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

Optimal. Leaf size=21 \[ \log \left (\frac {1}{3} \left (-5+x^2\right ) \left (3-16 \log ^2(2)\right )+\log (9)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 12, 1587} \begin {gather*} \log \left (-3 x^2-16 \left (5-x^2\right ) \log ^2(2)+3 (5-\log (9))\right ) \end {gather*}

Antiderivative was successfully verified.

[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 1587

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]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 1.10 \begin {gather*} \log \left (-15+80 \log ^2(2)+x^2 \left (3-16 \log ^2(2)\right )+\log (729)\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.69, size = 23, normalized size = 1.10 \begin {gather*} \log \left (16 \, {\left (x^{2} - 5\right )} \log \relax (2)^{2} - 3 \, x^{2} - 6 \, \log \relax (3) + 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.19, size = 28, normalized size = 1.33 \begin {gather*} \log \left ({\left | 16 \, x^{2} \log \relax (2)^{2} - 3 \, x^{2} - 80 \, \log \relax (2)^{2} - 6 \, \log \relax (3) + 15 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.29, size = 25, normalized size = 1.19

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.35, size = 23, normalized size = 1.10 \begin {gather*} \log \left (16 \, {\left (x^{2} - 5\right )} \log \relax (2)^{2} - 3 \, x^{2} - 6 \, \log \relax (3) + 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.26, size = 25, normalized size = 1.19 \begin {gather*} \ln \left (\left (16\,{\ln \relax (2)}^2-3\right )\,x^2-\ln \left (729\right )-80\,{\ln \relax (2)}^2+15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.42, size = 42, normalized size = 2.00 \begin {gather*} \frac {\left (-6 + 32 \log {\relax (2 )}^{2}\right ) \log {\left (x^{2} \left (-3 + 16 \log {\relax (2 )}^{2}\right ) - 80 \log {\relax (2 )}^{2} - 6 \log {\relax (3 )} + 15 \right )}}{2 \left (-3 + 16 \log {\relax (2 )}^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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