Integrand size = 58, antiderivative size = 27 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=1-\frac {1+\frac {7}{x}}{4+\frac {-5+x}{\log (3)}}+5 \log (64) \] Output:
1+30*ln(2)-(7/x+1)/(4+(-5+x)/ln(3))
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=-\frac {(7+x) \log (3)}{x (-5+x+\log (81))} \] Input:
Integrate[((-35 + 14*x + x^2)*Log[3] + 28*Log[3]^2)/(25*x^2 - 10*x^3 + x^4 + (-40*x^2 + 8*x^3)*Log[3] + 16*x^2*Log[3]^2),x]
Output:
-(((7 + x)*Log[3])/(x*(-5 + x + Log[81])))
Time = 0.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 2026, 2007, 2082, 1195, 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.
\(\displaystyle \int \frac {\left (x^2+14 x-35\right ) \log (3)+28 \log ^2(3)}{x^4-10 x^3+25 x^2+16 x^2 \log ^2(3)+\left (8 x^3-40 x^2\right ) \log (3)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (x^2+14 x-35\right ) \log (3)+28 \log ^2(3)}{x^4-10 x^3+x^2 \left (25+16 \log ^2(3)\right )+\left (8 x^3-40 x^2\right ) \log (3)}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2+14 x-35\right ) \log (3)+28 \log ^2(3)}{x^2 \left (x^2-2 x (5-\log (81))+(\log (81)-5)^2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (x^2+14 x-35\right ) \log (3)+28 \log ^2(3)}{x^2 (x-5+\log (81))^2}dx\) |
\(\Big \downarrow \) 2082 |
\(\displaystyle \int \frac {x^2 \log (3)+14 x \log (3)-7 \log (3) (5-\log (81))}{x^2 (x-5+\log (81))^2}dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {7 \log (3)}{x^2 (\log (81)-5)}+\frac {\log (3) (\log (81)-12)}{(\log (81)-5) (x-5+\log (81))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (3) (12-\log (81))}{(5-\log (81)) (-x+5-\log (81))}+\frac {7 \log (3)}{x (5-\log (81))}\) |
Input:
Int[((-35 + 14*x + x^2)*Log[3] + 28*Log[3]^2)/(25*x^2 - 10*x^3 + x^4 + (-4 0*x^2 + 8*x^3)*Log[3] + 16*x^2*Log[3]^2),x]
Output:
(7*Log[3])/(x*(5 - Log[81])) + (Log[3]*(12 - Log[81]))/((5 - Log[81])*(5 - x - Log[81]))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m* ExpandToSum[v, x]^n*ExpandToSum[w, x]^p, x] /; FreeQ[{m, n, p}, x] && Linea rQ[{u, v}, x] && QuadraticQ[w, x] && !(LinearMatchQ[{u, v}, x] && Quadrati cMatchQ[w, x])
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
gosper | \(-\frac {\left (x +7\right ) \ln \left (3\right )}{x \left (-5+4 \ln \left (3\right )+x \right )}\) | \(20\) |
norman | \(\frac {-x \ln \left (3\right )-7 \ln \left (3\right )}{x \left (-5+4 \ln \left (3\right )+x \right )}\) | \(24\) |
parallelrisch | \(\frac {-x \ln \left (3\right )-7 \ln \left (3\right )}{x \left (-5+4 \ln \left (3\right )+x \right )}\) | \(24\) |
risch | \(\frac {-x \ln \left (3\right )-7 \ln \left (3\right )}{x \left (-5+4 \ln \left (3\right )+x \right )}\) | \(25\) |
default | \(\ln \left (3\right ) \left (-\frac {4 \ln \left (3\right )-12}{\left (-5+4 \ln \left (3\right )\right ) \left (-5+4 \ln \left (3\right )+x \right )}-\frac {7}{\left (-5+4 \ln \left (3\right )\right ) x}\right )\) | \(43\) |
Input:
int((28*ln(3)^2+(x^2+14*x-35)*ln(3))/(16*x^2*ln(3)^2+(8*x^3-40*x^2)*ln(3)+ x^4-10*x^3+25*x^2),x,method=_RETURNVERBOSE)
Output:
-1/x*(x+7)*ln(3)/(-5+4*ln(3)+x)
Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=-\frac {{\left (x + 7\right )} \log \left (3\right )}{x^{2} + 4 \, x \log \left (3\right ) - 5 \, x} \] Input:
integrate((28*log(3)^2+(x^2+14*x-35)*log(3))/(16*x^2*log(3)^2+(8*x^3-40*x^ 2)*log(3)+x^4-10*x^3+25*x^2),x, algorithm="fricas")
Output:
-(x + 7)*log(3)/(x^2 + 4*x*log(3) - 5*x)
