right ) x +e^{4}+2 x^{2}+8 x +8\right )}{x^{2}+4 x +4} \] Input:
int(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*ex p(x)/(x^3+6*x^2+12*x+8),x)
Output:
(e**x*( - log(7)*x**3 - 4*log(7)*x**2 - 4*log(7)*x + e**4 + 2*x**2 + 8*x + 8))/(x**2 + 4*x + 4)