\(\int \frac {144-8 x+8 x \log (3)+(-108-48 x+16 x^2+(-36+8 x) \log (3)) \log (\frac {1}{9} (45-10 x))+(9 x^2-2 x^3) \log ^2(\frac {1}{9} (45-10 x))}{-144+32 x+(72 x-16 x^2) \log (\frac {1}{9} (45-10 x))+(-9 x^2+2 x^3) \log ^2(\frac {1}{9} (45-10 x))} \, dx\) [6615]

Optimal result

Integrand size = 115, antiderivative size = 29 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-25-x+\frac {x (3+\log (3))}{-x+\frac {4}{\log \left (5-\frac {10 x}{9}\right )}} \]

[Out]

(3+ln(3))*x/(4/ln(-10/9*x+5)-x)-x-25

Rubi [F]

\[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=\int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx \]

[In]

Int[(144 - 8*x + 8*x*Log[3] + (-108 - 48*x + 16*x^2 + (-36 + 8*x)*Log[3])*Log[(45 - 10*x)/9] + (9*x^2 - 2*x^3)
*Log[(45 - 10*x)/9]^2)/(-144 + 32*x + (72*x - 16*x^2)*Log[(45 - 10*x)/9] + (-9*x^2 + 2*x^3)*Log[(45 - 10*x)/9]
^2),x]

[Out]

-x + 4*(3 + Log[3])*Defer[Int][(-4 + x*Log[5 - (10*x)/9])^(-2), x] + 16*(3 + Log[3])*Defer[Int][1/(x*(-4 + x*L
og[5 - (10*x)/9])^2), x] + 36*(3 + Log[3])*Defer[Int][1/((-9 + 2*x)*(-4 + x*Log[5 - (10*x)/9])^2), x] + 4*(3 +
Log[3])*Defer[Int][1/(x*(-4 + x*Log[5 - (10*x)/9])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {144+x (-8+8 \log (3))+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx \\ & = \int \frac {-144-8 x (-1+\log (3))-4 (-9+2 x) (3+2 x+\log (3)) \log \left (5-\frac {10 x}{9}\right )-(9-2 x) x^2 \log ^2\left (5-\frac {10 x}{9}\right )}{(9-2 x) \left (4-x \log \left (5-\frac {10 x}{9}\right )\right )^2} \, dx \\ & = \int \left (-1+\frac {8 \left (-18+4 x+x^2\right ) (3+\log (3))}{x (-9+2 x) \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2}+\frac {4 (3+\log (3))}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )}\right ) \, dx \\ & = -x+(4 (3+\log (3))) \int \frac {1}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )} \, dx+(8 (3+\log (3))) \int \frac {-18+4 x+x^2}{x (-9+2 x) \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2} \, dx \\ & = -x+(4 (3+\log (3))) \int \frac {1}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )} \, dx+(8 (3+\log (3))) \int \left (\frac {1}{2 \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2}+\frac {2}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2}+\frac {9}{2 (-9+2 x) \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2}\right ) \, dx \\ & = -x+(4 (3+\log (3))) \int \frac {1}{\left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2} \, dx+(4 (3+\log (3))) \int \frac {1}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )} \, dx+(16 (3+\log (3))) \int \frac {1}{x \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2} \, dx+(36 (3+\log (3))) \int \frac {1}{(-9+2 x) \left (-4+x \log \left (5-\frac {10 x}{9}\right )\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-x-\frac {4 (3+\log (3))}{-4+x \log \left (5-\frac {10 x}{9}\right )} \]

[In]

Integrate[(144 - 8*x + 8*x*Log[3] + (-108 - 48*x + 16*x^2 + (-36 + 8*x)*Log[3])*Log[(45 - 10*x)/9] + (9*x^2 -
2*x^3)*Log[(45 - 10*x)/9]^2)/(-144 + 32*x + (72*x - 16*x^2)*Log[(45 - 10*x)/9] + (-9*x^2 + 2*x^3)*Log[(45 - 10
*x)/9]^2),x]

[Out]

-x - (4*(3 + Log[3]))/(-4 + x*Log[5 - (10*x)/9])

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17

method	result	size
norman	\(\frac {4 x -x^{2} \ln \left (-\frac {10 x}{9}+5\right )-12-4 \ln \left (3\right )}{\ln \left (-\frac {10 x}{9}+5\right ) x -4}\)	\(34\)
risch	\(-x -\frac {12}{\ln \left (-\frac {10 x}{9}+5\right ) x -4}-\frac {4 \ln \left (3\right )}{\ln \left (-\frac {10 x}{9}+5\right ) x -4}\)	\(35\)
parallelrisch	\(-\frac {-96+4 x^{2} \ln \left (-\frac {10 x}{9}+5\right )+36 \ln \left (-\frac {10 x}{9}+5\right ) x +16 \ln \left (3\right )-16 x}{4 \left (\ln \left (-\frac {10 x}{9}+5\right ) x -4\right )}\)	\(44\)
derivativedivides	\(\frac {-\frac {405 \ln \left (-\frac {10 x}{9}+5\right )}{2}+\frac {81 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )^{2}}{10}-40 x +480}{9 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )-45 \ln \left (-\frac {10 x}{9}+5\right )+40}+\frac {40 \ln \left (3\right )}{9 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )-45 \ln \left (-\frac {10 x}{9}+5\right )+40}\)	\(86\)
default	\(\frac {-\frac {405 \ln \left (-\frac {10 x}{9}+5\right )}{2}+\frac {81 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )^{2}}{10}-40 x +480}{9 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )-45 \ln \left (-\frac {10 x}{9}+5\right )+40}+\frac {40 \ln \left (3\right )}{9 \ln \left (-\frac {10 x}{9}+5\right ) \left (-\frac {10 x}{9}+5\right )-45 \ln \left (-\frac {10 x}{9}+5\right )+40}\)	\(86\)

