Integrand size = 97, antiderivative size = 27 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^3}{4 \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )} \] Output:
exp(3)/(4*ln(2*x^3)+4*exp(x+9)+16*x^2+20)
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^3}{4 \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )} \] Input:
Integrate[(-(E^(12 + x)*x) + E^3*(-3 - 8*x^2))/(100*x + 4*E^(18 + 2*x)*x + 160*x^3 + 64*x^5 + E^(9 + x)*(40*x + 32*x^3) + (40*x + 8*E^(9 + x)*x + 32 *x^3)*Log[2*x^3] + 4*x*Log[2*x^3]^2),x]
Output:
E^3/(4*(5 + E^(9 + x) + 4*x^2 + Log[2*x^3]))
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {e^3 \left (-8 x^2-3\right )-e^{x+12} x}{64 x^5+160 x^3+e^{x+9} \left (32 x^3+40 x\right )+4 x \log ^2\left (2 x^3\right )+\left (32 x^3+8 e^{x+9} x+40 x\right ) \log \left (2 x^3\right )+4 e^{2 x+18} x+100 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^3 \left (-8 x^2-e^{x+9} x-3\right )}{4 x \left (\log \left (2 x^3\right )+4 x^2+e^{x+9}+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} e^3 \int -\frac {8 x^2+e^{x+9} x+3}{x \left (4 x^2+e^{x+9}+\log \left (2 x^3\right )+5\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} e^3 \int \frac {8 x^2+e^{x+9} x+3}{x \left (4 x^2+e^{x+9}+\log \left (2 x^3\right )+5\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {e^3}{4 \left (\log \left (2 x^3\right )+4 x^2+e^{x+9}+5\right )}\) |
Input:
Int[(-(E^(12 + x)*x) + E^3*(-3 - 8*x^2))/(100*x + 4*E^(18 + 2*x)*x + 160*x ^3 + 64*x^5 + E^(9 + x)*(40*x + 32*x^3) + (40*x + 8*E^(9 + x)*x + 32*x^3)* Log[2*x^3] + 4*x*Log[2*x^3]^2),x]
Output:
E^3/(4*(5 + E^(9 + x) + 4*x^2 + Log[2*x^3]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{3}}{4 \ln \left (2 x^{3}\right )+4 \,{\mathrm e}^{x +9}+16 x^{2}+20}\) | \(24\) |
risch | \(\frac {{\mathrm e}^{3}}{-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-2 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+4 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-2 i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+20+16 x^{2}+4 \ln \left (2\right )+4 \,{\mathrm e}^{x +9}+12 \ln \left (x \right )}\) | \(151\) |
Input:
int((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*ln(2*x^3)^2+(8*x*exp(x+9)+ 32*x^3+40*x)*ln(2*x^3)+4*x*exp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^5+160*x^ 3+100*x),x,method=_RETURNVERBOSE)
Output:
1/4*exp(3)/(4*x^2+exp(x+9)+ln(2*x^3)+5)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^{6}}{4 \, {\left ({\left (4 \, x^{2} + 5\right )} e^{3} + e^{3} \log \left (2 \, x^{3}\right ) + e^{\left (x + 12\right )}\right )}} \] Input:
integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*ex p(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*exp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^ 5+160*x^3+100*x),x, algorithm="fricas")
Output:
1/4*e^6/((4*x^2 + 5)*e^3 + e^3*log(2*x^3) + e^(x + 12))
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^{3}}{16 x^{2} + 4 e^{x + 9} + 4 \log {\left (2 x^{3} \right )} + 20} \] Input:
integrate((-x*exp(3)*exp(x+9)+(-8*x**2-3)*exp(3))/(4*x*ln(2*x**3)**2+(8*x* exp(x+9)+32*x**3+40*x)*ln(2*x**3)+4*x*exp(x+9)**2+(32*x**3+40*x)*exp(x+9)+ 64*x**5+160*x**3+100*x),x)
Output:
exp(3)/(16*x**2 + 4*exp(x + 9) + 4*log(2*x**3) + 20)
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^{3}}{4 \, {\left (4 \, x^{2} + e^{\left (x + 9\right )} + \log \left (2\right ) + 3 \, \log \left (x\right ) + 5\right )}} \] Input:
integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*ex p(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*exp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^ 5+160*x^3+100*x),x, algorithm="maxima")
Output:
1/4*e^3/(4*x^2 + e^(x + 9) + log(2) + 3*log(x) + 5)
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^{3}}{4 \, {\left (4 \, x^{2} + e^{\left (x + 9\right )} + \log \left (2\right ) + 3 \, \log \left (x\right ) + 5\right )}} \] Input:
integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*ex p(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*exp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^ 5+160*x^3+100*x),x, algorithm="giac")
Output:
1/4*e^3/(4*x^2 + e^(x + 9) + log(2) + 3*log(x) + 5)
Time = 4.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {{\mathrm {e}}^3}{4\,\left ({\mathrm {e}}^{x+9}+\ln \left (2\,x^3\right )+4\,x^2+5\right )} \] Input:
int(-(exp(3)*(8*x^2 + 3) + x*exp(x + 9)*exp(3))/(100*x + exp(x + 9)*(40*x + 32*x^3) + 4*x*exp(2*x + 18) + log(2*x^3)*(40*x + 8*x*exp(x + 9) + 32*x^3 ) + 160*x^3 + 64*x^5 + 4*x*log(2*x^3)^2),x)
Output:
exp(3)/(4*(exp(x + 9) + log(2*x^3) + 4*x^2 + 5))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-e^{12+x} x+e^3 \left (-3-8 x^2\right )}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} \left (40 x+32 x^3\right )+\left (40 x+8 e^{9+x} x+32 x^3\right ) \log \left (2 x^3\right )+4 x \log ^2\left (2 x^3\right )} \, dx=\frac {e^{3}}{4 e^{x} e^{9}+4 \,\mathrm {log}\left (2 x^{3}\right )+16 x^{2}+20} \] Input:
int((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*exp(x+9) +32*x^3+40*x)*log(2*x^3)+4*x*exp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^5+160* x^3+100*x),x)
Output:
e**3/(4*(e**x*e**9 + log(2*x**3) + 4*x**2 + 5))