\(\int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx\) [2333]

Optimal result

Integrand size = 30, antiderivative size = 20 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {1}{3} x (-e+x) \left (4+\frac {8 x}{\log (\log (2))}\right ) \]

[Out]

x*(x-exp(1))*(8/3*x/ln(ln(2))+4/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12} \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {8 x^3}{3 \log (\log (2))}-\frac {8 e x^2}{3 \log (\log (2))}+\frac {1}{3} (e-2 x)^2 \]

[In]

Int[(-16*E*x + 24*x^2 + (-4*E + 8*x)*Log[Log[2]])/(3*Log[Log[2]]),x]

[Out]

(E - 2*x)^2/3 - (8*E*x^2)/(3*Log[Log[2]]) + (8*x^3)/(3*Log[Log[2]])

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 {\int \left (-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))\right ) \, dx}{3 \log (\log (2))} \\ & = \frac {1}{3} (e-2 x)^2-\frac {8 e x^2}{3 \log (\log (2))}+\frac {8 x^3}{3 \log (\log (2))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {-8 e x^2+8 x^3-4 e x \log (\log (2))+4 x^2 \log (\log (2))}{3 \log (\log (2))} \]

[In]

Integrate[(-16*E*x + 24*x^2 + (-4*E + 8*x)*Log[Log[2]])/(3*Log[Log[2]]),x]

[Out]

(-8*E*x^2 + 8*x^3 - 4*E*x*Log[Log[2]] + 4*x^2*Log[Log[2]])/(3*Log[Log[2]])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60

method	result	size
gosper	\(-\frac {4 x \left (\ln \left (\ln \left (2\right )\right ) {\mathrm e}+2 x \,{\mathrm e}-x \ln \left (\ln \left (2\right )\right )-2 x^{2}\right )}{3 \ln \left (\ln \left (2\right )\right )}\)	\(32\)
risch	\(-\frac {4 x \,{\mathrm e}}{3}-\frac {8 x^{2} {\mathrm e}}{3 \ln \left (\ln \left (2\right )\right )}+\frac {4 x^{2}}{3}+\frac {8 x^{3}}{3 \ln \left (\ln \left (2\right )\right )}\)	\(34\)
default	\(\frac {-4 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -8 x^{2} {\mathrm e}+4 x^{2} \ln \left (\ln \left (2\right )\right )+8 x^{3}}{3 \ln \left (\ln \left (2\right )\right )}\)	\(37\)
norman	\(-\frac {4 x \,{\mathrm e}}{3}+\frac {8 x^{3}}{3 \ln \left (\ln \left (2\right )\right )}-\frac {4 \left (2 \,{\mathrm e}-\ln \left (\ln \left (2\right )\right )\right ) x^{2}}{3 \ln \left (\ln \left (2\right )\right )}\)	\(37\)
parallelrisch	\(\frac {-4 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -8 x^{2} {\mathrm e}+4 x^{2} \ln \left (\ln \left (2\right )\right )+8 x^{3}}{3 \ln \left (\ln \left (2\right )\right )}\)	\(37\)

[In]

int(1/3*((-4*exp(1)+8*x)*ln(ln(2))-16*x*exp(1)+24*x^2)/ln(ln(2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \left (2\right )\right )\right )}}{3 \, \log \left (\log \left (2\right )\right )} \]

[In]

integrate(1/3*((-4*exp(1)+8*x)*log(log(2))-16*x*exp(1)+24*x^2)/log(log(2)),x, algorithm="fricas")

[Out]

4/3*(2*x^3 - 2*x^2*e + (x^2 - x*e)*log(log(2)))/log(log(2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (19) = 38\).

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {8 x^{3}}{3 \log {\left (\log {\left (2 \right )} \right )}} + \frac {x^{2} \left (- 8 e + 4 \log {\left (\log {\left (2 \right )} \right )}\right )}{3 \log {\left (\log {\left (2 \right )} \right )}} - \frac {4 e x}{3} \]

[In]

integrate(1/3*((-4*exp(1)+8*x)*ln(ln(2))-16*x*exp(1)+24*x**2)/ln(ln(2)),x)

[Out]

8*x**3/(3*log(log(2))) + x**2*(-8*E + 4*log(log(2)))/(3*log(log(2))) - 4*E*x/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \left (2\right )\right )\right )}}{3 \, \log \left (\log \left (2\right )\right )} \]

[In]

integrate(1/3*((-4*exp(1)+8*x)*log(log(2))-16*x*exp(1)+24*x^2)/log(log(2)),x, algorithm="maxima")

[Out]

4/3*(2*x^3 - 2*x^2*e + (x^2 - x*e)*log(log(2)))/log(log(2))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \left (2\right )\right )\right )}}{3 \, \log \left (\log \left (2\right )\right )} \]

[In]

integrate(1/3*((-4*exp(1)+8*x)*log(log(2))-16*x*exp(1)+24*x^2)/log(log(2)),x, algorithm="giac")

[Out]

4/3*(2*x^3 - 2*x^2*e + (x^2 - x*e)*log(log(2)))/log(log(2))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-16 e x+24 x^2+(-4 e+8 x) \log (\log (2))}{3 \log (\log (2))} \, dx=\frac {4\,x\,\left (2\,x+\ln \left (\ln \left (2\right )\right )\right )\,\left (x-\mathrm {e}\right )}{3\,\ln \left (\ln \left (2\right )\right )} \]

[In]

int(((log(log(2))*(8*x - 4*exp(1)))/3 - (16*x*exp(1))/3 + 8*x^2)/log(log(2)),x)

[Out]

(4*x*(2*x + log(log(2)))*(x - exp(1)))/(3*log(log(2)))