[In]

int(((-2*x^3+9*x^2)*ln(-10/9*x+5)^2+((8*x-36)*ln(3)+16*x^2-48*x-108)*ln(-10/9*x+5)+8*x*ln(3)-8*x+144)/((2*x^3-
9*x^2)*ln(-10/9*x+5)^2+(-16*x^2+72*x)*ln(-10/9*x+5)+32*x-144),x,method=_RETURNVERBOSE)

[Out]

(4*x-x^2*ln(-10/9*x+5)-12-4*ln(3))/(ln(-10/9*x+5)*x-4)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-\frac {x^{2} \log \left (-\frac {10}{9} \, x + 5\right ) - 4 \, x + 4 \, \log \left (3\right ) + 12}{x \log \left (-\frac {10}{9} \, x + 5\right ) - 4} \]

[In]

integrate(((-2*x^3+9*x^2)*log(-10/9*x+5)^2+((8*x-36)*log(3)+16*x^2-48*x-108)*log(-10/9*x+5)+8*x*log(3)-8*x+144
)/((2*x^3-9*x^2)*log(-10/9*x+5)^2+(-16*x^2+72*x)*log(-10/9*x+5)+32*x-144),x, algorithm="fricas")

[Out]

-(x^2*log(-10/9*x + 5) - 4*x + 4*log(3) + 12)/(x*log(-10/9*x + 5) - 4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=- x + \frac {-12 - 4 \log {\left (3 \right )}}{x \log {\left (5 - \frac {10 x}{9} \right )} - 4} \]

[In]

integrate(((-2*x**3+9*x**2)*ln(-10/9*x+5)**2+((8*x-36)*ln(3)+16*x**2-48*x-108)*ln(-10/9*x+5)+8*x*ln(3)-8*x+144
)/((2*x**3-9*x**2)*ln(-10/9*x+5)**2+(-16*x**2+72*x)*ln(-10/9*x+5)+32*x-144),x)

[Out]

-x + (-12 - 4*log(3))/(x*log(5 - 10*x/9) - 4)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.03 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-\frac {{\left (i \, \pi + \log \left (5\right ) - 2 \, \log \left (3\right )\right )} x^{2} + x^{2} \log \left (2 \, x - 9\right ) - 4 \, x + 4 \, \log \left (3\right ) + 12}{{\left (i \, \pi + \log \left (5\right ) - 2 \, \log \left (3\right )\right )} x + x \log \left (2 \, x - 9\right ) - 4} \]

[In]

integrate(((-2*x^3+9*x^2)*log(-10/9*x+5)^2+((8*x-36)*log(3)+16*x^2-48*x-108)*log(-10/9*x+5)+8*x*log(3)-8*x+144
)/((2*x^3-9*x^2)*log(-10/9*x+5)^2+(-16*x^2+72*x)*log(-10/9*x+5)+32*x-144),x, algorithm="maxima")

[Out]

-((I*pi + log(5) - 2*log(3))*x^2 + x^2*log(2*x - 9) - 4*x + 4*log(3) + 12)/((I*pi + log(5) - 2*log(3))*x + x*l
og(2*x - 9) - 4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-x - \frac {4 \, {\left (\log \left (3\right ) + 3\right )}}{x \log \left (-\frac {10}{9} \, x + 5\right ) - 4} \]

[In]

integrate(((-2*x^3+9*x^2)*log(-10/9*x+5)^2+((8*x-36)*log(3)+16*x^2-48*x-108)*log(-10/9*x+5)+8*x*log(3)-8*x+144
)/((2*x^3-9*x^2)*log(-10/9*x+5)^2+(-16*x^2+72*x)*log(-10/9*x+5)+32*x-144),x, algorithm="giac")

[Out]

-x - 4*(log(3) + 3)/(x*log(-10/9*x + 5) - 4)

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {144-8 x+8 x \log (3)+\left (-108-48 x+16 x^2+(-36+8 x) \log (3)\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (9 x^2-2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )}{-144+32 x+\left (72 x-16 x^2\right ) \log \left (\frac {1}{9} (45-10 x)\right )+\left (-9 x^2+2 x^3\right ) \log ^2\left (\frac {1}{9} (45-10 x)\right )} \, dx=-x-\frac {\ln \left (81\right )+12}{x\,\ln \left (5-\frac {10\,x}{9}\right )-4} \]

[In]

int((log(5 - (10*x)/9)^2*(9*x^2 - 2*x^3) - 8*x + 8*x*log(3) - log(5 - (10*x)/9)*(48*x - log(3)*(8*x - 36) - 16
*x^2 + 108) + 144)/(32*x - log(5 - (10*x)/9)^2*(9*x^2 - 2*x^3) + log(5 - (10*x)/9)*(72*x - 16*x^2) - 144),x)

[Out]

- x - (log(81) + 12)/(x*log(5 - (10*x)/9) - 4)