Integrand size = 61, antiderivative size = 26 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=i \pi +\frac {e \log ^2(3+x (25+x))}{x}+\log (3+\log (2)) \] Output:
ln(-3-ln(2))+ln(3+x*(x+25))^2*exp(1)/x
Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {e \log ^2\left (3+25 x+x^2\right )}{x} \] Input:
Integrate[(E*(50*x + 4*x^2)*Log[3 + 25*x + x^2] + E*(-3 - 25*x - x^2)*Log[ 3 + 25*x + x^2]^2)/(3*x^2 + 25*x^3 + x^4),x]
Output:
(E*Log[3 + 25*x + x^2]^2)/x
Time = 1.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, 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.
\(\displaystyle \int \frac {e \left (-x^2-25 x-3\right ) \log ^2\left (x^2+25 x+3\right )+e \left (4 x^2+50 x\right ) \log \left (x^2+25 x+3\right )}{x^4+25 x^3+3 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e \left (-x^2-25 x-3\right ) \log ^2\left (x^2+25 x+3\right )+e \left (4 x^2+50 x\right ) \log \left (x^2+25 x+3\right )}{x^2 \left (x^2+25 x+3\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2 e (2 x+25) \log \left (x^2+25 x+3\right )}{x \left (x^2+25 x+3\right )}-\frac {e \log ^2\left (x^2+25 x+3\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e \log ^2\left (x^2+25 x+3\right )}{x}\) |
Input:
Int[(E*(50*x + 4*x^2)*Log[3 + 25*x + x^2] + E*(-3 - 25*x - x^2)*Log[3 + 25 *x + x^2]^2)/(3*x^2 + 25*x^3 + x^4),x]
Output:
(E*Log[3 + 25*x + x^2]^2)/x
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 = 0.53 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69
method | result | size |
norman | \(\frac {{\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}}{x}\) | \(18\) |
risch | \(\frac {{\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}}{x}\) | \(18\) |
parallelrisch | \(\frac {{\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}}{x}\) | \(18\) |
orering | \(-\frac {\left (28 x^{5}+1700 x^{4}+41250 x^{3}+500000 x^{2}+2745517 x +278550\right ) \left (\left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\left (4 x^{2}+50 x \right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )\right )}{\left (4 x^{4}+200 x^{3}+3750 x^{2}+31250 x +379483\right ) \left (x^{4}+25 x^{3}+3 x^{2}\right )}-\frac {3 \left (x^{2}+25 x +3\right ) \left (8 x^{4}+350 x^{3}+5625 x^{2}+31400 x +1875\right ) \left (\frac {\left (-2 x -25\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\frac {2 \left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right ) \left (2 x +25\right )}{x^{2}+25 x +3}+\left (8 x +50\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )+\frac {\left (4 x^{2}+50 x \right ) {\mathrm e} \left (2 x +25\right )}{x^{2}+25 x +3}}{x^{4}+25 x^{3}+3 x^{2}}-\frac {\left (\left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\left (4 x^{2}+50 x \right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )\right ) \left (4 x^{3}+75 x^{2}+6 x \right )}{\left (x^{4}+25 x^{3}+3 x^{2}\right )^{2}}\right )}{4 x^{4}+200 x^{3}+3750 x^{2}+31250 x +379483}-\frac {\left (x^{2}+25 x +3\right )^{2} x \left (4 x^{2}+100 x +625\right ) \left (\frac {-2 \,{\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\frac {4 \left (-2 x -25\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right ) \left (2 x +25\right )}{x^{2}+25 x +3}+\frac {2 \left (-x^{2}-25 x -3\right ) {\mathrm e} \left (2 x +25\right )^{2}}{\left (x^{2}+25 x +3\right )^{2}}+\frac {4 \left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )}{x^{2}+25 x +3}-\frac {2 \left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right ) \left (2 x +25\right )^{2}}{\left (x^{2}+25 x +3\right )^{2}}+8 \,{\mathrm e} \ln \left (x^{2}+25 x +3\right )+\frac {2 \left (8 x +50\right ) {\mathrm e} \left (2 x +25\right )}{x^{2}+25 x +3}+\frac {2 \left (4 x^{2}+50 x \right ) {\mathrm e}}{x^{2}+25 x +3}-\frac {\left (4 x^{2}+50 x \right ) {\mathrm e} \left (2 x +25\right )^{2}}{\left (x^{2}+25 x +3\right )^{2}}}{x^{4}+25 x^{3}+3 x^{2}}-\frac {2 \left (\left (-2 x -25\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\frac {2 \left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right ) \left (2 x +25\right )}{x^{2}+25 x +3}+\left (8 x +50\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )+\frac {\left (4 x^{2}+50 x \right ) {\mathrm e} \left (2 x +25\right )}{x^{2}+25 x +3}\right ) \left (4 x^{3}+75 x^{2}+6 x \right )}{\left (x^{4}+25 x^{3}+3 x^{2}\right )^{2}}+\frac {2 \left (\left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\left (4 x^{2}+50 x \right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )\right ) \left (4 x^{3}+75 x^{2}+6 x \right )^{2}}{\left (x^{4}+25 x^{3}+3 x^{2}\right )^{3}}-\frac {\left (\left (-x^{2}-25 x -3\right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )^{2}+\left (4 x^{2}+50 x \right ) {\mathrm e} \ln \left (x^{2}+25 x +3\right )\right ) \left (12 x^{2}+150 x +6\right )}{\left (x^{4}+25 x^{3}+3 x^{2}\right )^{2}}\right )}{4 x^{4}+200 x^{3}+3750 x^{2}+31250 x +379483}\) | \(957\) |
Input:
int(((-x^2-25*x-3)*exp(1)*ln(x^2+25*x+3)^2+(4*x^2+50*x)*exp(1)*ln(x^2+25*x +3))/(x^4+25*x^3+3*x^2),x,method=_RETURNVERBOSE)
Output:
exp(1)*ln(x^2+25*x+3)^2/x
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {e \log \left (x^{2} + 25 \, x + 3\right )^{2}}{x} \] Input:
integrate(((-x^2-25*x-3)*exp(1)*log(x^2+25*x+3)^2+(4*x^2+50*x)*exp(1)*log( x^2+25*x+3))/(x^4+25*x^3+3*x^2),x, algorithm="fricas")
Output:
e*log(x^2 + 25*x + 3)^2/x
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {e \log {\left (x^{2} + 25 x + 3 \right )}^{2}}{x} \] Input:
integrate(((-x**2-25*x-3)*exp(1)*ln(x**2+25*x+3)**2+(4*x**2+50*x)*exp(1)*l n(x**2+25*x+3))/(x**4+25*x**3+3*x**2),x)
Output:
E*log(x**2 + 25*x + 3)**2/x
Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {e \log \left (x^{2} + 25 \, x + 3\right )^{2}}{x} \] Input:
integrate(((-x^2-25*x-3)*exp(1)*log(x^2+25*x+3)^2+(4*x^2+50*x)*exp(1)*log( x^2+25*x+3))/(x^4+25*x^3+3*x^2),x, algorithm="maxima")
Output:
e*log(x^2 + 25*x + 3)^2/x
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {e \log \left (x^{2} + 25 \, x + 3\right )^{2}}{x} \] Input:
integrate(((-x^2-25*x-3)*exp(1)*log(x^2+25*x+3)^2+(4*x^2+50*x)*exp(1)*log( x^2+25*x+3))/(x^4+25*x^3+3*x^2),x, algorithm="giac")
Output:
e*log(x^2 + 25*x + 3)^2/x
Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {\mathrm {e}\,{\ln \left (x^2+25\,x+3\right )}^2}{x} \] Input:
int((exp(1)*log(25*x + x^2 + 3)*(50*x + 4*x^2) - exp(1)*log(25*x + x^2 + 3 )^2*(25*x + x^2 + 3))/(3*x^2 + 25*x^3 + x^4),x)
Output:
(exp(1)*log(25*x + x^2 + 3)^2)/x
Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {e \left (50 x+4 x^2\right ) \log \left (3+25 x+x^2\right )+e \left (-3-25 x-x^2\right ) \log ^2\left (3+25 x+x^2\right )}{3 x^2+25 x^3+x^4} \, dx=\frac {\mathrm {log}\left (x^{2}+25 x +3\right )^{2} e}{x} \] Input:
int(((-x^2-25*x-3)*exp(1)*log(x^2+25*x+3)^2+(4*x^2+50*x)*exp(1)*log(x^2+25 *x+3))/(x^4+25*x^3+3*x^2),x)
Output:
(log(x**2 + 25*x + 3)**2*e)/x