Integrand size = 121, antiderivative size = 33 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\frac {x \left (2 x+x^2+\frac {x}{1+2 x}\right )}{4 \log \left (x^2\right )}}}{x^2} \] Output:
exp(1/4*x/ln(x^2)*(x/(1+2*x)+x^2+2*x))/x^2
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\frac {x^2 \left (3+5 x+2 x^2\right )}{4 (1+2 x) \log \left (x^2\right )}}}{x^2} \] Input:
Integrate[(E^((3*x^2 + 5*x^3 + 2*x^4)/((4 + 8*x)*Log[x^2]))*(-6*x^2 - 22*x ^3 - 24*x^4 - 8*x^5 + (6*x^2 + 21*x^3 + 28*x^4 + 12*x^5)*Log[x^2] + (-8 - 32*x - 32*x^2)*Log[x^2]^2))/((4*x^3 + 16*x^4 + 16*x^5)*Log[x^2]^2),x]
Output:
E^((x^2*(3 + 5*x + 2*x^2))/(4*(1 + 2*x)*Log[x^2]))/x^2
Leaf count is larger than twice the leaf count of optimal. \(197\) vs. \(2(33)=66\).
Time = 1.36 (sec) , antiderivative size = 197, normalized size of antiderivative = 5.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2026, 2007, 2726}
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.
\(\displaystyle \int \frac {e^{\frac {2 x^4+5 x^3+3 x^2}{(8 x+4) \log \left (x^2\right )}} \left (-8 x^5-24 x^4-22 x^3-6 x^2+\left (-32 x^2-32 x-8\right ) \log ^2\left (x^2\right )+\left (12 x^5+28 x^4+21 x^3+6 x^2\right ) \log \left (x^2\right )\right )}{\left (16 x^5+16 x^4+4 x^3\right ) \log ^2\left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {2 x^4+5 x^3+3 x^2}{(8 x+4) \log \left (x^2\right )}} \left (-8 x^5-24 x^4-22 x^3-6 x^2+\left (-32 x^2-32 x-8\right ) \log ^2\left (x^2\right )+\left (12 x^5+28 x^4+21 x^3+6 x^2\right ) \log \left (x^2\right )\right )}{x^3 \left (16 x^2+16 x+4\right ) \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {2 x^4+5 x^3+3 x^2}{(8 x+4) \log \left (x^2\right )}} \left (-8 x^5-24 x^4-22 x^3-6 x^2+\left (-32 x^2-32 x-8\right ) \log ^2\left (x^2\right )+\left (12 x^5+28 x^4+21 x^3+6 x^2\right ) \log \left (x^2\right )\right )}{x^3 (4 x+2)^2 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {\left (8 x^5+24 x^4+22 x^3+6 x^2-\left (12 x^5+28 x^4+21 x^3+6 x^2\right ) \log \left (x^2\right )\right ) \exp \left (\frac {2 x^4+5 x^3+3 x^2}{4 (2 x+1) \log \left (x^2\right )}\right )}{x^3 (2 x+1)^2 \left (-\frac {8 x^3+15 x^2+6 x}{(2 x+1) \log \left (x^2\right )}+\frac {2 \left (2 x^4+5 x^3+3 x^2\right )}{x (2 x+1) \log ^2\left (x^2\right )}+\frac {2 \left (2 x^4+5 x^3+3 x^2\right )}{(2 x+1)^2 \log \left (x^2\right )}\right ) \log ^2\left (x^2\right )}\) |
Input:
Int[(E^((3*x^2 + 5*x^3 + 2*x^4)/((4 + 8*x)*Log[x^2]))*(-6*x^2 - 22*x^3 - 2 4*x^4 - 8*x^5 + (6*x^2 + 21*x^3 + 28*x^4 + 12*x^5)*Log[x^2] + (-8 - 32*x - 32*x^2)*Log[x^2]^2))/((4*x^3 + 16*x^4 + 16*x^5)*Log[x^2]^2),x]
Output:
(E^((3*x^2 + 5*x^3 + 2*x^4)/(4*(1 + 2*x)*Log[x^2]))*(6*x^2 + 22*x^3 + 24*x ^4 + 8*x^5 - (6*x^2 + 21*x^3 + 28*x^4 + 12*x^5)*Log[x^2]))/(x^3*(1 + 2*x)^ 2*((2*(3*x^2 + 5*x^3 + 2*x^4))/(x*(1 + 2*x)*Log[x^2]^2) - (6*x + 15*x^2 + 8*x^3)/((1 + 2*x)*Log[x^2]) + (2*(3*x^2 + 5*x^3 + 2*x^4))/((1 + 2*x)^2*Log [x^2]))*Log[x^2]^2)
