\(\int \frac {-48 x^2+(-40-80 x+8 x^2) \log (5)}{25 x^2-10 x^3+x^4+(50 x-20 x^2+2 x^3) \log (5)+(25-10 x+x^2) \log ^2(5)} \, dx\) [360]

Optimal result

Integrand size = 66, antiderivative size = 21 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=-9+\frac {2 x (4+4 x)}{(-5+x) (x+\log (5))} \] Output:

2*(4+4*x)/(-5+x)*x/(ln(5)+x)-9
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=\frac {8 (-x (-6+\log (5))+5 \log (5))}{(-5+x) (x+\log (5))} \] Input:

Integrate[(-48*x^2 + (-40 - 80*x + 8*x^2)*Log[5])/(25*x^2 - 10*x^3 + x^4 + 
 (50*x - 20*x^2 + 2*x^3)*Log[5] + (25 - 10*x + x^2)*Log[5]^2),x]
 

Output:

(8*(-(x*(-6 + Log[5])) + 5*Log[5]))/((-5 + x)*(x + Log[5]))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(21)=42\).

Time = 0.35 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, 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.

\(\displaystyle \int \frac {\left (8 x^2-80 x-40\right ) \log (5)-48 x^2}{x^4-10 x^3+25 x^2+\left (x^2-10 x+25\right ) \log ^2(5)+\left (2 x^3-20 x^2+50 x\right ) \log (5)} \, dx\)

\(\Big \downarrow \) 2459

\(\displaystyle \int \frac {-8 \left (30+\log ^2(5)-\log (5)\right ) \left (x+\frac {1}{4} (2 \log (5)-10)\right )-8 (6-\log (5)) \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2-2 (6-\log (5)) (5+\log (5))^2}{\left (x+\frac {1}{4} (2 \log (5)-10)\right )^4-\frac {1}{2} (5+\log (5))^2 \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2+\frac {1}{16} (5+\log (5))^4}d\left (x+\frac {1}{4} (2 \log (5)-10)\right )\)

\(\Big \downarrow \) 1380

\(\displaystyle \int -\frac {32 \left (4 \left (30+\log ^2(5)-\log (5)\right ) \left (x+\frac {1}{4} (2 \log (5)-10)\right )+4 (6-\log (5)) \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2+(6-\log (5)) (5+\log (5))^2\right )}{\left (4 \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2-(5+\log (5))^2\right )^2}d\left (x+\frac {1}{4} (2 \log (5)-10)\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -32 \int \frac {4 (6-\log (5)) \left (x+\frac {1}{4} (-10+2 \log (5))\right )^2+4 \left (30-\log (5)+\log ^2(5)\right ) \left (x+\frac {1}{4} (-10+2 \log (5))\right )+(6-\log (5)) (5+\log (5))^2}{\left (4 \left (x+\frac {1}{4} (-10+2 \log (5))\right )^2-(5+\log (5))^2\right )^2}d\left (x+\frac {1}{4} (-10+2 \log (5))\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle -32 \left (\frac {\int 0d\left (x+\frac {1}{4} (-10+2 \log (5))\right )}{2 (5+\log (5))^2}-\frac {2 (6-\log (5)) \left (x+\frac {1}{4} (2 \log (5)-10)\right )+30+\log ^2(5)-\log (5)}{2 \left (4 \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2-(5+\log (5))^2\right )}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {16 \left (2 (6-\log (5)) \left (x+\frac {1}{4} (2 \log (5)-10)\right )+30+\log ^2(5)-\log (5)\right )}{4 \left (x+\frac {1}{4} (2 \log (5)-10)\right )^2-(5+\log (5))^2}\)

Input:

Int[(-48*x^2 + (-40 - 80*x + 8*x^2)*Log[5])/(25*x^2 - 10*x^3 + x^4 + (50*x 
 - 20*x^2 + 2*x^3)*Log[5] + (25 - 10*x + x^2)*Log[5]^2),x]
 

Output:

(16*(30 - Log[5] + Log[5]^2 + 2*(6 - Log[5])*(x + (-10 + 2*Log[5])/4)))/(- 
(5 + Log[5])^2 + 4*(x + (-10 + 2*Log[5])/4)^2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1380

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]
 

rule 2345

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]
 

rule 2459

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])
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24

method	result	size
norman	\(\frac {\left (-8 \ln \left (5\right )+48\right ) x +40 \ln \left (5\right )}{\left (-5+x \right ) \left (\ln \left (5\right )+x \right )}\)	\(26\)
gosper	\(-\frac {8 \left (x \ln \left (5\right )-5 \ln \left (5\right )-6 x \right )}{x \ln \left (5\right )+x^{2}-5 \ln \left (5\right )-5 x}\)	\(32\)
risch	\(\frac {\left (-8 \ln \left (5\right )+48\right ) x +40 \ln \left (5\right )}{x \ln \left (5\right )+x^{2}-5 \ln \left (5\right )-5 x}\)	\(32\)
parallelrisch	\(\frac {-8 x \ln \left (5\right )+40 \ln \left (5\right )+48 x}{x \ln \left (5\right )+x^{2}-5 \ln \left (5\right )-5 x}\)	\(32\)
default	\(-\frac {8 \ln \left (5\right ) \left (\ln \left (5\right )-1\right )}{\left (5+\ln \left (5\right )\right ) \left (\ln \left (5\right )+x \right )}+\frac {240}{\left (5+\ln \left (5\right )\right ) \left (-5+x \right )}\)	\(35\)

