Integrand size = 74, antiderivative size = 21 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {3}{-e+4 \left (-\frac {5}{32 x^2}+x\right )+\log (4)} \] Output:
-3/(2*ln(2)-exp(1)-5/8/x^2+4*x)
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {24 x^2}{-5-8 e x^2+32 x^3+8 x^2 \log (4)} \] Input:
Integrate[(240*x + 768*x^4)/(25 - 320*x^3 + 64*E^2*x^4 + 1024*x^6 + E*(80* x^2 - 512*x^5) + (-80*x^2 - 128*E*x^4 + 512*x^5)*Log[4] + 64*x^4*Log[4]^2) ,x]
Output:
(-24*x^2)/(-5 - 8*E*x^2 + 32*x^3 + 8*x^2*Log[4])
Time = 10.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 2027, 2462, 7239, 27, 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.
\(\displaystyle \int \frac {768 x^4+240 x}{1024 x^6+64 e^2 x^4+64 x^4 \log ^2(4)-320 x^3+e \left (80 x^2-512 x^5\right )+\left (512 x^5-128 e x^4-80 x^2\right ) \log (4)+25} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {768 x^4+240 x}{1024 x^6+64 x^4 \left (e^2+\log ^2(4)\right )-320 x^3+e \left (80 x^2-512 x^5\right )+\left (512 x^5-128 e x^4-80 x^2\right ) \log (4)+25}dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {x \left (768 x^3+240\right )}{1024 x^6+64 x^4 \left (e^2+\log ^2(4)\right )-320 x^3+e \left (80 x^2-512 x^5\right )+\left (512 x^5-128 e x^4-80 x^2\right ) \log (4)+25}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {6 \left (8 x^2 (e-\log (4))^2+60 x+5 (e-\log (4))\right )}{\left (-32 x^3+8 x^2 (e-\log (4))+5\right )^2}+\frac {6 (-4 x-e+\log (4))}{-32 x^3+8 x^2 (e-\log (4))+5}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {48 x \left (16 x^3+5\right )}{\left (-32 x^3+8 x^2 (e-\log (4))+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 48 \int \frac {x \left (16 x^3+5\right )}{\left (-32 x^3+8 (e-\log (4)) x^2+5\right )^2}dx\) |
\(\Big \downarrow \) 2021 |
\(\displaystyle \frac {24 x^2}{-32 x^3+8 x^2 (e-\log (4))+5}\) |
Input:
Int[(240*x + 768*x^4)/(25 - 320*x^3 + 64*E^2*x^4 + 1024*x^6 + E*(80*x^2 - 512*x^5) + (-80*x^2 - 128*E*x^4 + 512*x^5)*Log[4] + 64*x^4*Log[4]^2),x]
Output:
(24*x^2)/(5 - 32*x^3 + 8*x^2*(E - Log[4]))
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[(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[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33
method | result | size |
risch | \(-\frac {3 x^{2}}{2 \left (x^{2} \ln \left (2\right )-\frac {x^{2} {\mathrm e}}{2}+2 x^{3}-\frac {5}{16}\right )}\) | \(28\) |
gosper | \(-\frac {24 x^{2}}{16 x^{2} \ln \left (2\right )-8 x^{2} {\mathrm e}+32 x^{3}-5}\) | \(29\) |
norman | \(-\frac {24 x^{2}}{16 x^{2} \ln \left (2\right )-8 x^{2} {\mathrm e}+32 x^{3}-5}\) | \(29\) |
parallelrisch | \(-\frac {24 x^{2}}{16 x^{2} \ln \left (2\right )-8 x^{2} {\mathrm e}+32 x^{3}-5}\) | \(29\) |
