Integrand size = 96, antiderivative size = 27 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {x^2}{4 \left (\frac {e^4}{\frac {2}{x}+x}+\log ^2(2)\right )} \]
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(27)=54\).
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {-e^{12} x-e^8 \left (2+x^2\right ) \log ^2(2)+x^2 \left (2+x^2\right ) \log ^6(2)}{4 \left (e^4 x \log ^6(2)+\left (2+x^2\right ) \log ^8(2)\right )} \]
Integrate[((E^4*x*(2*x + 3*x^3))/(2 + x^2) + (4*x + 2*x^3)*Log[2]^2)/((E^8 *x^2*(8 + 4*x^2))/(2 + x^2)^2 + (E^4*x*(16 + 8*x^2)*Log[2]^2)/(2 + x^2) + (8 + 4*x^2)*Log[2]^4),x]
(-(E^12*x) - E^8*(2 + x^2)*Log[2]^2 + x^2*(2 + x^2)*Log[2]^6)/(4*(E^4*x*Lo g[2]^6 + (2 + x^2)*Log[2]^8))
Leaf count is larger than twice the leaf count of optimal. \(133\) vs. \(2(27)=54\).
Time = 0.60 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7239, 27, 2175, 25, 2019, 17}
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.
\(\displaystyle \int \frac {\left (2 x^3+4 x\right ) \log ^2(2)+\frac {e^4 x \left (3 x^3+2 x\right )}{x^2+2}}{\frac {e^8 \left (4 x^2+8\right ) x^2}{\left (x^2+2\right )^2}+\left (4 x^2+8\right ) \log ^4(2)+\frac {e^4 \left (8 x^2+16\right ) x \log ^2(2)}{x^2+2}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x \left (2 x^4 \log ^2(2)+3 e^4 x^3+8 x^2 \log ^2(2)+2 e^4 x+8 \log ^2(2)\right )}{4 \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {x \left (2 \log ^2(2) x^4+3 e^4 x^3+8 \log ^2(2) x^2+2 e^4 x+8 \log ^2(2)\right )}{\left (\log ^2(2) x^2+e^4 x+2 \log ^2(2)\right )^2}dx\) |
\(\Big \downarrow \) 2175 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int -\frac {2 \left (e^8-8 \log ^4(2)\right ) x^3+\frac {e^4 \left (e^8-8 \log ^4(2)\right ) x^2}{\log ^2(2)}+\left (12 e^8-\frac {e^{16}}{\log ^4(2)}-32 \log ^4(2)\right ) x-\frac {2 e^4 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}}{\log ^2(2) x^2+e^4 x+2 \log ^2(2)}dx}{e^8-8 \log ^4(2)}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{\left (e^8-8 \log ^4(2)\right ) \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {2 \left (e^8-8 \log ^4(2)\right ) x^3+\frac {e^4 \left (e^8-8 \log ^4(2)\right ) x^2}{\log ^2(2)}+\left (12 e^8-\frac {e^{16}}{\log ^4(2)}-32 \log ^4(2)\right ) x-\frac {2 e^4 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}}{\log ^2(2) x^2+e^4 x+2 \log ^2(2)}dx}{e^8-8 \log ^4(2)}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{\left (e^8-8 \log ^4(2)\right ) \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )}\right )\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \left (\left (\frac {2 e^8}{\log ^2(2)}-16 \log ^2(2)\right ) x-\frac {e^{12}}{\log ^4(2)}+8 e^4\right )dx}{e^8-8 \log ^4(2)}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{\left (e^8-8 \log ^4(2)\right ) \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{4} \left (\frac {\log ^2(2) \left (\frac {2 x \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}+e^4 \left (8-\frac {e^8}{\log ^4(2)}\right )\right )^2}{4 \left (e^8-8 \log ^4(2)\right )^2}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{\left (e^8-8 \log ^4(2)\right ) \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )}\right )\) |
Int[((E^4*x*(2*x + 3*x^3))/(2 + x^2) + (4*x + 2*x^3)*Log[2]^2)/((E^8*x^2*( 8 + 4*x^2))/(2 + x^2)^2 + (E^4*x*(16 + 8*x^2)*Log[2]^2)/(2 + x^2) + (8 + 4 *x^2)*Log[2]^4),x]
