\(\int \frac {9 x^2+x^2 \log (5)+(90 x+18 x^2+2 x^2 \log (5)) \log (\frac {-45-9 x-x \log (5)}{\log (5)})}{45+9 x+x \log (5)} \, dx\) [1264]

Optimal result

Integrand size = 57, antiderivative size = 20 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^2 \log \left (-x+\frac {9 (-5-x)}{\log (5)}\right ) \] Output:

x^2*ln(9/ln(5)*(-x-5)-x)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^2 \log \left (-\frac {45+x (9+\log (5))}{\log (5)}\right ) \] Input:

Integrate[(9*x^2 + x^2*Log[5] + (90*x + 18*x^2 + 2*x^2*Log[5])*Log[(-45 - 
9*x - x*Log[5])/Log[5]])/(45 + 9*x + x*Log[5]),x]
 

Output:

x^2*Log[-((45 + x*(9 + Log[5]))/Log[5])]
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 6, 7293, 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.

\(\displaystyle \int \frac {9 x^2+x^2 \log (5)+\left (18 x^2+2 x^2 \log (5)+90 x\right ) \log \left (\frac {-9 x+x (-\log (5))-45}{\log (5)}\right )}{9 x+x \log (5)+45} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {9 x^2+x^2 \log (5)+\left (18 x^2+2 x^2 \log (5)+90 x\right ) \log \left (\frac {-9 x+x (-\log (5))-45}{\log (5)}\right )}{x (9+\log (5))+45}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {x^2 (9+\log (5))+\left (18 x^2+2 x^2 \log (5)+90 x\right ) \log \left (\frac {-9 x+x (-\log (5))-45}{\log (5)}\right )}{x (9+\log (5))+45}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2 (9+\log (5))}{x (9+\log (5))+45}+2 x \log \left (-\frac {x (9+\log (5))}{\log (5)}-\frac {45}{\log (5)}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle x^2 \log \left (-\frac {x (9+\log (5))}{\log (5)}-\frac {45}{\log (5)}\right )\)

Input:

Int[(9*x^2 + x^2*Log[5] + (90*x + 18*x^2 + 2*x^2*Log[5])*Log[(-45 - 9*x - 
x*Log[5])/Log[5]])/(45 + 9*x + x*Log[5]),x]
 

Output:

x^2*Log[-45/Log[5] - (x*(9 + Log[5]))/Log[5]]
 

Defintions of rubi rules used

rule 6

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
norman	\(x^{2} \ln \left (\frac {-x \ln \left (5\right )-9 x -45}{\ln \left (5\right )}\right )\)	\(21\)
risch	\(x^{2} \ln \left (\frac {-x \ln \left (5\right )-9 x -45}{\ln \left (5\right )}\right )\)	\(21\)
parallelrisch	\(\ln \left (-\frac {x \ln \left (5\right )+9 x +45}{\ln \left (5\right )}\right ) x^{2}\)	\(21\)
parts	\(\left (\ln \left (5\right )+9\right ) \left (\frac {\frac {x^{2} \ln \left (5\right )}{2}+\frac {9 x^{2}}{2}-45 x}{\left (\ln \left (5\right )+9\right )^{2}}+\frac {2025 \ln \left (x \ln \left (5\right )+9 x +45\right )}{\left (\ln \left (5\right )+9\right )^{3}}\right )+\frac {2 \ln \left (5\right )^{2} \left (\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2} \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{2}-\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{4}+\frac {45 \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right ) \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )-\frac {45 \left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}+\frac {2025}{\ln \left (5\right )}}{\ln \left (5\right )}\right )}{\left (-\ln \left (5\right )-9\right )^{2}}\)	\(196\)
derivativedivides	\(\frac {\ln \left (5\right ) \left (-\frac {2 \ln \left (5\right ) \left (\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2} \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{2}-\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{4}\right )}{\ln \left (5\right )+9}-\frac {90 \left (\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right ) \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )-\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}+\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {\ln \left (5\right ) \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{2 \left (\ln \left (5\right )+9\right )}-\frac {90 \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {2025 \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right ) \left (\ln \left (5\right )+9\right )}\right )}{-\ln \left (5\right )-9}\)	\(248\)
default	\(\frac {\ln \left (5\right ) \left (-\frac {2 \ln \left (5\right ) \left (\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2} \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{2}-\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{4}\right )}{\ln \left (5\right )+9}-\frac {90 \left (\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right ) \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )-\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}+\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {\ln \left (5\right ) \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{2 \left (\ln \left (5\right )+9\right )}-\frac {90 \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {2025 \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right ) \left (\ln \left (5\right )+9\right )}\right )}{-\ln \left (5\right )-9}\)	\(248\)