default | \(\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25+1024 \textit {\_Z}^{6}+\left (1024 \ln \left (2\right )-512 \,{\mathrm e}\right ) \textit {\_Z}^{5}+\left (256 \ln \left (2\right )^{2}-256 \ln \left (2\right ) {\mathrm e}+64 \,{\mathrm e}^{2}\right ) \textit {\_Z}^{4}-320 \textit {\_Z}^{3}+\left (-160 \ln \left (2\right )+80 \,{\mathrm e}\right ) \textit {\_Z}^{2}\right )}{\sum }\frac {\left (16 \textit {\_R}^{4}+5 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{32 \textit {\_R}^{3} \ln \left (2\right )^{2}-32 \ln \left (2\right ) {\mathrm e} \textit {\_R}^{3}+160 \textit {\_R}^{4} \ln \left (2\right )+8 \textit {\_R}^{3} {\mathrm e}^{2}-80 \textit {\_R}^{4} {\mathrm e}+192 \textit {\_R}^{5}-10 \textit {\_R} \ln \left (2\right )+5 \textit {\_R} \,{\mathrm e}-30 \textit {\_R}^{2}}\right )}{2}\) | \(144\) |
Input:
int((768*x^4+240*x)/(256*x^4*ln(2)^2+2*(-128*x^4*exp(1)+512*x^5-80*x^2)*ln (2)+64*x^4*exp(1)^2+(-512*x^5+80*x^2)*exp(1)+1024*x^6-320*x^3+25),x,method =_RETURNVERBOSE)
Output:
-3/2*x^2/(x^2*ln(2)-1/2*x^2*exp(1)+2*x^3-5/16)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {24 \, x^{2}}{32 \, x^{3} - 8 \, x^{2} e + 16 \, x^{2} \log \left (2\right ) - 5} \] Input:
integrate((768*x^4+240*x)/(256*x^4*log(2)^2+2*(-128*x^4*exp(1)+512*x^5-80* x^2)*log(2)+64*x^4*exp(1)^2+(-512*x^5+80*x^2)*exp(1)+1024*x^6-320*x^3+25), x, algorithm="fricas")
Output:
-24*x^2/(32*x^3 - 8*x^2*e + 16*x^2*log(2) - 5)
Time = 0.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=- \frac {24 x^{2}}{32 x^{3} + x^{2} \left (- 8 e + 16 \log {\left (2 \right )}\right ) - 5} \] Input:
integrate((768*x**4+240*x)/(256*x**4*ln(2)**2+2*(-128*x**4*exp(1)+512*x**5 -80*x**2)*ln(2)+64*x**4*exp(1)**2+(-512*x**5+80*x**2)*exp(1)+1024*x**6-320 *x**3+25),x)
Output:
-24*x**2/(32*x**3 + x**2*(-8*E + 16*log(2)) - 5)
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {24 \, x^{2}}{32 \, x^{3} - 8 \, x^{2} {\left (e - 2 \, \log \left (2\right )\right )} - 5} \] Input:
integrate((768*x^4+240*x)/(256*x^4*log(2)^2+2*(-128*x^4*exp(1)+512*x^5-80* x^2)*log(2)+64*x^4*exp(1)^2+(-512*x^5+80*x^2)*exp(1)+1024*x^6-320*x^3+25), x, algorithm="maxima")
Output:
-24*x^2/(32*x^3 - 8*x^2*(e - 2*log(2)) - 5)
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {24 \, x^{2}}{32 \, x^{3} - 8 \, x^{2} e + 16 \, x^{2} \log \left (2\right ) - 5} \] Input:
integrate((768*x^4+240*x)/(256*x^4*log(2)^2+2*(-128*x^4*exp(1)+512*x^5-80* x^2)*log(2)+64*x^4*exp(1)^2+(-512*x^5+80*x^2)*exp(1)+1024*x^6-320*x^3+25), x, algorithm="giac")
Output:
-24*x^2/(32*x^3 - 8*x^2*e + 16*x^2*log(2) - 5)
Time = 3.77 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=\frac {24\,x^2}{-32\,x^3+\left (8\,\mathrm {e}-16\,\ln \left (2\right )\right )\,x^2+5} \] Input:
int((240*x + 768*x^4)/(256*x^4*log(2)^2 - 2*log(2)*(128*x^4*exp(1) + 80*x^ 2 - 512*x^5) + exp(1)*(80*x^2 - 512*x^5) + 64*x^4*exp(2) - 320*x^3 + 1024* x^6 + 25),x)
Output:
(24*x^2)/(x^2*(8*exp(1) - 16*log(2)) - 32*x^3 + 5)
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {240 x+768 x^4}{25-320 x^3+64 e^2 x^4+1024 x^6+e \left (80 x^2-512 x^5\right )+\left (-80 x^2-128 e x^4+512 x^5\right ) \log (4)+64 x^4 \log ^2(4)} \, dx=-\frac {24 x^{2}}{16 \,\mathrm {log}\left (2\right ) x^{2}-8 e \,x^{2}+32 x^{3}-5} \] Input:
int((768*x^4+240*x)/(256*x^4*log(2)^2+2*(-128*x^4*exp(1)+512*x^5-80*x^2)*l og(2)+64*x^4*exp(1)^2+(-512*x^5+80*x^2)*exp(1)+1024*x^6-320*x^3+25),x)
Output:
( - 24*x**2)/(16*log(2)*x**2 - 8*e*x**2 + 32*x**3 - 5)