(-((E^4*x*(E^4*x*(8 - E^8/Log[2]^4) - (2*(E^8 - 8*Log[2]^4))/Log[2]^2))/(( E^4*x + 2*Log[2]^2 + x^2*Log[2]^2)*(E^8 - 8*Log[2]^4))) + (Log[2]^2*(E^4*( 8 - E^8/Log[2]^4) + (2*x*(E^8 - 8*Log[2]^4))/Log[2]^2)^2)/(4*(E^8 - 8*Log[ 2]^4)^2))/4
3.18.25.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, a + b*x + c*x^2, x], R = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[Polyno mialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*((R*b - 2*a*S + (2*c*R - b*S)*x)/((p + 1)*(b^2 - 4*a*c))), x ] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2 )^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Qx + S*(2*a*e*m + b*d *(2*p + 3)) - R*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*R - b*S)*(m + 2*p + 3)*x , x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a *c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (Inte gerQ[p] || !IntegerQ[m] || !RationalQ[a, b, c, d, e]) && !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.78 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {x^{2}}{4 \ln \left (2\right )^{2}+4 \,{\mathrm e}^{-\ln \left (\frac {x^{2}+2}{x}\right )+4}}\) | \(28\) |
norman | \(\frac {-3 x^{2}-\frac {{\mathrm e}^{4} x^{3}}{\ln \left (2\right )^{2}}+\frac {x^{6}}{4}-\frac {2 \,{\mathrm e}^{4} x}{\ln \left (2\right )^{2}}-4}{\left (x^{2}+2\right ) \left (x^{2} \ln \left (2\right )^{2}+x \,{\mathrm e}^{4}+2 \ln \left (2\right )^{2}\right )}\) | \(62\) |
risch | \(-\frac {-\ln \left (2\right )^{6} x^{4}-2 \ln \left (2\right )^{6} x^{2}+\ln \left (2\right )^{2} {\mathrm e}^{8} x^{2}+x \,{\mathrm e}^{12}+2 \ln \left (2\right )^{2} {\mathrm e}^{8}}{4 \ln \left (2\right )^{6} \left (x^{2} \ln \left (2\right )^{2}+x \,{\mathrm e}^{4}+2 \ln \left (2\right )^{2}\right )}\) | \(69\) |
default | \(-\frac {-x^{2} \ln \left (2\right )^{2}+x \,{\mathrm e}^{4}}{4 \ln \left (2\right )^{4}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\ln \left (2\right )^{4} \textit {\_Z}^{4}+2 \,{\mathrm e}^{4} \ln \left (2\right )^{2} \textit {\_Z}^{3}+\left (4 \ln \left (2\right )^{4}+{\mathrm e}^{8}\right ) \textit {\_Z}^{2}+4 \textit {\_Z} \,{\mathrm e}^{4} \ln \left (2\right )^{2}+4 \ln \left (2\right )^{4}\right )}{\sum }\frac {\left ({\mathrm e}^{4} \left (-2 \ln \left (2\right )^{4}+{\mathrm e}^{8}\right ) \textit {\_R}^{2}+4 \,{\mathrm e}^{8} \textit {\_R} \ln \left (2\right )^{2}+4 \,{\mathrm e}^{4} \ln \left (2\right )^{4}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} \ln \left (2\right )^{4}+3 \,{\mathrm e}^{4} \ln \left (2\right )^{2} \textit {\_R}^{2}+4 \textit {\_R} \ln \left (2\right )^{4}+2 \,{\mathrm e}^{4} \ln \left (2\right )^{2}+\textit {\_R} \,{\mathrm e}^{8}}}{8 \ln \left (2\right )^{4}}\) | \(162\) |
int(((3*x^3+2*x)*exp(-ln((x^2+2)/x)+4)+(2*x^3+4*x)*ln(2)^2)/((4*x^2+8)*exp (-ln((x^2+2)/x)+4)^2+(8*x^2+16)*ln(2)^2*exp(-ln((x^2+2)/x)+4)+(4*x^2+8)*ln (2)^4),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {{\left (x^{4} + 2 \, x^{2}\right )} \log \left (2\right )^{6} - {\left (x^{2} + 2\right )} e^{8} \log \left (2\right )^{2} - x e^{12}}{4 \, {\left ({\left (x^{2} + 2\right )} \log \left (2\right )^{8} + x e^{4} \log \left (2\right )^{6}\right )}} \]
integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^ 2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+( 4*x^2+8)*log(2)^4),x, algorithm=\
1/4*((x^4 + 2*x^2)*log(2)^6 - (x^2 + 2)*e^8*log(2)^2 - x*e^12)/((x^2 + 2)* log(2)^8 + x*e^4*log(2)^6)
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (19) = 38\).
