\(\int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+(80-16 x^2) \log ^2(2)+3 \log (9)} \, dx\) [10180]

Optimal result

Integrand size = 37, antiderivative size = 21 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left (\frac {1}{3} \left (-5+x^2\right ) \left (3-16 \log ^2(2)\right )+\log (9)\right ) \]

[Out]

ln(1/3*(x^2-5)*(3-16*ln(2)^2)+2*ln(3))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 12, 1601} \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left (-3 x^2-16 \left (5-x^2\right ) \log ^2(2)+3 (5-\log (9))\right ) \]

[In]

Int[(6*x - 32*x*Log[2]^2)/(-15 + 3*x^2 + (80 - 16*x^2)*Log[2]^2 + 3*Log[9]),x]

[Out]

Log[-3*x^2 - 16*(5 - x^2)*Log[2]^2 + 3*(5 - Log[9])]

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 12

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

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (6-32 \log ^2(2)\right )}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx \\ & = \left (2 \left (3-16 \log ^2(2)\right )\right ) \int \frac {x}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx \\ & = \log \left (-3 x^2-16 \left (5-x^2\right ) \log ^2(2)+3 (5-\log (9))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left (-15+80 \log ^2(2)+x^2 \left (3-16 \log ^2(2)\right )+\log (729)\right ) \]

[In]

Integrate[(6*x - 32*x*Log[2]^2)/(-15 + 3*x^2 + (80 - 16*x^2)*Log[2]^2 + 3*Log[9]),x]

[Out]

Log[-15 + 80*Log[2]^2 + x^2*(3 - 16*Log[2]^2) + Log[729]]

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19

method	result	size
derivativedivides	\(\ln \left (6 \ln \left (3\right )+\left (-16 x^{2}+80\right ) \ln \left (2\right )^{2}+3 x^{2}-15\right )\)	\(25\)
risch	\(\ln \left (x^{2} \left (16 \ln \left (2\right )^{2}-3\right )-80 \ln \left (2\right )^{2}-6 \ln \left (3\right )+15\right )\)	\(26\)
default	\(\ln \left (16 x^{2} \ln \left (2\right )^{2}-80 \ln \left (2\right )^{2}-3 x^{2}-6 \ln \left (3\right )+15\right )\)	\(28\)
norman	\(\ln \left (16 x^{2} \ln \left (2\right )^{2}-80 \ln \left (2\right )^{2}-3 x^{2}-6 \ln \left (3\right )+15\right )\)	\(28\)
parallelrisch	\(\ln \left (\frac {16 x^{2} \ln \left (2\right )^{2}-80 \ln \left (2\right )^{2}-3 x^{2}-6 \ln \left (3\right )+15}{16 \ln \left (2\right )^{2}-3}\right )\)	\(39\)
meijerg	\(-\frac {\left (-32 \ln \left (2\right )^{2}+6\right ) \ln \left (1-\frac {x^{2} \left (16 \ln \left (2\right )^{2}-3\right )}{80 \ln \left (2\right )^{2}+6 \ln \left (3\right )-15}\right )}{2 \left (16 \ln \left (2\right )^{2}-3\right )}\)	\(51\)

[In]

int((-32*x*ln(2)^2+6*x)/(6*ln(3)+(-16*x^2+80)*ln(2)^2+3*x^2-15),x,method=_RETURNVERBOSE)

[Out]

ln(6*ln(3)+(-16*x^2+80)*ln(2)^2+3*x^2-15)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left (16 \, {\left (x^{2} - 5\right )} \log \left (2\right )^{2} - 3 \, x^{2} - 6 \, \log \left (3\right ) + 15\right ) \]

[In]

integrate((-32*x*log(2)^2+6*x)/(6*log(3)+(-16*x^2+80)*log(2)^2+3*x^2-15),x, algorithm="fricas")

[Out]

log(16*(x^2 - 5)*log(2)^2 - 3*x^2 - 6*log(3) + 15)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\frac {\left (-6 + 32 \log {\left (2 \right )}^{2}\right ) \log {\left (x^{2} \left (-3 + 16 \log {\left (2 \right )}^{2}\right ) - 80 \log {\left (2 \right )}^{2} - 6 \log {\left (3 \right )} + 15 \right )}}{2 \left (-3 + 16 \log {\left (2 \right )}^{2}\right )} \]

[In]

integrate((-32*x*ln(2)**2+6*x)/(6*ln(3)+(-16*x**2+80)*ln(2)**2+3*x**2-15),x)

[Out]

(-6 + 32*log(2)**2)*log(x**2*(-3 + 16*log(2)**2) - 80*log(2)**2 - 6*log(3) + 15)/(2*(-3 + 16*log(2)**2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left (16 \, {\left (x^{2} - 5\right )} \log \left (2\right )^{2} - 3 \, x^{2} - 6 \, \log \left (3\right ) + 15\right ) \]

[In]

integrate((-32*x*log(2)^2+6*x)/(6*log(3)+(-16*x^2+80)*log(2)^2+3*x^2-15),x, algorithm="maxima")

[Out]

log(16*(x^2 - 5)*log(2)^2 - 3*x^2 - 6*log(3) + 15)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\log \left ({\left | 16 \, x^{2} \log \left (2\right )^{2} - 3 \, x^{2} - 80 \, \log \left (2\right )^{2} - 6 \, \log \left (3\right ) + 15 \right |}\right ) \]

[In]

integrate((-32*x*log(2)^2+6*x)/(6*log(3)+(-16*x^2+80)*log(2)^2+3*x^2-15),x, algorithm="giac")

[Out]

log(abs(16*x^2*log(2)^2 - 3*x^2 - 80*log(2)^2 - 6*log(3) + 15))

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {6 x-32 x \log ^2(2)}{-15+3 x^2+\left (80-16 x^2\right ) \log ^2(2)+3 \log (9)} \, dx=\ln \left (\left (16\,{\ln \left (2\right )}^2-3\right )\,x^2-\ln \left (729\right )-80\,{\ln \left (2\right )}^2+15\right ) \]

[In]

int((6*x - 32*x*log(2)^2)/(6*log(3) - log(2)^2*(16*x^2 - 80) + 3*x^2 - 15),x)

[Out]

log(x^2*(16*log(2)^2 - 3) - log(729) - 80*log(2)^2 + 15)