3.95.66 \(\int \frac {10 x-2 x^2-10 x \log (x)+((10 x-2 x^2) \log (x)+(450-180 x+18 x^2) \log ^2(x)) \log (\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)})+((65 x+77 x^2-18 x^3) \log (x)+(2925+2880 x-1503 x^2+162 x^3) \log ^2(x)) \log ^2(\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)})}{((5 x-x^2) \log (x)+(225-90 x+9 x^2) \log ^2(x)) \log ^2(\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)})} \, dx\)

Optimal. Leaf size=28 \[ x+(2+3 x)^2+\frac {2 x}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \]

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Rubi [F]  time = 6.44, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {10 x-2 x^2-10 x \log (x)+\left (\left (10 x-2 x^2\right ) \log (x)+\left (450-180 x+18 x^2\right ) \log ^2(x)\right ) \log \left (\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)}\right )+\left (\left (65 x+77 x^2-18 x^3\right ) \log (x)+\left (2925+2880 x-1503 x^2+162 x^3\right ) \log ^2(x)\right ) \log ^2\left (\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)}\right )}{\left (\left (5 x-x^2\right ) \log (x)+\left (225-90 x+9 x^2\right ) \log ^2(x)\right ) \log ^2\left (\frac {x+(45-9 x) \log (x)}{(-5+x) \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*x - 2*x^2 - 10*x*Log[x] + ((10*x - 2*x^2)*Log[x] + (450 - 180*x + 18*x^2)*Log[x]^2)*Log[(x + (45 - 9*x
)*Log[x])/((-5 + x)*Log[x])] + ((65*x + 77*x^2 - 18*x^3)*Log[x] + (2925 + 2880*x - 1503*x^2 + 162*x^3)*Log[x]^
2)*Log[(x + (45 - 9*x)*Log[x])/((-5 + x)*Log[x])]^2)/(((5*x - x^2)*Log[x] + (225 - 90*x + 9*x^2)*Log[x]^2)*Log
[(x + (45 - 9*x)*Log[x])/((-5 + x)*Log[x])]^2),x]

[Out]

13*x + 9*x^2 - 10*Defer[Int][1/((-x - 45*Log[x] + 9*x*Log[x])*Log[-9 + x/((-5 + x)*Log[x])]^2), x] - 50*Defer[
Int][1/((-5 + x)*(-x - 45*Log[x] + 9*x*Log[x])*Log[-9 + x/((-5 + x)*Log[x])]^2), x] - 2*Defer[Int][x/(Log[x]*(
-x - 45*Log[x] + 9*x*Log[x])*Log[-9 + x/((-5 + x)*Log[x])]^2), x] + 2*Defer[Int][Log[-9 + x/((-5 + x)*Log[x])]
^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x-2 x^2-10 x \log (x)+2 (-5+x) \log (x) (-x+9 (-5+x) \log (x)) \log \left (-9+\frac {x}{(-5+x) \log (x)}\right )+\left (-65-77 x+18 x^2\right ) \log (x) (-x+9 (-5+x) \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}{(5-x) \log (x) (x-9 (-5+x) \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=\int \left (13+18 x-\frac {2 x (-5+x+5 \log (x))}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}+\frac {2}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )}\right ) \, dx\\ &=13 x+9 x^2-2 \int \frac {x (-5+x+5 \log (x))}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=13 x+9 x^2-2 \int \left (\frac {-5+x+5 \log (x)}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}+\frac {5 (-5+x+5 \log (x))}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}\right ) \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=13 x+9 x^2-2 \int \frac {-5+x+5 \log (x)}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \frac {-5+x+5 \log (x)}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=13 x+9 x^2-2 \int \left (\frac {5}{(-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}-\frac {5}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}+\frac {x}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}\right ) \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \left (\frac {5}{(-5+x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}-\frac {5}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}+\frac {x}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}\right ) \, dx\\ &=13 x+9 x^2-2 \int \frac {x}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \frac {1}{(-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+10 \int \frac {1}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \frac {x}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-50 \int \frac {1}{(-5+x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+50 \int \frac {1}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=13 x+9 x^2-2 \int \frac {x}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \left (\frac {1}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}+\frac {5}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )}\right ) \, dx-10 \int \frac {1}{(-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+10 \int \frac {1}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-50 \int \frac {1}{(-5+x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+50 \int \frac {1}{(-5+x) \log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ &=13 x+9 x^2-2 \int \frac {x}{\log (x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx+2 \int \frac {1}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-10 \int \frac {1}{(-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx-50 \int \frac {1}{(-5+x) (-x-45 \log (x)+9 x \log (x)) \log ^2\left (-9+\frac {x}{(-5+x) \log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 28, normalized size = 1.00 \begin {gather*} 13 x+9 x^2+\frac {2 x}{\log \left (-9+\frac {x}{(-5+x) \log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x - 2*x^2 - 10*x*Log[x] + ((10*x - 2*x^2)*Log[x] + (450 - 180*x + 18*x^2)*Log[x]^2)*Log[(x + (45
- 9*x)*Log[x])/((-5 + x)*Log[x])] + ((65*x + 77*x^2 - 18*x^3)*Log[x] + (2925 + 2880*x - 1503*x^2 + 162*x^3)*L
og[x]^2)*Log[(x + (45 - 9*x)*Log[x])/((-5 + x)*Log[x])]^2)/(((5*x - x^2)*Log[x] + (225 - 90*x + 9*x^2)*Log[x]^
2)*Log[(x + (45 - 9*x)*Log[x])/((-5 + x)*Log[x])]^2),x]

