Integrand size = 95, antiderivative size = 25 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {4 e^5}{x-\frac {(-3+\log (2))^2}{4 e x^3}} \] Output:
4/(x-1/4*(ln(2)-3)^2/x^3/exp(1))*exp(5)
Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(25)=50\).
Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {4 e^6 x^3 \left (36-18 \log (2)+4 \log ^2(2)-\log (64)\right )}{\left (9+\log ^2(2)-\log (64)\right ) \left (-9+4 e x^4-\log ^2(2)+\log (64)\right )} \] Input:
Integrate[(E^5*(-432*E*x^2 - 64*E^2*x^6) + 288*E^6*x^2*Log[2] - 48*E^6*x^2 *Log[2]^2)/(81 - 72*E*x^4 + 16*E^2*x^8 + (-108 + 48*E*x^4)*Log[2] + (54 - 8*E*x^4)*Log[2]^2 - 12*Log[2]^3 + Log[2]^4),x]
Output:
(4*E^6*x^3*(36 - 18*Log[2] + 4*Log[2]^2 - Log[64]))/((9 + Log[2]^2 - Log[6 4])*(-9 + 4*E*x^4 - Log[2]^2 + Log[64]))
Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 2454, 1380, 6, 25, 2021}
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.
\(\displaystyle \int \frac {-48 e^6 x^2 \log ^2(2)+288 e^6 x^2 \log (2)+e^5 \left (-64 e^2 x^6-432 e x^2\right )}{16 e^2 x^8-72 e x^4+\left (54-8 e x^4\right ) \log ^2(2)+\left (48 e x^4-108\right ) \log (2)+81+\log ^4(2)-12 \log ^3(2)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^6 x^2 \left (288 \log (2)-48 \log ^2(2)\right )+e^5 \left (-64 e^2 x^6-432 e x^2\right )}{16 e^2 x^8-72 e x^4+\left (54-8 e x^4\right ) \log ^2(2)+\left (48 e x^4-108\right ) \log (2)+81+\log ^4(2)-12 \log ^3(2)}dx\) |
\(\Big \downarrow \) 2454 |
\(\displaystyle \int \frac {e^6 x^2 \left (288 \log (2)-48 \log ^2(2)\right )+e^5 \left (-64 e^2 x^6-432 e x^2\right )}{16 e^2 x^8-8 e x^4 (3-\log (2))^2+(\log (2)-3)^4}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 16 e^2 \int -\frac {4 e^7 x^6-3 e^6 (6-\log (2)) \log (2) x^2+27 e^6 x^2}{e^2 \left (4 e x^4-(3-\log (2))^2\right )^2}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle 16 e^2 \int -\frac {4 e^7 x^6+e^6 (27-3 (6-\log (2)) \log (2)) x^2}{e^2 \left (4 e x^4-(3-\log (2))^2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -16 e^2 \int \frac {4 e^5 x^6+3 e^4 (3-\log (2))^2 x^2}{\left (4 e x^4-(3-\log (2))^2\right )^2}dx\) |
\(\Big \downarrow \) 2021 |
\(\displaystyle \frac {16 e^6 x^3}{4 e x^4-(3-\log (2))^2}\) |
Input:
Int[(E^5*(-432*E*x^2 - 64*E^2*x^6) + 288*E^6*x^2*Log[2] - 48*E^6*x^2*Log[2 ]^2)/(81 - 72*E*x^4 + 16*E^2*x^8 + (-108 + 48*E*x^4)*Log[2] + (54 - 8*E*x^ 4)*Log[2]^2 - 12*Log[2]^3 + Log[2]^4),x]
Output:
(16*E^6*x^3)/(4*E*x^4 - (3 - Log[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[(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[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x ]}, Simp[Coeff[Pp, x, p]*x^(p - q + 1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x, q]*Pp , Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; Free Q[m, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && NeQ[m, -1]
Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[ p, x] && PolyQ[Pq, x] && TrinomialQ[u, x] && !TrinomialMatchQ[u, x]
Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{6} x^{3}}{x^{4} {\mathrm e}-\frac {\ln \left (2\right )^{2}}{4}+\frac {3 \ln \left (2\right )}{2}-\frac {9}{4}}\) | \(28\) |
gosper | \(\frac {16 x^{3} {\mathrm e}^{6}}{4 x^{4} {\mathrm e}-\ln \left (2\right )^{2}+6 \ln \left (2\right )-9}\) | \(29\) |
