Integrand size = 150, antiderivative size = 23 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=\frac {2}{\left (\frac {1}{e^3}+x\right ) \left (\frac {x}{1-5 x}+\log (4)\right )} \]
Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=-\frac {2 e^3 (1-5 x)}{\left (1+e^3 x\right ) (-\log (4)+x (-1+5 \log (4)))} \]
Integrate[(-2*E^3 + E^6*(-4*x + 10*x^2) + E^6*(-2 + 20*x - 50*x^2)*Log[4]) /(x^2 + 2*E^3*x^3 + E^6*x^4 + (2*x - 10*x^2 + E^3*(4*x^2 - 20*x^3) + E^6*( 2*x^3 - 10*x^4))*Log[4] + (1 - 10*x + 25*x^2 + E^3*(2*x - 20*x^2 + 50*x^3) + E^6*(x^2 - 10*x^3 + 25*x^4))*Log[4]^2),x]
Leaf count is larger than twice the leaf count of optimal. \(199\) vs. \(2(23)=46\).
Time = 0.72 (sec) , antiderivative size = 199, normalized size of antiderivative = 8.65, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2459, 1380, 27, 2345, 24}
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.
| \(\displaystyle \int \frac {e^6 \left (10 x^2-4 x\right )+e^6 \left (-50 x^2+20 x-2\right ) \log (4)-2 e^3}{e^6 x^4+2 e^3 x^3+x^2+\left (25 x^2+e^3 \left (50 x^3-20 x^2+2 x\right )+e^6 \left (25 x^4-10 x^3+x^2\right )-10 x+1\right ) \log ^2(4)+\left (-10 x^2+e^6 \left (2 x^3-10 x^4\right )+e^3 \left (4 x^2-20 x^3\right )+2 x\right ) \log (4)} \, dx\) |
| \(\Big \downarrow \) 2459 |
| \(\displaystyle \int \frac {10 e^6 (1-5 \log (4)) \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )^2-2 e^3 \left (5+e^3 (2-5 \log (4))-25 \log (4)\right ) \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )+\frac {5 \left (1-5 \log (4)-e^3 \log (4)\right )^2}{2 (1-5 \log (4))}}{e^6 (1-5 \log (4))^2 \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )^4-\frac {1}{2} \left (1-5 \log (4)-e^3 \log (4)\right )^2 \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )^2+\frac {\left (1-5 \log (4)-e^3 \log (4)\right )^4}{16 e^6 (1-5 \log (4))^2}}d\left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle e^6 (1-5 \log (4))^2 \int -\frac {8 \left (-20 e^6 (1-5 \log (4)) \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )^2+4 e^3 \left (5+e^3 (2-5 \log (4))-25 \log (4)\right ) \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )-\frac {5 \left (1-5 \log (4)-e^3 \log (4)\right )^2}{1-5 \log (4)}\right )}{\left (4 e^6 (1-5 \log (4))^2 \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )^2-\left (1-5 \log (4)-e^3 \log (4)\right )^2\right )^2}d\left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -8 e^6 (1-5 \log (4))^2 \int \frac {-20 e^6 (1-5 \log (4)) \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )^2+4 e^3 \left (5+e^3 (2-5 \log (4))-25 \log (4)\right ) \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )-\frac {5 \left (1-5 \log (4)-e^3 \log (4)\right )^2}{1-5 \log (4)}}{\left (4 e^6 (1-5 \log (4))^2 \left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )^2-\left (1-5 \log (4)-e^3 \log (4)\right )^2\right )^2}d\left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle -8 e^6 (1-5 \log (4))^2 \left (\frac {\int 