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[(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[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 1.87 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (3+2 x \right ) \left (1+x \right )}{4 \left (1+2 x \right ) \ln \left (x^{2}\right )}}}{x^{2}}\) | \(32\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (2 x^{2}+5 x +3\right )}{\left (8 x +4\right ) \ln \left (x^{2}\right )}}}{x^{2}}\) | \(33\) |
Input:
int(((-32*x^2-32*x-8)*ln(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*ln(x^2)-8*x^5 -24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x^3+3*x^2)/(8*x+4)/ln(x^2))/(16*x^5+16* x^4+4*x^3)/ln(x^2)^2,x,method=_RETURNVERBOSE)
Output:
1/x^2*exp(1/4*x^2*(3+2*x)*(1+x)/(1+2*x)/ln(x^2))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\left (\frac {2 \, x^{4} + 5 \, x^{3} + 3 \, x^{2}}{4 \, {\left (2 \, x + 1\right )} \log \left (x^{2}\right )}\right )}}{x^{2}} \] Input:
integrate(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^ 2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(1 6*x^5+16*x^4+4*x^3)/log(x^2)^2,x, algorithm="fricas")
Output:
e^(1/4*(2*x^4 + 5*x^3 + 3*x^2)/((2*x + 1)*log(x^2)))/x^2
Exception generated. \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((-32*x**2-32*x-8)*ln(x**2)**2+(12*x**5+28*x**4+21*x**3+6*x**2)* ln(x**2)-8*x**5-24*x**4-22*x**3-6*x**2)*exp((2*x**4+5*x**3+3*x**2)/(8*x+4) /ln(x**2))/(16*x**5+16*x**4+4*x**3)/ln(x**2)**2,x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Exception generated. \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^ 2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(1 6*x^5+16*x^4+4*x^3)/log(x^2)^2,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\left (\frac {2 \, x^{4} + 5 \, x^{3} + 3 \, x^{2}}{4 \, {\left (2 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )\right )}}\right )}}{x^{2}} \] Input:
integrate(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^ 2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(1 6*x^5+16*x^4+4*x^3)/log(x^2)^2,x, algorithm="giac")
Output:
e^(1/4*(2*x^4 + 5*x^3 + 3*x^2)/(2*x*log(x^2) + log(x^2)))/x^2
Time = 8.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.09 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{\frac {x^4}{2\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}\,{\mathrm {e}}^{\frac {3\,x^2}{4\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}\,{\mathrm {e}}^{\frac {5\,x^3}{4\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}}{x^2} \] Input:
int(-(exp((3*x^2 + 5*x^3 + 2*x^4)/(log(x^2)*(8*x + 4)))*(log(x^2)^2*(32*x + 32*x^2 + 8) + 6*x^2 + 22*x^3 + 24*x^4 + 8*x^5 - log(x^2)*(6*x^2 + 21*x^3 + 28*x^4 + 12*x^5)))/(log(x^2)^2*(4*x^3 + 16*x^4 + 16*x^5)),x)
Output:
(exp(x^4/(2*(log(x^2) + 2*x*log(x^2))))*exp((3*x^2)/(4*(log(x^2) + 2*x*log (x^2))))*exp((5*x^3)/(4*(log(x^2) + 2*x*log(x^2)))))/x^2
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{\left (4 x^3+16 x^4+16 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\frac {2 x^{4}+5 x^{3}+3 x^{2}}{8 \,\mathrm {log}\left (x^{2}\right ) x +4 \,\mathrm {log}\left (x^{2}\right )}}}{x^{2}} \] Input:
int(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^2)-8*x ^5-24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(16*x^5+ 16*x^4+4*x^3)/log(x^2)^2,x)
Output:
e**((2*x**4 + 5*x**3 + 3*x**2)/(8*log(x**2)*x + 4*log(x**2)))/x**2