Integrand size = 101, antiderivative size = 25 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 e^{-\frac {3 x^2}{-4 e^x x+x \log (4)}}}{x} \] Output:
4/x/exp(3*x^2/(2*x*ln(2)-4*exp(x+ln(x))))
Time = 1.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 e^{\frac {3 x}{4 e^x-\log (4)}}}{x} \] Input:
Integrate[(E^((3*x^2)/(4*E^x*x - x*Log[4]))*(-64*E^(2*x)*x^2 - 12*x^3*Log[ 4] - 4*x^2*Log[4]^2 + E^x*x*(48*x^2 - 48*x^3 + 32*x*Log[4])))/(16*E^(2*x)* x^4 - 8*E^x*x^4*Log[4] + x^4*Log[4]^2),x]
Output:
(4*E^((3*x)/(4*E^x - Log[4])))/x
Leaf count is larger than twice the leaf count of optimal. \(112\) vs. \(2(25)=50\).
Time = 2.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7292, 27, 25, 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.
\(\displaystyle \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-12 x^3 \log (4)-64 e^{2 x} x^2-4 x^2 \log ^2(4)+e^x x \left (-48 x^3+48 x^2+32 x \log (4)\right )\right )}{16 e^{2 x} x^4+x^4 \log ^2(4)-8 e^x x^4 \log (4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-12 e^x x^2+12 e^x x-16 e^{2 x}-3 x \log (4)+8 e^x \log (4)-\log ^2(4)\right )}{x^2 \left (4 e^x-\log (4)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (12 e^x x^2-12 e^x x+3 \log (4) x+16 e^{2 x}+\log ^2(4)-8 e^x \log (4)\right )}{x^2 \left (4 e^x-\log (4)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (12 e^x x^2-12 e^x x+3 \log (4) x+16 e^{2 x}+\log ^2(4)-8 e^x \log (4)\right )}{x^2 \left (4 e^x-\log (4)\right )^2}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {4 e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-4 e^x x^2+4 e^x x-x \log (4)\right )}{x^2 \left (4 e^x-\log (4)\right )^2 \left (\frac {x^2 \left (4 e^x x+4 e^x-\log (4)\right )}{\left (4 e^x x-x \log (4)\right )^2}-\frac {2 x}{4 e^x x-x \log (4)}\right )}\) |
Input:
Int[(E^((3*x^2)/(4*E^x*x - x*Log[4]))*(-64*E^(2*x)*x^2 - 12*x^3*Log[4] - 4 *x^2*Log[4]^2 + E^x*x*(48*x^2 - 48*x^3 + 32*x*Log[4])))/(16*E^(2*x)*x^4 - 8*E^x*x^4*Log[4] + x^4*Log[4]^2),x]
Output:
(-4*E^((3*x^2)/(4*E^x*x - x*Log[4]))*(4*E^x*x - 4*E^x*x^2 - x*Log[4]))/(x^ 2*(4*E^x - Log[4])^2*((x^2*(4*E^x + 4*E^x*x - Log[4]))/(4*E^x*x - x*Log[4] )^2 - (2*x)/(4*E^x*x - x*Log[4])))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 = 11.53 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{\frac {3 x}{2 \left (-\ln \left (2\right )+2 \,{\mathrm e}^{x}\right )}}}{x}\) | \(21\) |
parallelrisch | \(\frac {\left (64 x \ln \left (2\right )-128 \,{\mathrm e}^{x +\ln \left (x \right )}\right ) {\mathrm e}^{\frac {3 x^{2}}{2 \left (-x \ln \left (2\right )+2 \,{\mathrm e}^{x} x \right )}}}{16 x \left (x \ln \left (2\right )-2 \,{\mathrm e}^{x +\ln \left (x \right )}\right )}\) | \(56\) |
Input:
int((-64*exp(x+ln(x))^2+(64*x*ln(2)-48*x^3+48*x^2)*exp(x+ln(x))-16*x^2*ln( 2)^2-24*x^3*ln(2))/(16*x^2*exp(x+ln(x))^2-16*x^3*ln(2)*exp(x+ln(x))+4*x^4* ln(2)^2)/exp(-3*x^2/(4*exp(x+ln(x))-2*x*ln(2))),x,method=_RETURNVERBOSE)
Output:
4/x*exp(3/2*x/(-ln(2)+2*exp(x)))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 \, e^{\left (-\frac {3 \, x^{2}}{2 \, {\left (x \log \left (2\right ) - 2 \, e^{\left (x + \log \left (x\right )\right )}\right )}}\right )}}{x} \] Input:
integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-1 6*x^2*log(2)^2-24*x^3*log(2))/(16*x^2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+ log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algori thm="fricas")
Output:
4*e^(-3/2*x^2/(x*log(2) - 2*e^(x + log(x))))/x