Input:

int(((8*x^2-80*x-40)*ln(5)-48*x^2)/((x^2-10*x+25)*ln(5)^2+(2*x^3-20*x^2+50 
*x)*ln(5)+x^4-10*x^3+25*x^2),x,method=_RETURNVERBOSE)
 

Output:

((-8*ln(5)+48)*x+40*ln(5))/(-5+x)/(ln(5)+x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=-\frac {8 \, {\left ({\left (x - 5\right )} \log \left (5\right ) - 6 \, x\right )}}{x^{2} + {\left (x - 5\right )} \log \left (5\right ) - 5 \, x} \] Input:

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-2 
0*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2),x, algorithm="fricas")
 

Output:

-8*((x - 5)*log(5) - 6*x)/(x^2 + (x - 5)*log(5) - 5*x)
 

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=- \frac {x \left (-48 + 8 \log {\left (5 \right )}\right ) - 40 \log {\left (5 \right )}}{x^{2} + x \left (-5 + \log {\left (5 \right )}\right ) - 5 \log {\left (5 \right )}} \] Input:

integrate(((8*x**2-80*x-40)*ln(5)-48*x**2)/((x**2-10*x+25)*ln(5)**2+(2*x** 
3-20*x**2+50*x)*ln(5)+x**4-10*x**3+25*x**2),x)
 

Output:

-(x*(-48 + 8*log(5)) - 40*log(5))/(x**2 + x*(-5 + log(5)) - 5*log(5))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=-\frac {8 \, {\left (x {\left (\log \left (5\right ) - 6\right )} - 5 \, \log \left (5\right )\right )}}{x^{2} + x {\left (\log \left (5\right ) - 5\right )} - 5 \, \log \left (5\right )} \] Input:

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-2 
0*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2),x, algorithm="maxima")
 

Output:

-8*(x*(log(5) - 6) - 5*log(5))/(x^2 + x*(log(5) - 5) - 5*log(5))
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=-\frac {8 \, {\left (x \log \left (5\right ) - 6 \, x - 5 \, \log \left (5\right )\right )}}{x^{2} + x \log \left (5\right ) - 5 \, x - 5 \, \log \left (5\right )} \] Input:

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-2 
0*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2),x, algorithm="giac")
 

Output:

-8*(x*log(5) - 6*x - 5*log(5))/(x^2 + x*log(5) - 5*x - 5*log(5))
 

Mupad [B] (verification not implemented)

Time = 4.91 (sec) , antiderivative size = 684, normalized size of antiderivative = 32.57 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=\text {Too large to display} \] Input:

int(-(log(5)*(80*x - 8*x^2 + 40) + 48*x^2)/(log(5)^2*(x^2 - 10*x + 25) + l 
og(5)*(50*x - 20*x^2 + 2*x^3) + 25*x^2 - 10*x^3 + x^4),x)
 

Output:

symsum(log(1843200*x*log(5) - 2880000*root(110592000*log(5) + 33177600*log 
(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456* 
log(5)^6 - 92160000, z, k)*log(5) - 2880000*root(110592000*log(5) + 331776 
00*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 1 
47456*log(5)^6 - 92160000, z, k)*x - 2304000*log(5) - 1136000*root(1105920 
00*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 26 
54208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^2 - 140800*root( 
110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5) 
^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^3 + 3840* 
root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*l 
og(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^4 + 
5120*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367 
616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5) 
^5 + 640*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 1 
6367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*lo 
g(5)^6 - 1904640*x*log(5)^2 + 122880*x*log(5)^3 - 61440*x*log(5)^4 + 39168 
00*log(5)^2 - 1858560*log(5)^3 + 261120*log(5)^4 - 15360*log(5)^5 - 115200 
0*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616 
*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x*log(5) 
- 118400*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 ...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.67 \[ \int \frac {-48 x^2+\left (-40-80 x+8 x^2\right ) \log (5)}{25 x^2-10 x^3+x^4+\left (50 x-20 x^2+2 x^3\right ) \log (5)+\left (25-10 x+x^2\right ) \log ^2(5)} \, dx=\frac {8 \,\mathrm {log}\left (5\right ) x^{2}+40 \,\mathrm {log}\left (5\right )-48 x^{2}}{x \mathrm {log}\left (5\right )^{2}-5 \mathrm {log}\left (5\right )^{2}+\mathrm {log}\left (5\right ) x^{2}-10 \,\mathrm {log}\left (5\right ) x +25 \,\mathrm {log}\left (5\right )-5 x^{2}+25 x} \] Input:

int(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-20*x^2+ 
50*x)*log(5)+x^4-10*x^3+25*x^2),x)
 

Output:

(8*(log(5)*x**2 + 5*log(5) - 6*x**2))/(log(5)**2*x - 5*log(5)**2 + log(5)* 
x**2 - 10*log(5)*x + 25*log(5) - 5*x**2 + 25*x)