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

Optimal. Leaf size=20 \[ \frac {1}{3} x (-e+x) \left (4+\frac {8 x}{\log (\log (2))}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.85, number of steps used = 2, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12} \begin {gather*} \frac {8 x^3}{3 \log (\log (2))}-\frac {8 e x^2}{3 \log (\log (2))}+\frac {1}{3} (e-2 x)^2 \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.80 \begin {gather*} \frac {-8 e x^2+8 x^3-4 e x \log (\log (2))+4 x^2 \log (\log (2))}{3 \log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 0.85, size = 33, normalized size = 1.65 \begin {gather*} \frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \relax (2)\right )\right )}}{3 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.15, size = 33, normalized size = 1.65 \begin {gather*} \frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \relax (2)\right )\right )}}{3 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.05, size = 32, normalized size = 1.60

method	result	size
gosper	\(-\frac {4 x \left (\ln \left (\ln \relax (2)\right ) {\mathrm e}+2 x \,{\mathrm e}-x \ln \left (\ln \relax (2)\right )-2 x^{2}\right )}{3 \ln \left (\ln \relax (2)\right )}\)	\(32\)
risch	\(-\frac {4 x \,{\mathrm e}}{3}-\frac {8 x^{2} {\mathrm e}}{3 \ln \left (\ln \relax (2)\right )}+\frac {4 x^{2}}{3}+\frac {8 x^{3}}{3 \ln \left (\ln \relax (2)\right )}\)	\(34\)
default	\(\frac {\ln \left (\ln \relax (2)\right ) \left (-4 x \,{\mathrm e}+4 x^{2}\right )-8 x^{2} {\mathrm e}+8 x^{3}}{3 \ln \left (\ln \relax (2)\right )}\)	\(36\)
norman	\(-\frac {4 x \,{\mathrm e}}{3}+\frac {8 x^{3}}{3 \ln \left (\ln \relax (2)\right )}-\frac {4 \left (2 \,{\mathrm e}-\ln \left (\ln \relax (2)\right )\right ) x^{2}}{3 \ln \left (\ln \relax (2)\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.37, size = 33, normalized size = 1.65 \begin {gather*} \frac {4 \, {\left (2 \, x^{3} - 2 \, x^{2} e + {\left (x^{2} - x e\right )} \log \left (\log \relax (2)\right )\right )}}{3 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 1.35, size = 21, normalized size = 1.05 \begin {gather*} \frac {4\,x\,\left (2\,x+\ln \left (\ln \relax (2)\right )\right )\,\left (x-\mathrm {e}\right )}{3\,\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [B]  time = 0.07, size = 41, normalized size = 2.05 \begin {gather*} \frac {8 x^{3}}{3 \log {\left (\log {\relax (2 )} \right )}} + \frac {x^{2} \left (- 8 e + 4 \log {\left (\log {\relax (2 )} \right )}\right )}{3 \log {\left (\log {\relax (2 )} \right )}} - \frac {4 e x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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