norman | \(\frac {16 x^{3} {\mathrm e}^{5} {\mathrm e}}{4 x^{4} {\mathrm e}-\ln \left (2\right )^{2}+6 \ln \left (2\right )-9}\) | \(31\) |
parallelrisch | \(\frac {16 x^{3} {\mathrm e}^{5} {\mathrm e}}{4 x^{4} {\mathrm e}-\ln \left (2\right )^{2}+6 \ln \left (2\right )-9}\) | \(31\) |
Input:
int((-48*x^2*exp(1)*exp(5)*ln(2)^2+288*x^2*exp(1)*exp(5)*ln(2)+(-64*x^6*ex p(1)^2-432*x^2*exp(1))*exp(5))/(ln(2)^4-12*ln(2)^3+(-8*x^4*exp(1)+54)*ln(2 )^2+(48*x^4*exp(1)-108)*ln(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81),x,method=_ RETURNVERBOSE)
Output:
4*exp(6)*x^3/(x^4*exp(1)-1/4*ln(2)^2+3/2*ln(2)-9/4)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {16 \, x^{3} e^{6}}{4 \, x^{4} e - \log \left (2\right )^{2} + 6 \, \log \left (2\right ) - 9} \] Input:
integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-6 4*x^6*exp(1)^2-432*x^2*exp(1))*exp(5))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1 )+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81 ),x, algorithm="fricas")
Output:
16*x^3*e^6/(4*x^4*e - log(2)^2 + 6*log(2) - 9)
Time = 0.85 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {16 x^{3} e^{6}}{4 e x^{4} - 9 - \log {\left (2 \right )}^{2} + 6 \log {\left (2 \right )}} \] Input:
integrate((-48*x**2*exp(1)*exp(5)*ln(2)**2+288*x**2*exp(1)*exp(5)*ln(2)+(- 64*x**6*exp(1)**2-432*x**2*exp(1))*exp(5))/(ln(2)**4-12*ln(2)**3+(-8*x**4* exp(1)+54)*ln(2)**2+(48*x**4*exp(1)-108)*ln(2)+16*x**8*exp(1)**2-72*x**4*e xp(1)+81),x)
Output:
16*x**3*exp(6)/(4*E*x**4 - 9 - log(2)**2 + 6*log(2))
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {16 \, x^{3} e^{6}}{4 \, x^{4} e - \log \left (2\right )^{2} + 6 \, \log \left (2\right ) - 9} \] Input:
integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-6 4*x^6*exp(1)^2-432*x^2*exp(1))*exp(5))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1 )+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81 ),x, algorithm="maxima")
Output:
16*x^3*e^6/(4*x^4*e - log(2)^2 + 6*log(2) - 9)
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {16 \, x^{3} e^{6}}{4 \, x^{4} e - \log \left (2\right )^{2} + 6 \, \log \left (2\right ) - 9} \] Input:
integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-6 4*x^6*exp(1)^2-432*x^2*exp(1))*exp(5))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1 )+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81 ),x, algorithm="giac")
Output:
16*x^3*e^6/(4*x^4*e - log(2)^2 + 6*log(2) - 9)
Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=\frac {16\,x^3\,{\mathrm {e}}^6}{4\,\mathrm {e}\,x^4+\ln \left (64\right )-{\ln \left (2\right )}^2-9} \] Input:
int(-(exp(5)*(432*x^2*exp(1) + 64*x^6*exp(2)) - 288*x^2*exp(6)*log(2) + 48 *x^2*exp(6)*log(2)^2)/(16*x^8*exp(2) - 72*x^4*exp(1) - log(2)^2*(8*x^4*exp (1) - 54) + log(2)*(48*x^4*exp(1) - 108) - 12*log(2)^3 + log(2)^4 + 81),x)
Output:
(16*x^3*exp(6))/(log(64) + 4*x^4*exp(1) - log(2)^2 - 9)
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx=-\frac {16 e^{6} x^{3}}{\mathrm {log}\left (2\right )^{2}-6 \,\mathrm {log}\left (2\right )-4 e \,x^{4}+9} \] Input:
int((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-64*x^6* exp(1)^2-432*x^2*exp(1))*exp(5))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1)+54)* log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81),x)
Output:
( - 16*e**6*x**3)/(log(2)**2 - 6*log(2) - 4*e*x**4 + 9)