Integrand size = 113, antiderivative size = 36 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=3 \left (-\frac {x}{5}+\frac {\log (3-x)}{x}-x \left (e^{2 x}-x^2+\log (5+x)\right )\right ) \]
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.75 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=\frac {3}{5} \left (\frac {5}{3} \log (3-x)-\frac {5}{3} \log (-3+x)\right )+\frac {3}{5} \left (-x-5 e^{2 x} x+5 x^3+\frac {5 \log (3-x)}{x}-5 x \log (5+x)\right ) \]
Integrate[(75*x + 60*x^2 + 39*x^3 - 693*x^4 + 90*x^5 + 45*x^6 + E^(2*x)*(2 25*x^2 + 420*x^3 - 75*x^4 - 30*x^5) + (225 - 30*x - 15*x^2)*Log[3 - x] + ( 225*x^2 - 30*x^3 - 15*x^4)*Log[5 + x])/(-75*x^2 + 10*x^3 + 5*x^4),x]
(3*((5*Log[3 - x])/3 - (5*Log[-3 + x])/3))/5 + (3*(-x - 5*E^(2*x)*x + 5*x^ 3 + (5*Log[3 - x])/x - 5*x*Log[5 + x]))/5
Time = 1.53 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.67, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2026, 7279, 2009}
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.
\(\displaystyle \int \frac {45 x^6+90 x^5-693 x^4+39 x^3+60 x^2+\left (-15 x^2-30 x+225\right ) \log (3-x)+\left (-15 x^4-30 x^3+225 x^2\right ) \log (x+5)+e^{2 x} \left (-30 x^5-75 x^4+420 x^3+225 x^2\right )+75 x}{5 x^4+10 x^3-75 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {45 x^6+90 x^5-693 x^4+39 x^3+60 x^2+\left (-15 x^2-30 x+225\right ) \log (3-x)+\left (-15 x^4-30 x^3+225 x^2\right ) \log (x+5)+e^{2 x} \left (-30 x^5-75 x^4+420 x^3+225 x^2\right )+75 x}{x^2 \left (5 x^2+10 x-75\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {9 x^4}{(x-3) (x+5)}+\frac {18 x^3}{(x-3) (x+5)}-\frac {693 x^2}{5 (x-3) (x+5)}-\frac {3 \log (3-x)}{x^2}+\frac {39 x}{5 (x-3) (x+5)}-3 e^{2 x} (2 x+1)+\frac {12}{(x-3) (x+5)}+\frac {15}{(x-3) (x+5) x}-3 \log (x+5)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 x^3-\frac {3 x}{5}+\frac {3 e^{2 x}}{2}-\frac {3}{2} e^{2 x} (2 x+1)-3 (x+5) \log (x+5)+15 \log (x+5)+\frac {3 \log (3-x)}{x}\) |
Int[(75*x + 60*x^2 + 39*x^3 - 693*x^4 + 90*x^5 + 45*x^6 + E^(2*x)*(225*x^2 + 420*x^3 - 75*x^4 - 30*x^5) + (225 - 30*x - 15*x^2)*Log[3 - x] + (225*x^ 2 - 30*x^3 - 15*x^4)*Log[5 + x])/(-75*x^2 + 10*x^3 + 5*x^4),x]
(3*E^(2*x))/2 - (3*x)/5 + 3*x^3 - (3*E^(2*x)*(1 + 2*x))/2 + (3*Log[3 - x]) /x + 15*Log[5 + x] - 3*(5 + x)*Log[5 + x]
3.1.86.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 1.39 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {3 \ln \left (-x +3\right )}{x}-3 x \ln \left (5+x \right )+3 x^{3}-\frac {3 x}{5}-3 x \,{\mathrm e}^{2 x}\) | \(35\) |
default | \(-3 x \,{\mathrm e}^{2 x}+\ln \left (-x \right )+\frac {\ln \left (-x +3\right ) \left (-x +3\right )}{x}+3 x^{3}-\frac {3 x}{5}-\ln \left (x \right )+\ln \left (-3+x \right )+15 \ln \left (5+x \right )-3 \left (5+x \right ) \ln \left (5+x \right )+15\) | \(60\) |
parts | \(-3 x \,{\mathrm e}^{2 x}+\ln \left (-x \right )+\frac {\ln \left (-x +3\right ) \left (-x +3\right )}{x}+3 x^{3}-\frac {3 x}{5}-\ln \left (x \right )+\ln \left (-3+x \right )+15 \ln \left (5+x \right )-3 \left (5+x \right ) \ln \left (5+x \right )+15\) | \(60\) |
parallelrisch | \(-\frac {-60 x^{4}+60 \,{\mathrm e}^{2 x} x^{2}+60 \ln \left (5+x \right ) x^{2}-15 \ln \left (-3+x \right ) x +12 x^{2}+15 x \ln \left (-x +3\right )+3453 x -60 \ln \left (-x +3\right )}{20 x}\) | \(62\) |
