\(\int \frac {(-6 x^2-48 \log (2)) \log (3)}{x^2} \, dx\) [8250]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=\frac {\left (x+x^2-(6 (-8+x)+x) (x+\log (2))\right ) \log (3)}{x} \]

[Out]

(x+x^2-(7*x-48)*(ln(2)+x))*ln(3)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14} \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=\frac {48 \log (2) \log (3)}{x}-6 x \log (3) \]

[In]

Int[((-6*x^2 - 48*Log[2])*Log[3])/x^2,x]

[Out]

-6*x*Log[3] + (48*Log[2]*Log[3])/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \log (3) \int \frac {-6 x^2-48 \log (2)}{x^2} \, dx \\ & = \log (3) \int \left (-6-\frac {48 \log (2)}{x^2}\right ) \, dx \\ & = -6 x \log (3)+\frac {48 \log (2) \log (3)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=-6 \log (3) \left (x-\frac {\log (256)}{x}\right ) \]

[In]

Integrate[((-6*x^2 - 48*Log[2])*Log[3])/x^2,x]

[Out]

-6*Log[3]*(x - Log[256]/x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
default	\(6 \ln \left (3\right ) \left (-x +\frac {8 \ln \left (2\right )}{x}\right )\)	\(16\)
risch	\(-6 x \ln \left (3\right )+\frac {48 \ln \left (3\right ) \ln \left (2\right )}{x}\)	\(16\)
parallelrisch	\(\frac {\ln \left (3\right ) \left (-6 x^{2}+48 \ln \left (2\right )\right )}{x}\)	\(17\)
gosper	\(\frac {6 \ln \left (3\right ) \left (-x^{2}+8 \ln \left (2\right )\right )}{x}\)	\(18\)
norman	\(\frac {-6 x^{2} \ln \left (3\right )+48 \ln \left (2\right ) \ln \left (3\right )}{x}\)	\(19\)

[In]

int((-48*ln(2)-6*x^2)*ln(3)/x^2,x,method=_RETURNVERBOSE)

[Out]

6*ln(3)*(-x+8*ln(2)/x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=-\frac {6 \, {\left (x^{2} - 8 \, \log \left (2\right )\right )} \log \left (3\right )}{x} \]

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="fricas")

[Out]

-6*(x^2 - 8*log(2))*log(3)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=- 6 x \log {\left (3 \right )} + \frac {48 \log {\left (2 \right )} \log {\left (3 \right )}}{x} \]

[In]

integrate((-48*ln(2)-6*x**2)*ln(3)/x**2,x)

[Out]

-6*x*log(3) + 48*log(2)*log(3)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=-6 \, {\left (x - \frac {8 \, \log \left (2\right )}{x}\right )} \log \left (3\right ) \]

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="maxima")

[Out]

-6*(x - 8*log(2)/x)*log(3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=-6 \, {\left (x - \frac {8 \, \log \left (2\right )}{x}\right )} \log \left (3\right ) \]

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="giac")

[Out]

-6*(x - 8*log(2)/x)*log(3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {\left (-6 x^2-48 \log (2)\right ) \log (3)}{x^2} \, dx=\frac {6\,\ln \left (3\right )\,\left (8\,\ln \left (2\right )-x^2\right )}{x} \]

[In]

int(-(log(3)*(48*log(2) + 6*x^2))/x^2,x)

[Out]

(6*log(3)*(8*log(2) - x^2))/x