Integrand size = 79, antiderivative size = 30 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=\frac {8 x \left (3-x^2\right ) \log (2)}{e-(-e+5 (3-x)) x} \]
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-8 \left (\frac {x}{5}+\frac {-15 e+e^2+150 x-35 e x+e^2 x}{25 \left (e-15 x+e x+5 x^2\right )}\right ) \log (2) \]
Integrate[((-120*x^2 + 240*x^3 - 40*x^4 + E*(24 - 24*x^2 - 16*x^3))*Log[2] )/(225*x^2 - 150*x^3 + 25*x^4 + E^2*(1 + 2*x + x^2) + E*(-30*x - 20*x^2 + 10*x^3)),x]
Leaf count is larger than twice the leaf count of optimal. \(130\) vs. \(2(30)=60\).
Time = 0.49 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {27, 27, 2459, 1380, 27, 2345, 27, 281, 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.
\(\displaystyle \int \frac {\left (-40 x^4+240 x^3-120 x^2+e \left (-16 x^3-24 x^2+24\right )\right ) \log (2)}{25 x^4-150 x^3+225 x^2+e^2 \left (x^2+2 x+1\right )+e \left (10 x^3-20 x^2-30 x\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (2) \int -\frac {8 \left (5 x^4-30 x^3+15 x^2-e \left (-2 x^3-3 x^2+3\right )\right )}{25 x^4-150 x^3+225 x^2+e^2 \left (x^2+2 x+1\right )-10 e \left (-x^3+2 x^2+3 x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -8 \log (2) \int \frac {5 x^4-30 x^3+15 x^2-e \left (-2 x^3-3 x^2+3\right )}{25 x^4-150 x^3+225 x^2+e^2 \left (x^2+2 x+1\right )-10 e \left (-x^3+2 x^2+3 x\right )}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle -8 \log (2) \int \frac {5 \left (x+\frac {1}{100} (-150+10 e)\right )^4-\frac {3}{10} (5-e) (35-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2-\frac {1}{25} (15-e) \left (150-45 e+e^2\right ) \left (x+\frac {1}{100} (-150+10 e)\right )-\frac {3 (5-e)^2 (25-e) (45-e)}{2000}}{25 \left (x+\frac {1}{100} (-150+10 e)\right )^4-\frac {1}{2} (5-e) (45-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2+\frac {1}{400} \left (225-50 e+e^2\right )^2}d\left (x+\frac {1}{100} (-150+10 e)\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -200 \log (2) \int -\frac {-10000 \left (x+\frac {1}{100} (-150+10 e)\right )^4+600 (5-e) (35-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2+80 (15-e) \left (150-45 e+e^2\right ) \left (x+\frac {1}{100} (-150+10 e)\right )+3 (5-e)^2 (25-e) (45-e)}{125 \left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (-150+10 e)\right )^2\right )^2}d\left (x+\frac {1}{100} (-150+10 e)\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8}{5} \log (2) \int \frac {-10000 \left (x+\frac {1}{100} (-150+10 e)\right )^4+600 (5-e) (35-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2+80 (15-e) \left (150-45 e+e^2\right ) \left (x+\frac {1}{100} (-150+10 e)\right )+3 (5-e)^2 (25-e) (45-e)}{\left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (-150+10 e)\right )^2\right )^2}d\left (x+\frac {1}{100} (-150+10 e)\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {8}{5} \log (2) \left (\frac {\int \frac {2 \left (\left (225-50 e+e^2\right )^2-100 (5-e) (45-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2\right )}{(5-e) (45-e)-100 \left (x+\frac {1}{100} (-150+10 e)\right )^2}d\left (x+\frac {1}{100} (-150+10 e)\right )}{2 \left (225-50 e+e^2\right )}-\frac {2 (5-e) (45-e) \left (10 (5-e) (30-e) \left (x+\frac {1}{100} (10 e-150)\right )+(15-e) \left (150-45 e+e^2\right )\right )}{5 \left (225-50 e+e^2\right ) \left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (10 e-150)\right )^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8}{5} \log (2) \left (\frac {\int \frac {\left (225-50 e+e^2\right )^2-100 (5-e) (45-e) \left (x+\frac {1}{100} (-150+10 e)\right )^2}{(5-e) (45-e)-100 \left (x+\frac {1}{100} (-150+10 e)\right )^2}d\left (x+\frac {1}{100} (-150+10 e)\right )}{225-50 e+e^2}-\frac {2 (5-e) (45-e) \left (10 (5-e) (30-e) \left (x+\frac {1}{100} (10 e-150)\right )+(15-e) \left (150-45 e+e^2\right )\right )}{5 \left (225-50 e+e^2\right ) \left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (10 e-150)\right )^2\right )}\right )\) |
\(\Big \downarrow \) 281 |
\(\displaystyle -\frac {8}{5} \log (2) \left (\frac {(5-e) (45-e) \int 1d\left (x+\frac {1}{100} (-150+10 e)\right )}{225-50 e+e^2}-\frac {2 (5-e) (45-e) \left (10 (5-e) (30-e) \left (x+\frac {1}{100} (10 e-150)\right )+(15-e) \left (150-45 e+e^2\right )\right )}{5 \left (225-50 e+e^2\right ) \left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (10 e-150)\right )^2\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {8}{5} \left (\frac {(5-e) (45-e) \left (x+\frac {1}{100} (10 e-150)\right )}{225-50 e+e^2}-\frac {2 (5-e) (45-e) \left (10 (5-e) (30-e) \left (x+\frac {1}{100} (10 e-150)\right )+(15-e) \left (150-45 e+e^2\right )\right )}{5 \left (225-50 e+e^2\right ) \left ((5-e) (45-e)-100 \left (x+\frac {1}{100} (10 e-150)\right )^2\right )}\right ) \log (2)\) |
Int[((-120*x^2 + 240*x^3 - 40*x^4 + E*(24 - 24*x^2 - 16*x^3))*Log[2])/(225 *x^2 - 150*x^3 + 25*x^4 + E^2*(1 + 2*x + x^2) + E*(-30*x - 20*x^2 + 10*x^3 )),x]
(-8*(((5 - E)*(45 - E)*((-150 + 10*E)/100 + x))/(225 - 50*E + E^2) - (2*(5 - E)*(45 - E)*((15 - E)*(150 - 45*E + E^2) + 10*(5 - E)*(30 - E)*((-150 + 10*E)/100 + x)))/(5*(225 - 50*E + E^2)*((5 - E)*(45 - E) - 100*((-150 + 1 0*E)/100 + x)^2)))*Log[2])/5
3.2.59.3.1 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_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
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 = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(-\frac {8 x \left (x^{2}-3\right ) \ln \left (2\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(28\) |
parallelrisch | \(\frac {\ln \left (2\right ) \left (-40 x^{3}+120 x \right )}{5 x \,{\mathrm e}+25 x^{2}+5 \,{\mathrm e}-75 x}\) | \(31\) |
norman | \(\frac {24 x \ln \left (2\right )-8 x^{3} \ln \left (2\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(32\) |
risch | \(-\frac {8 x \ln \left (2\right )}{5}+\frac {\ln \left (2\right ) \left (\frac {\left (-\frac {8 \,{\mathrm e}^{2}}{5}+56 \,{\mathrm e}-240\right ) x}{5}-\frac {8 \,{\mathrm e}^{2}}{25}+\frac {24 \,{\mathrm e}}{5}\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(49\) |