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=\frac {- x \log {\left (3 \right )} - 7 \log {\left (3 \right )}}{x^{2} + x \left (-5 + 4 \log {\left (3 \right )}\right )} \] Input:
integrate((28*ln(3)**2+(x**2+14*x-35)*ln(3))/(16*x**2*ln(3)**2+(8*x**3-40* x**2)*ln(3)+x**4-10*x**3+25*x**2),x)
Output:
(-x*log(3) - 7*log(3))/(x**2 + x*(-5 + 4*log(3)))
Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=-\frac {x \log \left (3\right ) + 7 \, \log \left (3\right )}{x^{2} + x {\left (4 \, \log \left (3\right ) - 5\right )}} \] Input:
integrate((28*log(3)^2+(x^2+14*x-35)*log(3))/(16*x^2*log(3)^2+(8*x^3-40*x^ 2)*log(3)+x^4-10*x^3+25*x^2),x, algorithm="maxima")
Output:
-(x*log(3) + 7*log(3))/(x^2 + x*(4*log(3) - 5))
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=-\frac {x \log \left (3\right ) + 7 \, \log \left (3\right )}{x^{2} + 4 \, x \log \left (3\right ) - 5 \, x} \] Input:
integrate((28*log(3)^2+(x^2+14*x-35)*log(3))/(16*x^2*log(3)^2+(8*x^3-40*x^ 2)*log(3)+x^4-10*x^3+25*x^2),x, algorithm="giac")
Output:
-(x*log(3) + 7*log(3))/(x^2 + 4*x*log(3) - 5*x)
Time = 3.27 (sec) , antiderivative size = 5106, normalized size of antiderivative = 189.11 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=\text {Too large to display} \] Input:
int((28*log(3)^2 + log(3)*(14*x + x^2 - 35))/(16*x^2*log(3)^2 - log(3)*(40 *x^2 - 8*x^3) + 25*x^2 - 10*x^3 + x^4),x)
Output:
log((2240*log(3)^4*log(81) - 1680*log(3)^3*log(81) - 8400*log(3)^2*log(81) + 33600*log(3)^3 - 29120*log(3)^4 + 6272*log(3)^5 + 2240*log(3)^2*log(81) ^2 - 280*log(3)^2*log(81)^3 + 168*log(3)^3*log(81)^2 + 14*log(3)^2*log(81) ^4 - 224*log(3)^4*log(81)^2)/(150*log(81)^2 - 500*log(81) - 20*log(81)^3 + log(81)^4 + 625) - ((14*log(3))/(log(81) - 5)^2 - (14*log(3)*(4*log(3) - 5)^2)/(log(81) - 5)^4)*((37500*log(3) - 71250*log(3)*log(81) + 32125*log(3 )*log(81)^2 - 14000*log(3)^2*log(81) - 6000*log(3)*log(81)^3 + 67200*log(3 )^3*log(81) + 550*log(3)*log(81)^4 - 17920*log(3)^4*log(81) - 30*log(3)*lo g(81)^5 + log(3)*log(81)^6 + 122500*log(3)^2 - 168000*log(3)^3 + 44800*log (3)^4 - 12600*log(3)^2*log(81)^2 + 2800*log(3)^2*log(81)^3 - 6720*log(3)^3 *log(81)^2 - 140*log(3)^2*log(81)^4 + 1792*log(3)^4*log(81)^2)/(150*log(81 )^2 - 500*log(81) - 20*log(81)^3 + log(81)^4 + 625) + ((125000*log(3) + 18 7500*log(81) - 150000*log(3)*log(81) + 75000*log(3)*log(81)^2 - 20000*log( 3)*log(81)^3 + 3000*log(3)*log(81)^4 - 240*log(3)*log(81)^5 + 8*log(3)*log (81)^6 - 93750*log(81)^2 + 25000*log(81)^3 - 3750*log(81)^4 + 300*log(81)^ 5 - 10*log(81)^6 - 156250)/(150*log(81)^2 - 500*log(81) - 20*log(81)^3 + l og(81)^4 + 625) + (x*(12500*log(81) - 200000*log(3) + 160000*log(3)*log(81 ) - 48000*log(3)*log(81)^2 - 64000*log(3)^2*log(81) + 6400*log(3)*log(81)^ 3 - 320*log(3)*log(81)^4 + 80000*log(3)^2 - 26250*log(81)^2 + 11000*log(81 )^3 - 2050*log(81)^4 + 180*log(81)^5 - 6*log(81)^6 + 19200*log(3)^2*log...
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {\left (-35+14 x+x^2\right ) \log (3)+28 \log ^2(3)}{25 x^2-10 x^3+x^4+\left (-40 x^2+8 x^3\right ) \log (3)+16 x^2 \log ^2(3)} \, dx=\frac {\mathrm {log}\left (3\right ) \left (-28 \,\mathrm {log}\left (3\right )+x^{2}+35\right )}{x \left (16 \mathrm {log}\left (3\right )^{2}+4 \,\mathrm {log}\left (3\right ) x -40 \,\mathrm {log}\left (3\right )-5 x +25\right )} \] Input:
int((28*log(3)^2+(x^2+14*x-35)*log(3))/(16*x^2*log(3)^2+(8*x^3-40*x^2)*log (3)+x^4-10*x^3+25*x^2),x)
Output:
(log(3)*( - 28*log(3) + x**2 + 35))/(x*(16*log(3)**2 + 4*log(3)*x - 40*log (3) - 5*x + 25))