0d\left (x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )}{2 \left (1-5 \log (4)-e^3 \log (4)\right )^2}-\frac {-10 e^3 (1-5 \log (4)) \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )+5-25 \log (4)+e^3 (2-5 \log (4))}{2 e^3 (1-5 \log (4))^2 \left (4 e^6 (1-5 \log (4))^2 \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )^2-\left (1-5 \log (4)-e^3 \log (4)\right )^2\right )}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {4 e^3 \left (-10 e^3 (1-5 \log (4)) \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )+5-25 \log (4)+e^3 (2-5 \log (4))\right )}{4 e^6 (1-5 \log (4))^2 \left (x+\frac {2 e^3+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)-20 e^3 \log (4)+2 e^6 \log (4)}{4 \left (e^6+25 e^6 \log ^2(4)-10 e^6 \log (4)\right )}\right )^2-\left (1-5 \log (4)-e^3 \log (4)\right )^2}\) |
Int[(-2*E^3 + E^6*(-4*x + 10*x^2) + E^6*(-2 + 20*x - 50*x^2)*Log[4])/(x^2 + 2*E^3*x^3 + E^6*x^4 + (2*x - 10*x^2 + E^3*(4*x^2 - 20*x^3) + E^6*(2*x^3 - 10*x^4))*Log[4] + (1 - 10*x + 25*x^2 + E^3*(2*x - 20*x^2 + 50*x^3) + E^6 *(x^2 - 10*x^3 + 25*x^4))*Log[4]^2),x]
(4*E^3*(5 + E^3*(2 - 5*Log[4]) - 25*Log[4] - 10*E^3*(1 - 5*Log[4])*(x + (2 *E^3 - 20*E^3*Log[4] + 2*E^6*Log[4] + 50*E^3*Log[4]^2 - 10*E^6*Log[4]^2)/( 4*(E^6 - 10*E^6*Log[4] + 25*E^6*Log[4]^2)))))/(-(1 - 5*Log[4] - E^3*Log[4] )^2 + 4*E^6*(1 - 5*Log[4])^2*(x + (2*E^3 - 20*E^3*Log[4] + 2*E^6*Log[4] + 50*E^3*Log[4]^2 - 10*E^6*Log[4]^2)/(4*(E^6 - 10*E^6*Log[4] + 25*E^6*Log[4] ^2)))^2)
3.28.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
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 = 0.44 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04
| method | result | size |
| risch | \(\frac {x \,{\mathrm e}^{3}-\frac {{\mathrm e}^{3}}{5}}{\ln \left (2\right ) {\mathrm e}^{3} x^{2}-\frac {{\mathrm e}^{3} \ln \left (2\right ) x}{5}-\frac {x^{2} {\mathrm e}^{3}}{10}+x \ln \left (2\right )-\frac {\ln \left (2\right )}{5}-\frac {x}{10}}\) | \(47\) |
| gosper | \(\frac {2 \left (5 x -1\right ) {\mathrm e}^{3}}{10 \ln \left (2\right ) {\mathrm e}^{3} x^{2}-2 \,{\mathrm e}^{3} \ln \left (2\right ) x -x^{2} {\mathrm e}^{3}+10 x \ln \left (2\right )-2 \ln \left (2\right )-x}\) | \(48\) |
| norman | \(\frac {\frac {\left (4 \ln \left (2\right ) {\mathrm e}^{6}+2 \,{\mathrm e}^{3}\right ) x}{2 \ln \left (2\right )}-\frac {\left (10 \ln \left (2\right )-1\right ) {\mathrm e}^{6} x^{2}}{\ln \left (2\right )}}{\left (10 x \ln \left (2\right )-2 \ln \left (2\right )-x \right ) \left (x \,{\mathrm e}^{3}+1\right )}\) | \(65\) |
| parallelrisch | \(-\frac {20 \ln \left (2\right ) {\mathrm e}^{6} x^{2}-4 \ln \left (2\right ) {\mathrm e}^{6} x -2 x^{2} {\mathrm e}^{6}-2 x \,{\mathrm e}^{3}}{2 \ln \left (2\right ) \left (10 \ln \left (2\right ) {\mathrm e}^{3} x^{2}-2 \,{\mathrm e}^{3} \ln \left (2\right ) x -x^{2} {\mathrm e}^{3}+10 x \ln \left (2\right )-2 \ln \left (2\right )-x \right )}\) | \(80\) |