default | \(\ln \left (2\right ) \left (-\frac {8 x}{5}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+\left (10 \,{\mathrm e}-150\right ) \textit {\_Z}^{3}+\left ({\mathrm e}^{2}-20 \,{\mathrm e}+225\right ) \textit {\_Z}^{2}+\left (2 \,{\mathrm e}^{2}-30 \,{\mathrm e}\right ) \textit {\_Z} +{\mathrm e}^{2}\right )}{\sum }\frac {\left (\left ({\mathrm e}^{2}-35 \,{\mathrm e}+150\right ) \textit {\_R}^{2}+2 \left ({\mathrm e}^{2}-15 \,{\mathrm e}\right ) \textit {\_R} +{\mathrm e}^{2}+15 \,{\mathrm e}\right ) \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{2} \textit {\_R} +15 \textit {\_R}^{2} {\mathrm e}+50 \textit {\_R}^{3}+{\mathrm e}^{2}-20 \textit {\_R} \,{\mathrm e}-225 \textit {\_R}^{2}-15 \,{\mathrm e}+225 \textit {\_R}}\right )}{5}\right )\) | \(129\) |
int(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*ln(2)/((x^2+2*x+1) *exp(1)^2+(10*x^3-20*x^2-30*x)*exp(1)+25*x^4-150*x^3+225*x^2),x,method=_RE TURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8 \, {\left (25 \, x^{3} - 75 \, x^{2} + {\left (x + 1\right )} e^{2} + 5 \, {\left (x^{2} - 6 \, x - 3\right )} e + 150 \, x\right )} \log \left (2\right )}{25 \, {\left (5 \, x^{2} + {\left (x + 1\right )} e - 15 \, x\right )}} \]
integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2 +2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x)*exp(1)+25*x^4-150*x^3+225*x^2),x, al gorithm=\
-8/25*(25*x^3 - 75*x^2 + (x + 1)*e^2 + 5*(x^2 - 6*x - 3)*e + 150*x)*log(2) /(5*x^2 + (x + 1)*e - 15*x)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).
Time = 0.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=- \frac {8 x \log {\left (2 \right )}}{5} - \frac {x \left (- 280 e \log {\left (2 \right )} + 8 e^{2} \log {\left (2 \right )} + 1200 \log {\left (2 \right )}\right ) - 120 e \log {\left (2 \right )} + 8 e^{2} \log {\left (2 \right )}}{125 x^{2} + x \left (-375 + 25 e\right ) + 25 e} \]
integrate(((-16*x**3-24*x**2+24)*exp(1)-40*x**4+240*x**3-120*x**2)*ln(2)/( (x**2+2*x+1)*exp(1)**2+(10*x**3-20*x**2-30*x)*exp(1)+25*x**4-150*x**3+225* x**2),x)
-8*x*log(2)/5 - (x*(-280*E*log(2) + 8*exp(2)*log(2) + 1200*log(2)) - 120*E *log(2) + 8*exp(2)*log(2))/(125*x**2 + x*(-375 + 25*E) + 25*E)
Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8}{25} \, {\left (5 \, x + \frac {x {\left (e^{2} - 35 \, e + 150\right )} + e^{2} - 15 \, e}{5 \, x^{2} + x {\left (e - 15\right )} + e}\right )} \log \left (2\right ) \]
integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2 +2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x)*exp(1)+25*x^4-150*x^3+225*x^2),x, al gorithm=\
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8}{25} \, {\left (5 \, x + \frac {x e^{2} - 35 \, x e + 150 \, x + e^{2} - 15 \, e}{5 \, x^{2} + x e - 15 \, x + e}\right )} \log \left (2\right ) \]
integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2 +2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x)*exp(1)+25*x^4-150*x^3+225*x^2),x, al gorithm=\
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {\frac {8\,{\mathrm {e}}^2\,\ln \left (2\right )}{5}-24\,\mathrm {e}\,\ln \left (2\right )+x\,\left (240\,\ln \left (2\right )-56\,\mathrm {e}\,\ln \left (2\right )+\frac {8\,{\mathrm {e}}^2\,\ln \left (2\right )}{5}\right )}{25\,x^2+\left (5\,\mathrm {e}-75\right )\,x+5\,\mathrm {e}}-\frac {8\,x\,\ln \left (2\right )}{5} \]
int(-(log(2)*(exp(1)*(24*x^2 + 16*x^3 - 24) + 120*x^2 - 240*x^3 + 40*x^4)) /(exp(2)*(2*x + x^2 + 1) - exp(1)*(30*x + 20*x^2 - 10*x^3) + 225*x^2 - 150 *x^3 + 25*x^4),x)