\(\int \frac {1}{9} (-8 x-6 x^2) \log ^2(\log (5)) \, dx\) [6449]

Optimal result

Integrand size = 18, antiderivative size = 19 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-3+\frac {1}{9} (-4-2 x) x^2 \log ^2(\log (5)) \]

[Out]

-1/9*x^2*(4+2*x)*ln(ln(5))^2-3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12} \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2}{9} x^3 \log ^2(\log (5))-\frac {4}{9} x^2 \log ^2(\log (5)) \]

[In]

Int[((-8*x - 6*x^2)*Log[Log[5]]^2)/9,x]

[Out]

(-4*x^2*Log[Log[5]]^2)/9 - (2*x^3*Log[Log[5]]^2)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \log ^2(\log (5)) \int \left (-8 x-6 x^2\right ) \, dx \\ & = -\frac {4}{9} x^2 \log ^2(\log (5))-\frac {2}{9} x^3 \log ^2(\log (5)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2}{9} \left (2 x^2+x^3\right ) \log ^2(\log (5)) \]

[In]

Integrate[((-8*x - 6*x^2)*Log[Log[5]]^2)/9,x]

[Out]

(-2*(2*x^2 + x^3)*Log[Log[5]]^2)/9

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
gosper	\(-\frac {2 \ln \left (\ln \left (5\right )\right )^{2} \left (2+x \right ) x^{2}}{9}\)	\(14\)
default	\(\frac {2 \ln \left (\ln \left (5\right )\right )^{2} \left (-x^{3}-2 x^{2}\right )}{9}\)	\(19\)
parallelrisch	\(\frac {\ln \left (\ln \left (5\right )\right )^{2} \left (-2 x^{3}-4 x^{2}\right )}{9}\)	\(19\)
norman	\(-\frac {4 x^{2} \ln \left (\ln \left (5\right )\right )^{2}}{9}-\frac {2 x^{3} \ln \left (\ln \left (5\right )\right )^{2}}{9}\)	\(22\)
risch	\(-\frac {4 x^{2} \ln \left (\ln \left (5\right )\right )^{2}}{9}-\frac {2 x^{3} \ln \left (\ln \left (5\right )\right )^{2}}{9}\)	\(22\)

[In]

int(1/9*(-6*x^2-8*x)*ln(ln(5))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2}{9} \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(1/9*(-6*x^2-8*x)*log(log(5))^2,x, algorithm="fricas")

[Out]

-2/9*(x^3 + 2*x^2)*log(log(5))^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=- \frac {2 x^{3} \log {\left (\log {\left (5 \right )} \right )}^{2}}{9} - \frac {4 x^{2} \log {\left (\log {\left (5 \right )} \right )}^{2}}{9} \]

[In]

integrate(1/9*(-6*x**2-8*x)*ln(ln(5))**2,x)

[Out]

-2*x**3*log(log(5))**2/9 - 4*x**2*log(log(5))**2/9

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2}{9} \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(1/9*(-6*x^2-8*x)*log(log(5))^2,x, algorithm="maxima")

[Out]

-2/9*(x^3 + 2*x^2)*log(log(5))^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2}{9} \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(1/9*(-6*x^2-8*x)*log(log(5))^2,x, algorithm="giac")

[Out]

-2/9*(x^3 + 2*x^2)*log(log(5))^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {1}{9} \left (-8 x-6 x^2\right ) \log ^2(\log (5)) \, dx=-\frac {2\,x^2\,{\ln \left (\ln \left (5\right )\right )}^2\,\left (x+2\right )}{9} \]

[In]

int(-(log(log(5))^2*(8*x + 6*x^2))/9,x)

[Out]

-(2*x^2*log(log(5))^2*(x + 2))/9