int(((-15*x^4-30*x^3+225*x^2)*ln(5+x)+(-15*x^2-30*x+225)*ln(-x+3)+(-30*x^5 -75*x^4+420*x^3+225*x^2)*exp(2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60*x^2+75*x )/(5*x^4+10*x^3-75*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=\frac {3 \, {\left (5 \, x^{4} - 5 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} \log \left (x + 5\right ) - x^{2} + 5 \, \log \left (-x + 3\right )\right )}}{5 \, x} \]
integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(-x+3)+ (-30*x^5-75*x^4+420*x^3+225*x^2)*exp(2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60* x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=3 x^{3} - 3 x e^{2 x} - 3 x \log {\left (x + 5 \right )} - \frac {3 x}{5} + \frac {3 \log {\left (3 - x \right )}}{x} \]
integrate(((-15*x**4-30*x**3+225*x**2)*ln(5+x)+(-15*x**2-30*x+225)*ln(-x+3 )+(-30*x**5-75*x**4+420*x**3+225*x**2)*exp(2*x)+45*x**6+90*x**5-693*x**4+3 9*x**3+60*x**2+75*x)/(5*x**4+10*x**3-75*x**2),x)
Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=\frac {3 \, {\left (5 \, x^{4} - 5 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} \log \left (x + 5\right ) - x^{2} + 5 \, \log \left (-x + 3\right )\right )}}{5 \, x} \]
integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(-x+3)+ (-30*x^5-75*x^4+420*x^3+225*x^2)*exp(2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60* x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (34) = 68\).
Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.44 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=\frac {3 \, {\left (5 \, {\left (x + 5\right )}^{4} e^{10} - 100 \, {\left (x + 5\right )}^{3} e^{10} - 5 \, {\left (x + 5\right )}^{2} e^{10} \log \left (x + 5\right ) + 749 \, {\left (x + 5\right )}^{2} e^{10} - 5 \, {\left (x + 5\right )}^{2} e^{\left (2 \, x + 10\right )} + 50 \, {\left (x + 5\right )} e^{10} \log \left (x + 5\right ) - 1870 \, {\left (x + 5\right )} e^{10} + 50 \, {\left (x + 5\right )} e^{\left (2 \, x + 10\right )} - 125 \, e^{10} \log \left (x + 5\right ) + 5 \, e^{10} \log \left (-x + 3\right ) - 125 \, e^{\left (2 \, x + 10\right )}\right )}}{5 \, {\left ({\left (x + 5\right )} e^{10} - 5 \, e^{10}\right )}} \]
integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(-x+3)+ (-30*x^5-75*x^4+420*x^3+225*x^2)*exp(2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60* x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm=\
3/5*(5*(x + 5)^4*e^10 - 100*(x + 5)^3*e^10 - 5*(x + 5)^2*e^10*log(x + 5) + 749*(x + 5)^2*e^10 - 5*(x + 5)^2*e^(2*x + 10) + 50*(x + 5)*e^10*log(x + 5 ) - 1870*(x + 5)*e^10 + 50*(x + 5)*e^(2*x + 10) - 125*e^10*log(x + 5) + 5* e^10*log(-x + 3) - 125*e^(2*x + 10))/((x + 5)*e^10 - 5*e^10)
Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.58 \[ \int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx=3\,x^3-3\,x\,{\mathrm {e}}^{2\,x}-3\,x\,\ln \left (x+5\right )-\frac {3\,x}{5}+\frac {\ln \left (3-x\right )\,\left (3\,x^3+6\,x^2-45\,x\right )}{x^2\,\left (x-3\right )\,\left (x+5\right )} \]
int((75*x - log(3 - x)*(30*x + 15*x^2 - 225) - log(x + 5)*(30*x^3 - 225*x^ 2 + 15*x^4) + exp(2*x)*(225*x^2 + 420*x^3 - 75*x^4 - 30*x^5) + 60*x^2 + 39 *x^3 - 693*x^4 + 90*x^5 + 45*x^6)/(10*x^3 - 75*x^2 + 5*x^4),x)