[Out]

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

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fricas [B]  time = 0.66, size = 63, normalized size = 2.25 \begin {gather*} \frac {{\left (9 \, x^{2} + 13 \, x\right )} \log \left (-\frac {9 \, {\left (x - 5\right )} \log \relax (x) - x}{{\left (x - 5\right )} \log \relax (x)}\right ) + 2 \, x}{\log \left (-\frac {9 \, {\left (x - 5\right )} \log \relax (x) - x}{{\left (x - 5\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^3-1503*x^2+2880*x+2925)*log(x)^2+(-18*x^3+77*x^2+65*x)*log(x))*log(((-9*x+45)*log(x)+x)/log
(x)/(x-5))^2+((18*x^2-180*x+450)*log(x)^2+(-2*x^2+10*x)*log(x))*log(((-9*x+45)*log(x)+x)/log(x)/(x-5))-10*x*lo
g(x)-2*x^2+10*x)/((9*x^2-90*x+225)*log(x)^2+(-x^2+5*x)*log(x))/log(((-9*x+45)*log(x)+x)/log(x)/(x-5))^2,x, alg
orithm="fricas")

[Out]

((9*x^2 + 13*x)*log(-(9*(x - 5)*log(x) - x)/((x - 5)*log(x))) + 2*x)/log(-(9*(x - 5)*log(x) - x)/((x - 5)*log(
x)))

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giac [A]  time = 0.63, size = 39, normalized size = 1.39 \begin {gather*} 9 \, x^{2} + 13 \, x - \frac {2 \, x}{\log \left (x \log \relax (x) - 5 \, \log \relax (x)\right ) - \log \left (-9 \, x \log \relax (x) + x + 45 \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^3-1503*x^2+2880*x+2925)*log(x)^2+(-18*x^3+77*x^2+65*x)*log(x))*log(((-9*x+45)*log(x)+x)/log
(x)/(x-5))^2+((18*x^2-180*x+450)*log(x)^2+(-2*x^2+10*x)*log(x))*log(((-9*x+45)*log(x)+x)/log(x)/(x-5))-10*x*lo
g(x)-2*x^2+10*x)/((9*x^2-90*x+225)*log(x)^2+(-x^2+5*x)*log(x))/log(((-9*x+45)*log(x)+x)/log(x)/(x-5))^2,x, alg
orithm="giac")

[Out]

9*x^2 + 13*x - 2*x/(log(x*log(x) - 5*log(x)) - log(-9*x*log(x) + x + 45*log(x)))

