\(\int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx\) [5095]

Optimal result

Integrand size = 54, antiderivative size = 18 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^2}{2 \log (x+x \log (15 x))} \]

[Out]

1/2*x^2/ln(x*ln(15*x)+x)

Rubi [F]

\[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx \]

[In]

Int[(-2*x - x*Log[15*x] + (2*x + 2*x*Log[15*x])*Log[x + x*Log[15*x]])/((2 + 2*Log[15*x])*Log[x + x*Log[15*x]]^
2),x]

[Out]

Defer[Int][x/((-1 - Log[15*x])*Log[x + x*Log[15*x]]^2), x] + Defer[Int][(x*Log[15*x])/((-1 - Log[15*x])*Log[x
+ x*Log[15*x]]^2), x]/2 + Defer[Int][x/Log[x + x*Log[15*x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{2 (1+\log (15 x)) \log ^2(x+x \log (15 x))} \, dx \\ & = \frac {1}{2} \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(1+\log (15 x)) \log ^2(x+x \log (15 x))} \, dx \\ & = \frac {1}{2} \int \left (\frac {x (-2-\log (15 x))}{(1+\log (15 x)) \log ^2(x+x \log (15 x))}+\frac {2 x}{\log (x+x \log (15 x))}\right ) \, dx \\ & = \frac {1}{2} \int \frac {x (-2-\log (15 x))}{(1+\log (15 x)) \log ^2(x+x \log (15 x))} \, dx+\int \frac {x}{\log (x+x \log (15 x))} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 x}{(-1-\log (15 x)) \log ^2(x+x \log (15 x))}+\frac {x \log (15 x)}{(-1-\log (15 x)) \log ^2(x+x \log (15 x))}\right ) \, dx+\int \frac {x}{\log (x+x \log (15 x))} \, dx \\ & = \frac {1}{2} \int \frac {x \log (15 x)}{(-1-\log (15 x)) \log ^2(x+x \log (15 x))} \, dx+\int \frac {x}{(-1-\log (15 x)) \log ^2(x+x \log (15 x))} \, dx+\int \frac {x}{\log (x+x \log (15 x))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^2}{2 \log (x (1+\log (15 x)))} \]

[In]

Integrate[(-2*x - x*Log[15*x] + (2*x + 2*x*Log[15*x])*Log[x + x*Log[15*x]])/((2 + 2*Log[15*x])*Log[x + x*Log[1
5*x]]^2),x]

[Out]

x^2/(2*Log[x*(1 + Log[15*x])])

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
norman	\(\frac {x^{2}}{2 \ln \left (x \ln \left (15 x \right )+x \right )}\)	\(17\)
parallelrisch	\(\frac {x^{2}}{2 \ln \left (x \left (\ln \left (15 x \right )+1\right )\right )}\)	\(17\)

[In]

int(((2*x*ln(15*x)+2*x)*ln(x*ln(15*x)+x)-x*ln(15*x)-2*x)/(2*ln(15*x)+2)/ln(x*ln(15*x)+x)^2,x,method=_RETURNVER
BOSE)

[Out]

1/2*x^2/ln(x*ln(15*x)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^{2}}{2 \, \log \left (x \log \left (15 \, x\right ) + x\right )} \]

[In]

integrate(((2*x*log(15*x)+2*x)*log(x*log(15*x)+x)-x*log(15*x)-2*x)/(2*log(15*x)+2)/log(x*log(15*x)+x)^2,x, alg
orithm="fricas")

[Out]

1/2*x^2/log(x*log(15*x) + x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^{2}}{2 \log {\left (x \log {\left (15 x \right )} + x \right )}} \]

[In]

integrate(((2*x*ln(15*x)+2*x)*ln(x*ln(15*x)+x)-x*ln(15*x)-2*x)/(2*ln(15*x)+2)/ln(x*ln(15*x)+x)**2,x)

[Out]

x**2/(2*log(x*log(15*x) + x))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^{2}}{2 \, {\left (\log \left (x\right ) + \log \left (\log \left (5\right ) + \log \left (3\right ) + \log \left (x\right ) + 1\right )\right )}} \]

[In]

integrate(((2*x*log(15*x)+2*x)*log(x*log(15*x)+x)-x*log(15*x)-2*x)/(2*log(15*x)+2)/log(x*log(15*x)+x)^2,x, alg
orithm="maxima")

[Out]

1/2*x^2/(log(x) + log(log(5) + log(3) + log(x) + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (16) = 32\).

Time = 0.30 (sec) , antiderivative size = 164, normalized size of antiderivative = 9.11 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^{2} \log \left (15\right ) \log \left (15 \, x\right ) + x^{2} \log \left (15 \, x\right ) \log \left (x\right ) + x^{2} \log \left (15\right ) + 2 \, x^{2} \log \left (15 \, x\right ) + x^{2} \log \left (x\right ) + 2 \, x^{2}}{2 \, {\left (\log \left (15\right ) \log \left (15 \, x\right ) \log \left (x\right ) + \log \left (15 \, x\right ) \log \left (x\right )^{2} + \log \left (15\right ) \log \left (15 \, x\right ) \log \left (\log \left (15 \, x\right ) + 1\right ) + \log \left (15 \, x\right ) \log \left (x\right ) \log \left (\log \left (15 \, x\right ) + 1\right ) + 2 \, \log \left (15\right ) \log \left (x\right ) + \log \left (15 \, x\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2} + 2 \, \log \left (15\right ) \log \left (\log \left (15 \, x\right ) + 1\right ) + \log \left (15 \, x\right ) \log \left (\log \left (15 \, x\right ) + 1\right ) + 2 \, \log \left (x\right ) \log \left (\log \left (15 \, x\right ) + 1\right ) + 2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (15 \, x\right ) + 1\right )\right )}} \]

[In]

integrate(((2*x*log(15*x)+2*x)*log(x*log(15*x)+x)-x*log(15*x)-2*x)/(2*log(15*x)+2)/log(x*log(15*x)+x)^2,x, alg
orithm="giac")

[Out]

1/2*(x^2*log(15)*log(15*x) + x^2*log(15*x)*log(x) + x^2*log(15) + 2*x^2*log(15*x) + x^2*log(x) + 2*x^2)/(log(1
5)*log(15*x)*log(x) + log(15*x)*log(x)^2 + log(15)*log(15*x)*log(log(15*x) + 1) + log(15*x)*log(x)*log(log(15*
x) + 1) + 2*log(15)*log(x) + log(15*x)*log(x) + 2*log(x)^2 + 2*log(15)*log(log(15*x) + 1) + log(15*x)*log(log(
15*x) + 1) + 2*log(x)*log(log(15*x) + 1) + 2*log(x) + 2*log(log(15*x) + 1))

Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2 x-x \log (15 x)+(2 x+2 x \log (15 x)) \log (x+x \log (15 x))}{(2+2 \log (15 x)) \log ^2(x+x \log (15 x))} \, dx=\frac {x^2}{2\,\ln \left (x+x\,\ln \left (15\,x\right )\right )} \]

[In]

int(-(2*x - log(x + x*log(15*x))*(2*x + 2*x*log(15*x)) + x*log(15*x))/(log(x + x*log(15*x))^2*(2*log(15*x) + 2
)),x)

[Out]

x^2/(2*log(x + x*log(15*x)))