int((2*(-50*x^2+20*x-2)*exp(3)^2*ln(2)+(10*x^2-4*x)*exp(3)^2-2*exp(3))/(4* ((25*x^4-10*x^3+x^2)*exp(3)^2+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10*x+1)*ln (2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+2*x)*ln(2) +x^4*exp(3)^2+2*x^3*exp(3)+x^2),x,method=_RETURNVERBOSE)
(x*exp(3)-1/5*exp(3))/(ln(2)*exp(3)*x^2-1/5*exp(3)*ln(2)*x-1/10*x^2*exp(3) +x*ln(2)-1/5*ln(2)-1/10*x)
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=-\frac {2 \, {\left (5 \, x - 1\right )} e^{3}}{x^{2} e^{3} - 2 \, {\left ({\left (5 \, x^{2} - x\right )} e^{3} + 5 \, x - 1\right )} \log \left (2\right ) + x} \]
integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp( 3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10* x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+2* x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 2.57 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=- \frac {- 10 x e^{3} + 2 e^{3}}{x^{2} \left (- e^{3} + 10 e^{3} \log {\left (2 \right )}\right ) + x \left (- 2 e^{3} \log {\left (2 \right )} - 1 + 10 \log {\left (2 \right )}\right ) - 2 \log {\left (2 \right )}} \]
integrate((2*(-50*x**2+20*x-2)*exp(3)**2*ln(2)+(10*x**2-4*x)*exp(3)**2-2*e xp(3))/(4*((25*x**4-10*x**3+x**2)*exp(3)**2+(50*x**3-20*x**2+2*x)*exp(3)+2 5*x**2-10*x+1)*ln(2)**2+2*((-10*x**4+2*x**3)*exp(3)**2+(-20*x**3+4*x**2)*e xp(3)-10*x**2+2*x)*ln(2)+x**4*exp(3)**2+2*x**3*exp(3)+x**2),x)
-(-10*x*exp(3) + 2*exp(3))/(x**2*(-exp(3) + 10*exp(3)*log(2)) + x*(-2*exp( 3)*log(2) - 1 + 10*log(2)) - 2*log(2))
Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=\frac {2 \, {\left (5 \, x e^{3} - e^{3}\right )}}{{\left (10 \, e^{3} \log \left (2\right ) - e^{3}\right )} x^{2} - {\left (2 \, {\left (e^{3} - 5\right )} \log \left (2\right ) + 1\right )} x - 2 \, \log \left (2\right )} \]
integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp( 3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10* x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+2* x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17 \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=\frac {2 \, {\left (5 \, x e^{3} - e^{3}\right )}}{10 \, x^{2} e^{3} \log \left (2\right ) - x^{2} e^{3} - 2 \, x e^{3} \log \left (2\right ) + 10 \, x \log \left (2\right ) - x - 2 \, \log \left (2\right )} \]
integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp( 3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10* x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+2* x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm=\
2*(5*x*e^3 - e^3)/(10*x^2*e^3*log(2) - x^2*e^3 - 2*x*e^3*log(2) + 10*x*log (2) - x - 2*log(2))
Timed out. \[ \int \frac {-2 e^3+e^6 \left (-4 x+10 x^2\right )+e^6 \left (-2+20 x-50 x^2\right ) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+\left (2 x-10 x^2+e^3 \left (4 x^2-20 x^3\right )+e^6 \left (2 x^3-10 x^4\right )\right ) \log (4)+\left (1-10 x+25 x^2+e^3 \left (2 x-20 x^2+50 x^3\right )+e^6 \left (x^2-10 x^3+25 x^4\right )\right ) \log ^2(4)} \, dx=\text {Hanged} \]