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 e^{\frac {3 x^{2}}{4 x e^{x} - 2 x \log {\left (2 \right )}}}}{x} \] Input:
integrate((-64*exp(x+ln(x))**2+(64*x*ln(2)-48*x**3+48*x**2)*exp(x+ln(x))-1 6*x**2*ln(2)**2-24*x**3*ln(2))/(16*x**2*exp(x+ln(x))**2-16*x**3*ln(2)*exp( x+ln(x))+4*x**4*ln(2)**2)/exp(-3*x**2/(4*exp(x+ln(x))-2*x*ln(2))),x)
Output:
4*exp(3*x**2/(4*x*exp(x) - 2*x*log(2)))/x
\[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\int { -\frac {2 \, {\left (3 \, x^{3} \log \left (2\right ) + 2 \, x^{2} \log \left (2\right )^{2} + 2 \, {\left (3 \, x^{3} - 3 \, x^{2} - 4 \, x \log \left (2\right )\right )} e^{\left (x + \log \left (x\right )\right )} + 8 \, e^{\left (2 \, x + 2 \, \log \left (x\right )\right )}\right )} e^{\left (-\frac {3 \, x^{2}}{2 \, {\left (x \log \left (2\right ) - 2 \, e^{\left (x + \log \left (x\right )\right )}\right )}}\right )}}{x^{4} \log \left (2\right )^{2} - 4 \, x^{3} e^{\left (x + \log \left (x\right )\right )} \log \left (2\right ) + 4 \, x^{2} e^{\left (2 \, x + 2 \, \log \left (x\right )\right )}} \,d x } \] Input:
integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-1 6*x^2*log(2)^2-24*x^3*log(2))/(16*x^2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+ log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algori thm="maxima")
Output:
-2*integrate(-(3*x^3*log(2) + 2*x^2*log(2)^2 + 8*x^2*e^(2*x) + 2*(3*x^3 - 3*x^2 - 4*x*log(2))*x*e^x)*e^(3/2*x^2/(2*x*e^x - x*log(2)))/(4*x^4*e^x*log (2) - x^4*log(2)^2 - 4*x^4*e^(2*x)), x)
Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 \, e^{\left (x - \frac {4 \, x e^{x} - 2 \, x \log \left (2\right ) - 3 \, x}{2 \, {\left (2 \, e^{x} - \log \left (2\right )\right )}}\right )}}{x} \] Input:
integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-1 6*x^2*log(2)^2-24*x^3*log(2))/(16*x^2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+ log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algori thm="giac")
Output:
4*e^(x - 1/2*(4*x*e^x - 2*x*log(2) - 3*x)/(2*e^x - log(2)))/x
Timed out. \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {3\,x^2}{4\,{\mathrm {e}}^{x+\ln \left (x\right )}-2\,x\,\ln \left (2\right )}}\,\left (64\,{\mathrm {e}}^{2\,x+2\,\ln \left (x\right )}+16\,x^2\,{\ln \left (2\right )}^2-{\mathrm {e}}^{x+\ln \left (x\right )}\,\left (-48\,x^3+48\,x^2+64\,\ln \left (2\right )\,x\right )+24\,x^3\,\ln \left (2\right )\right )}{4\,x^4\,{\ln \left (2\right )}^2+16\,x^2\,{\mathrm {e}}^{2\,x+2\,\ln \left (x\right )}-16\,x^3\,{\mathrm {e}}^{x+\ln \left (x\right )}\,\ln \left (2\right )} \,d x \] Input:
int(-(exp((3*x^2)/(4*exp(x + log(x)) - 2*x*log(2)))*(64*exp(2*x + 2*log(x) ) + 16*x^2*log(2)^2 - exp(x + log(x))*(64*x*log(2) + 48*x^2 - 48*x^3) + 24 *x^3*log(2)))/(4*x^4*log(2)^2 + 16*x^2*exp(2*x + 2*log(x)) - 16*x^3*exp(x + log(x))*log(2)),x)
Output:
int(-(exp((3*x^2)/(4*exp(x + log(x)) - 2*x*log(2)))*(64*exp(2*x + 2*log(x) ) + 16*x^2*log(2)^2 - exp(x + log(x))*(64*x*log(2) + 48*x^2 - 48*x^3) + 24 *x^3*log(2)))/(4*x^4*log(2)^2 + 16*x^2*exp(2*x + 2*log(x)) - 16*x^3*exp(x + log(x))*log(2)), x)
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx=\frac {4 e^{\frac {3 x}{4 e^{x}-2 \,\mathrm {log}\left (2\right )}}}{x} \] Input:
int((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-16*x^2* log(2)^2-24*x^3*log(2))/(16*x^2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+log(x) )+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x)
Output:
(4*e**((3*x)/(4*e**x - 2*log(2))))/x