\(\int \frac {9 x^2+x^2 \log (5)+(90 x+18 x^2+2 x^2 \log (5)) \log (\frac {-45-9 x-x \log (5)}{\log (5)})}{45+9 x+x \log (5)} \, dx\) [10264]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 6820, 14, 45, 2442} \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^2 \log \left (-\frac {x (9+\log (5))}{\log (5)}-\frac {45}{\log (5)}\right ) \]

[In]

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]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \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+x (9+\log (5))} \, dx \\ & = \int \frac {x^2 (9+\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+x (9+\log (5))} \, dx \\ & = \int x \left (\frac {x (9+\log (5))}{45+x (9+\log (5))}+2 \log \left (-\frac {45+x (9+\log (5))}{\log (5)}\right )\right ) \, dx \\ & = \int \left (\frac {x^2 (9+\log (5))}{45+x (9+\log (5))}+2 x \log \left (-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}\right )\right ) \, dx \\ & = 2 \int x \log \left (-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}\right ) \, dx+(9+\log (5)) \int \frac {x^2}{45+x (9+\log (5))} \, dx \\ & = x^2 \log \left (-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}\right )+(9+\log (5)) \int \left (-\frac {45}{(9+\log (5))^2}+\frac {x}{9+\log (5)}+\frac {2025}{(9+\log (5))^2 (45+x (9+\log (5)))}\right ) \, dx+\frac {(9+\log (5)) \int \frac {x^2}{-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}} \, dx}{\log (5)} \\ & = \frac {x^2}{2}-\frac {45 x}{9+\log (5)}+\frac {2025 \log (45+x (9+\log (5)))}{(9+\log (5))^2}+x^2 \log \left (-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}\right )+\frac {(9+\log (5)) \int \left (\frac {45 \log (5)}{(9+\log (5))^2}-\frac {x \log (5)}{9+\log (5)}+\frac {2025 \log (5)}{(9+\log (5))^2 (-45-x (9+\log (5)))}\right ) \, dx}{\log (5)} \\ & = x^2 \log \left (-\frac {45}{\log (5)}-\frac {x (9+\log (5))}{\log (5)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^2 \log \left (-\frac {45+x (9+\log (5))}{\log (5)}\right ) \]

[In]

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]

[Out]

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

Maple [A] (verified)

Time = 0.20 (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	\(\frac {\ln \left (5\right ) \left (-\frac {2 \ln \left (5\right ) \left (\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2} \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{2}-\frac {\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{4}\right )}{\ln \left (5\right )+9}-\frac {90 \left (\left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right ) \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )-\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}+\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {\ln \left (5\right ) \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )^{2}}{2 \left (\ln \left (5\right )+9\right )}-\frac {90 \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right )+9}-\frac {2025 \ln \left (\frac {\left (-\ln \left (5\right )-9\right ) x}{\ln \left (5\right )}-\frac {45}{\ln \left (5\right )}\right )}{\ln \left (5\right ) \left (\ln \left (5\right )+9\right )}\right )}{-\ln \left (5\right )-9}\)	\(248\)

[In]

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=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log \left (-\frac {x \log \left (5\right ) + 9 \, x + 45}{\log \left (5\right )}\right ) \]

[In]

integrate(((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, al
gorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log {\left (\frac {- 9 x - x \log {\left (5 \right )} - 45}{\log {\left (5 \right )}} \right )} \]

[In]

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)

[Out]

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 = 0.29 (sec) , antiderivative size = 577, normalized size of antiderivative = 28.85 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=\text {Too large to display} \]

[In]

integrate(((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, al
gorithm="maxima")

[Out]

((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*
log(5) + 81) + 4050*log(x*(log(5) + 9) + 45)/(log(5)^3 + 27*log(5)^2 + 243*log(5) + 729))*log(5) - 1/2*((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 + 12
150*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/(log(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*lo
g(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)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=x^{2} \log \left (-x \log \left (5\right ) - 9 \, x - 45\right ) - x^{2} \log \left (\log \left (5\right )\right ) \]

[In]

integrate(((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, al
gorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.71 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {9 x^2+x^2 \log (5)+\left (90 x+18 x^2+2 x^2 \log (5)\right ) \log \left (\frac {-45-9 x-x \log (5)}{\log (5)}\right )}{45+9 x+x \log (5)} \, dx=-x^2\,\left (\ln \left (\ln \left (5\right )\right )-\ln \left (-9\,x-x\,\ln \left (5\right )-45\right )\right ) \]

[In]

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)

[Out]

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