Time = 0.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.78 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {x^{2}}{4 \log {\left (2 \right )}^{2}} - \frac {x e^{4}}{4 \log {\left (2 \right )}^{4}} + \frac {x \left (- e^{12} + 2 e^{4} \log {\left (2 \right )}^{4}\right ) - 2 e^{8} \log {\left (2 \right )}^{2}}{4 x^{2} \log {\left (2 \right )}^{8} + 4 x e^{4} \log {\left (2 \right )}^{6} + 8 \log {\left (2 \right )}^{8}} \]
integrate(((3*x**3+2*x)*exp(-ln((x**2+2)/x)+4)+(2*x**3+4*x)*ln(2)**2)/((4* x**2+8)*exp(-ln((x**2+2)/x)+4)**2+(8*x**2+16)*ln(2)**2*exp(-ln((x**2+2)/x) +4)+(4*x**2+8)*ln(2)**4),x)
x**2/(4*log(2)**2) - x*exp(4)/(4*log(2)**4) + (x*(-exp(12) + 2*exp(4)*log( 2)**4) - 2*exp(8)*log(2)**2)/(4*x**2*log(2)**8 + 4*x*exp(4)*log(2)**6 + 8* log(2)**8)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (25) = 50\).
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.70 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=-\frac {2 \, e^{8} \log \left (2\right )^{2} - {\left (2 \, e^{4} \log \left (2\right )^{4} - e^{12}\right )} x}{4 \, {\left (x^{2} \log \left (2\right )^{8} + x e^{4} \log \left (2\right )^{6} + 2 \, \log \left (2\right )^{8}\right )}} + \frac {x^{2} \log \left (2\right )^{2} - x e^{4}}{4 \, \log \left (2\right )^{4}} \]
integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^ 2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+( 4*x^2+8)*log(2)^4),x, algorithm=\
-1/4*(2*e^8*log(2)^2 - (2*e^4*log(2)^4 - e^12)*x)/(x^2*log(2)^8 + x*e^4*lo g(2)^6 + 2*log(2)^8) + 1/4*(x^2*log(2)^2 - x*e^4)/log(2)^4
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.78 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {2 \, x e^{4} \log \left (2\right )^{4} - 2 \, e^{8} \log \left (2\right )^{2} - x e^{12}}{4 \, {\left (x^{2} \log \left (2\right )^{2} + x e^{4} + 2 \, \log \left (2\right )^{2}\right )} \log \left (2\right )^{6}} + \frac {x^{2} \log \left (2\right )^{6} - x e^{4} \log \left (2\right )^{4}}{4 \, \log \left (2\right )^{8}} \]
integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^ 2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+( 4*x^2+8)*log(2)^4),x, algorithm=\
1/4*(2*x*e^4*log(2)^4 - 2*e^8*log(2)^2 - x*e^12)/((x^2*log(2)^2 + x*e^4 + 2*log(2)^2)*log(2)^6) + 1/4*(x^2*log(2)^6 - x*e^4*log(2)^4)/log(2)^8
Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\frac {e^4 x \left (2 x+3 x^3\right )}{2+x^2}+\left (4 x+2 x^3\right ) \log ^2(2)}{\frac {e^8 x^2 \left (8+4 x^2\right )}{\left (2+x^2\right )^2}+\frac {e^4 x \left (16+8 x^2\right ) \log ^2(2)}{2+x^2}+\left (8+4 x^2\right ) \log ^4(2)} \, dx=\frac {x^4+2\,x^2}{4\,{\ln \left (2\right )}^2\,x^2+4\,{\mathrm {e}}^4\,x+8\,{\ln \left (2\right )}^2} \]