Input:

int(((2*x^2*ln(5)+18*x^2+90*x)*ln((-x*ln(5)-9*x-45)/ln(5))+x^2*ln(5)+9*x^2 
)/(x*ln(5)+9*x+45),x,method=_RETURNVERBOSE)
 

Output:

x^2*ln((-x*ln(5)-9*x-45)/ln(5))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log \left (-\frac {x \log \left (5\right ) + 9 \, x + 45}{\log \left (5\right )}\right ) \] Input:

integrate(((2*x^2*log(5)+18*x^2+90*x)*log((-x*log(5)-9*x-45)/log(5))+x^2*l 
og(5)+9*x^2)/(x*log(5)+9*x+45),x, algorithm="fricas")
 

Output:

x^2*log(-(x*log(5) + 9*x + 45)/log(5))
 

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log {\left (\frac {- 9 x - x \log {\left (5 \right )} - 45}{\log {\left (5 \right )}} \right )} \] Input:

integrate(((2*x**2*ln(5)+18*x**2+90*x)*ln((-x*ln(5)-9*x-45)/ln(5))+x**2*ln 
(5)+9*x**2)/(x*ln(5)+9*x+45),x)
 

Output:

x**2*log((-9*x - x*log(5) - 45)/log(5))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (18) = 36\).

Time = 0.12 (sec) , antiderivative size = 577, normalized size of antiderivative = 28.85 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=\text {Too large to display} \] Input:

integrate(((2*x^2*log(5)+18*x^2+90*x)*log((-x*log(5)-9*x-45)/log(5))+x^2*l 
og(5)+9*x^2)/(x*log(5)+9*x+45),x, algorithm="maxima")
 

Output:

((x^2*(log(5) + 9) - 90*x)/(log(5)^2 + 18*log(5) + 81) + 4050*log(x*(log(5 
) + 9) + 45)/(log(5)^3 + 27*log(5)^2 + 243*log(5) + 729))*log(5)*log(-x - 
9*x/log(5) - 45/log(5)) + 1/2*((x^2*(log(5) + 9) - 90*x)/(log(5)^2 + 18*lo 
g(5) + 81) + 4050*log(x*(log(5) + 9) + 45)/(log(5)^3 + 27*log(5)^2 + 243*l 
og(5) + 729))*log(5) - 1/2*((log(5)^3 + 18*log(5)^2 + 81*log(5))*x^2 + 405 
0*log(5)*log(x*(log(5) + 9) + 45)^2 - 270*(log(5)^2 + 9*log(5))*x + 12150* 
log(5)*log(x*(log(5) + 9) + 45))*(9/log(5) + 1)*log(5)/(log(5)^4 + 36*log( 
5)^3 + 486*log(5)^2 + 2916*log(5) + 6561) + 9*((x^2*(log(5) + 9) - 90*x)/( 
log(5)^2 + 18*log(5) + 81) + 4050*log(x*(log(5) + 9) + 45)/(log(5)^3 + 27* 
log(5)^2 + 243*log(5) + 729))*log(-x - 9*x/log(5) - 45/log(5)) + 90*(x/(lo 
g(5) + 9) - 45*log(x*(log(5) + 9) + 45)/(log(5)^2 + 18*log(5) + 81))*log(- 
x - 9*x/log(5) - 45/log(5)) - 9/2*((log(5)^3 + 18*log(5)^2 + 81*log(5))*x^ 
2 + 4050*log(5)*log(x*(log(5) + 9) + 45)^2 - 270*(log(5)^2 + 9*log(5))*x + 
 12150*log(5)*log(x*(log(5) + 9) + 45))*(9/log(5) + 1)/(log(5)^4 + 36*log( 
5)^3 + 486*log(5)^2 + 2916*log(5) + 6561) + 45*(45*log(5)*log(x*(log(5) + 
9) + 45)^2 - 2*(log(5)^2 + 9*log(5))*x + 90*log(5)*log(x*(log(5) + 9) + 45 
))*(9/log(5) + 1)/(log(5)^3 + 27*log(5)^2 + 243*log(5) + 729) + 9/2*(x^2*( 
log(5) + 9) - 90*x)/(log(5)^2 + 18*log(5) + 81) + 18225*log(x*(log(5) + 9) 
 + 45)/(log(5)^3 + 27*log(5)^2 + 243*log(5) + 729)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log \left (-x \log \left (5\right ) - 9 \, x - 45\right ) - x^{2} \log \left (\log \left (5\right )\right ) \] Input:

integrate(((2*x^2*log(5)+18*x^2+90*x)*log((-x*log(5)-9*x-45)/log(5))+x^2*l 
og(5)+9*x^2)/(x*log(5)+9*x+45),x, algorithm="giac")
 

Output:

x^2*log(-x*log(5) - 9*x - 45) - x^2*log(log(5))
 

Mupad [B] (verification not implemented)

Time = 7.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=-x^2\,\left (\ln \left (\ln \left (5\right )\right )-\ln \left (-9\,x-x\,\ln \left (5\right )-45\right )\right ) \] Input:

int((log(-(9*x + x*log(5) + 45)/log(5))*(90*x + 2*x^2*log(5) + 18*x^2) + x 
^2*log(5) + 9*x^2)/(9*x + x*log(5) + 45),x)
 

Output:

-x^2*(log(log(5)) - log(- 9*x - x*log(5) - 45))
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 5.60 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=\frac {2025 \,\mathrm {log}\left (\mathrm {log}\left (5\right ) x +9 x +45\right )+\mathrm {log}\left (\frac {-\mathrm {log}\left (5\right ) x -9 x -45}{\mathrm {log}\left (5\right )}\right ) \mathrm {log}\left (5\right )^{2} x^{2}+18 \,\mathrm {log}\left (\frac {-\mathrm {log}\left (5\right ) x -9 x -45}{\mathrm {log}\left (5\right )}\right ) \mathrm {log}\left (5\right ) x^{2}+81 \,\mathrm {log}\left (\frac {-\mathrm {log}\left (5\right ) x -9 x -45}{\mathrm {log}\left (5\right )}\right ) x^{2}-2025 \,\mathrm {log}\left (\frac {-\mathrm {log}\left (5\right ) x -9 x -45}{\mathrm {log}\left (5\right )}\right )}{\mathrm {log}\left (5\right )^{2}+18 \,\mathrm {log}\left (5\right )+81} \] Input:

int(((2*x^2*log(5)+18*x^2+90*x)*log((-x*log(5)-9*x-45)/log(5))+x^2*log(5)+ 
9*x^2)/(x*log(5)+9*x+45),x)
 

Output:

(2025*log(log(5)*x + 9*x + 45) + log(( - log(5)*x - 9*x - 45)/log(5))*log( 
5)**2*x**2 + 18*log(( - log(5)*x - 9*x - 45)/log(5))*log(5)*x**2 + 81*log( 
( - log(5)*x - 9*x - 45)/log(5))*x**2 - 2025*log(( - log(5)*x - 9*x - 45)/ 
log(5)))/(log(5)**2 + 18*log(5) + 81)