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maple [C]  time = 0.40, size = 395, normalized size = 14.11

method	result	size
risch	\(9 x^{2}+13 x +\frac {4 i x}{\pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{\ln \relax (x ) \left (x -5\right )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{\ln \relax (x ) \left (x -5\right )}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{\ln \relax (x ) \left (x -5\right )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-\pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{\ln \relax (x ) \left (x -5\right )}\right )^{3}-\pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}{\ln \relax (x ) \left (x -5\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-2 \pi +4 i \ln \relax (3)-2 i \ln \left (x -5\right )-2 i \ln \left (\ln \relax (x )\right )+2 i \ln \left (\left (\ln \relax (x )-\frac {1}{9}\right ) x -5 \ln \relax (x )\right )}\)	\(395\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((162*x^3-1503*x^2+2880*x+2925)*ln(x)^2+(-18*x^3+77*x^2+65*x)*ln(x))*ln(((-9*x+45)*ln(x)+x)/ln(x)/(x-5))^
2+((18*x^2-180*x+450)*ln(x)^2+(-2*x^2+10*x)*ln(x))*ln(((-9*x+45)*ln(x)+x)/ln(x)/(x-5))-10*x*ln(x)-2*x^2+10*x)/
((9*x^2-90*x+225)*ln(x)^2+(-x^2+5*x)*ln(x))/ln(((-9*x+45)*ln(x)+x)/ln(x)/(x-5))^2,x,method=_RETURNVERBOSE)

[Out]

9*x^2+13*x+4*I*x/(Pi*csgn(I/(x-5))*csgn(I*((ln(x)-1/9)*x-5*ln(x)))*csgn(I/(x-5)*((ln(x)-1/9)*x-5*ln(x)))-Pi*cs
gn(I/(x-5))*csgn(I/(x-5)*((ln(x)-1/9)*x-5*ln(x)))^2+2*Pi*csgn(I/ln(x)*((ln(x)-1/9)*x-5*ln(x))/(x-5))^2-Pi*csgn
(I*((ln(x)-1/9)*x-5*ln(x)))*csgn(I/(x-5)*((ln(x)-1/9)*x-5*ln(x)))^2+Pi*csgn(I/(x-5)*((ln(x)-1/9)*x-5*ln(x)))^3
-Pi*csgn(I/(x-5)*((ln(x)-1/9)*x-5*ln(x)))*csgn(I/ln(x)*((ln(x)-1/9)*x-5*ln(x))/(x-5))^2+Pi*csgn(I/(x-5)*((ln(x
)-1/9)*x-5*ln(x)))*csgn(I/ln(x)*((ln(x)-1/9)*x-5*ln(x))/(x-5))*csgn(I/ln(x))-Pi*csgn(I/ln(x)*((ln(x)-1/9)*x-5*
ln(x))/(x-5))^3-Pi*csgn(I/ln(x)*((ln(x)-1/9)*x-5*ln(x))/(x-5))^2*csgn(I/ln(x))-2*Pi+4*I*ln(3)-2*I*ln(x-5)-2*I*
ln(ln(x))+2*I*ln((ln(x)-1/9)*x-5*ln(x)))

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maxima [B]  time = 0.44, size = 78, normalized size = 2.79 \begin {gather*} \frac {{\left (9 \, x^{2} + 13 \, x\right )} \log \left (-9 \, {\left (x - 5\right )} \log \relax (x) + x\right ) - {\left (9 \, x^{2} + 13 \, x\right )} \log \left (x - 5\right ) - {\left (9 \, x^{2} + 13 \, x\right )} \log \left (\log \relax (x)\right ) + 2 \, x}{\log \left (-9 \, {\left (x - 5\right )} \log \relax (x) + x\right ) - \log \left (x - 5\right ) - \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x^3-1503*x^2+2880*x+2925)*log(x)^2+(-18*x^3+77*x^2+65*x)*log(x))*log(((-9*x+45)*log(x)+x)/log
(x)/(x-5))^2+((18*x^2-180*x+450)*log(x)^2+(-2*x^2+10*x)*log(x))*log(((-9*x+45)*log(x)+x)/log(x)/(x-5))-10*x*lo
g(x)-2*x^2+10*x)/((9*x^2-90*x+225)*log(x)^2+(-x^2+5*x)*log(x))/log(((-9*x+45)*log(x)+x)/log(x)/(x-5))^2,x, alg
orithm="maxima")

