Integrand size = 50, antiderivative size = 23 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=3+\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \] Output:
3+12*exp(1+exp(1))/(x^2-4*x*ln(2)^2+5)
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \] Input:
Integrate[(E^(1 + E)*(-24*x + 12*Log[4]^2))/(25 + 10*x^2 + x^4 + (-10*x - 2*x^3)*Log[4]^2 + x^2*Log[4]^4),x]
Output:
(12*E^(1 + E))/(5 + x^2 - x*Log[4]^2)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6, 27, 27, 2459, 27, 27, 1380, 27, 241}
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.
\(\displaystyle \int \frac {e^{1+e} \left (12 \log ^2(4)-24 x\right )}{x^4+\left (-2 x^3-10 x\right ) \log ^2(4)+10 x^2+x^2 \log ^4(4)+25} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{1+e} \left (12 \log ^2(4)-24 x\right )}{x^4+\left (-2 x^3-10 x\right ) \log ^2(4)+x^2 \left (10+\log ^4(4)\right )+25}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{1+e} \int -\frac {12 \left (2 x-\log ^2(4)\right )}{x^4+\left (10+\log ^4(4)\right ) x^2-2 \left (x^3+5 x\right ) \log ^2(4)+25}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -12 e^{1+e} \int \frac {2 x-\log ^2(4)}{x^4+\left (10+\log ^4(4)\right ) x^2-2 \left (x^3+5 x\right ) \log ^2(4)+25}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle -12 e^{1+e} \int \frac {2 \left (x-\frac {\log ^2(4)}{2}\right )}{\left (x-\frac {\log ^2(4)}{2}\right )^4+\frac {1}{2} \left (20-\log ^4(4)\right ) \left (x-\frac {\log ^2(4)}{2}\right )^2+\frac {1}{16} \left (20-\log ^4(4)\right )^2}d\left (x-\frac {\log ^2(4)}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -24 e^{1+e} \int \frac {16 \left (x-\frac {\log ^2(4)}{2}\right )}{16 \left (x-\frac {\log ^2(4)}{2}\right )^4+8 \left (20-\log ^4(4)\right ) \left (x-\frac {\log ^2(4)}{2}\right )^2+\left (-20+\log ^4(4)\right )^2}d\left (x-\frac {\log ^2(4)}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -384 e^{1+e} \int \frac {x-\frac {\log ^2(4)}{2}}{16 \left (x-\frac {\log ^2(4)}{2}\right )^4+8 \left (20-\log ^4(4)\right ) \left (x-\frac {\log ^2(4)}{2}\right )^2+\left (-20+\log ^4(4)\right )^2}d\left (x-\frac {\log ^2(4)}{2}\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -6144 e^{1+e} \int \frac {x-\frac {\log ^2(4)}{2}}{16 \left (4 \left (x-\frac {\log ^2(4)}{2}\right )^2-\log ^4(4)+20\right )^2}d\left (x-\frac {\log ^2(4)}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -384 e^{1+e} \int \frac {x-\frac {\log ^2(4)}{2}}{\left (4 \left (x-\frac {\log ^2(4)}{2}\right )^2-\log ^4(4)+20\right )^2}d\left (x-\frac {\log ^2(4)}{2}\right )\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {48 e^{1+e}}{4 \left (x-\frac {\log ^2(4)}{2}\right )^2+20-\log ^4(4)}\) |
Input:
Int[(E^(1 + E)*(-24*x + 12*Log[4]^2))/(25 + 10*x^2 + x^4 + (-10*x - 2*x^3) *Log[4]^2 + x^2*Log[4]^4),x]
Output:
(48*E^(1 + E))/(20 - Log[4]^4 + 4*(x - Log[4]^2/2)^2)
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {3 \,{\mathrm e}^{1+{\mathrm e}}}{x \ln \left (2\right )^{2}-\frac {x^{2}}{4}-\frac {5}{4}}\) | \(23\) |
gosper | \(-\frac {12 \,{\mathrm e}^{1+{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
norman | \(-\frac {12 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
parallelrisch | \(-\frac {12 \,{\mathrm e}^{1+{\mathrm e}}}{4 x \ln \left (2\right )^{2}-x^{2}-5}\) | \(24\) |
default | \(\frac {24 \,{\mathrm e}^{1+{\mathrm e}} \left (10-8 \ln \left (2\right )^{4}\right )}{\left (20-16 \ln \left (2\right )^{4}\right ) \left (x^{2}-4 x \ln \left (2\right )^{2}+5\right )}\) | \(40\) |
Input:
int((48*ln(2)^2-24*x)*exp(1+exp(1))/(16*x^2*ln(2)^4+4*(-2*x^3-10*x)*ln(2)^ 2+x^4+10*x^2+25),x,method=_RETURNVERBOSE)
Output:
-3/(x*ln(2)^2-1/4*x^2-5/4)*exp(1+exp(1))
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \] Input:
integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x )*log(2)^2+x^4+10*x^2+25),x, algorithm="fricas")
Output:
-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12 e e^{e}}{x^{2} - 4 x \log {\left (2 \right )}^{2} + 5} \] Input:
integrate((48*ln(2)**2-24*x)*exp(1+exp(1))/(16*x**2*ln(2)**4+4*(-2*x**3-10 *x)*ln(2)**2+x**4+10*x**2+25),x)
Output:
12*E*exp(E)/(x**2 - 4*x*log(2)**2 + 5)
Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \] Input:
integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x )*log(2)^2+x^4+10*x^2+25),x, algorithm="maxima")
Output:
-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \left (2\right )^{2} - x^{2} - 5} \] Input:
integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x )*log(2)^2+x^4+10*x^2+25),x, algorithm="giac")
Output:
-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)
Time = 3.93 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=\frac {12\,{\mathrm {e}}^{\mathrm {e}+1}}{x^2-4\,{\ln \left (2\right )}^2\,x+5} \] Input:
int(-(exp(exp(1) + 1)*(24*x - 48*log(2)^2))/(16*x^2*log(2)^4 - 4*log(2)^2* (10*x + 2*x^3) + 10*x^2 + x^4 + 25),x)
Output:
(12*exp(exp(1) + 1))/(x^2 - 4*x*log(2)^2 + 5)
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+10 x^2+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \log ^4(4)} \, dx=-\frac {12 e^{e} e}{4 \mathrm {log}\left (2\right )^{2} x -x^{2}-5} \] Input:
int((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x)*log( 2)^2+x^4+10*x^2+25),x)
Output:
( - 12*e**e*e)/(4*log(2)**2*x - x**2 - 5)