[Out]

((9*x^2 + 13*x)*log(-9*(x - 5)*log(x) + x) - (9*x^2 + 13*x)*log(x - 5) - (9*x^2 + 13*x)*log(log(x)) + 2*x)/(lo
g(-9*(x - 5)*log(x) + x) - log(x - 5) - log(log(x)))

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mupad [B]  time = 8.38, size = 184, normalized size = 6.57 \begin {gather*} 451\,x+90\,\ln \relax (x)+\frac {14000}{x+5}-\frac {\frac {2\,{\left (x-5\right )}^2\,\left (27\,x^3-305\,x^2+1075\,x-1125\right )}{25\,\left (x+5\right )}+\frac {2\,x\,\ln \relax (x)\,\left (36\,x^3-525\,x^2+2600\,x-4375\right )}{5\,\left (x+5\right )}}{x+5\,\ln \relax (x)-5}+\frac {2\,x+\frac {2\,\ln \left (\frac {x-\ln \relax (x)\,\left (9\,x-45\right )}{\ln \relax (x)\,\left (x-5\right )}\right )\,\ln \relax (x)\,\left (x-5\right )\,\left (x+45\,\ln \relax (x)-9\,x\,\ln \relax (x)\right )}{x+5\,\ln \relax (x)-5}}{\ln \left (\frac {x-\ln \relax (x)\,\left (9\,x-45\right )}{\ln \relax (x)\,\left (x-5\right )}\right )}-\ln \relax (x)\,\left (36\,x-\frac {18\,x^2}{5}\right )-37\,x^2+\frac {54\,x^3}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log((x - log(x)*(9*x - 45))/(log(x)*(x - 5)))*(log(x)^2*(18*x^2 - 180*x + 450) + log(x)*(10*x - 2*
x^2)) - 10*x*log(x) - 2*x^2 + log((x - log(x)*(9*x - 45))/(log(x)*(x - 5)))^2*(log(x)^2*(2880*x - 1503*x^2 + 1
62*x^3 + 2925) + log(x)*(65*x + 77*x^2 - 18*x^3)))/(log((x - log(x)*(9*x - 45))/(log(x)*(x - 5)))^2*(log(x)^2*
(9*x^2 - 90*x + 225) + log(x)*(5*x - x^2))),x)

[Out]

451*x + 90*log(x) + 14000/(x + 5) - ((2*(x - 5)^2*(1075*x - 305*x^2 + 27*x^3 - 1125))/(25*(x + 5)) + (2*x*log(
x)*(2600*x - 525*x^2 + 36*x^3 - 4375))/(5*(x + 5)))/(x + 5*log(x) - 5) + (2*x + (2*log((x - log(x)*(9*x - 45))
/(log(x)*(x - 5)))*log(x)*(x - 5)*(x + 45*log(x) - 9*x*log(x)))/(x + 5*log(x) - 5))/log((x - log(x)*(9*x - 45)
)/(log(x)*(x - 5))) - log(x)*(36*x - (18*x^2)/5) - 37*x^2 + (54*x^3)/25

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sympy [A]  time = 0.50, size = 29, normalized size = 1.04 \begin {gather*} 9 x^{2} + 13 x + \frac {2 x}{\log {\left (\frac {x + \left (45 - 9 x\right ) \log {\relax (x )}}{\left (x - 5\right ) \log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((162*x**3-1503*x**2+2880*x+2925)*ln(x)**2+(-18*x**3+77*x**2+65*x)*ln(x))*ln(((-9*x+45)*ln(x)+x)/ln
(x)/(x-5))**2+((18*x**2-180*x+450)*ln(x)**2+(-2*x**2+10*x)*ln(x))*ln(((-9*x+45)*ln(x)+x)/ln(x)/(x-5))-10*x*ln(
x)-2*x**2+10*x)/((9*x**2-90*x+225)*ln(x)**2+(-x**2+5*x)*ln(x))/ln(((-9*x+45)*ln(x)+x)/ln(x)/(x-5))**2,x)

[Out]

9*x**2 + 13*x + 2*x/log((x + (45 - 9*x)*log(x))/((x - 5)*log(x)))

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