3.21.2 \(\int \frac {20 e^2 x^2-20 e^2 x^2 \log (x \log (4))+(3+3 x^2) \log ^2(x \log (4))}{3 x^2 \log ^2(x \log (4))} \, dx\)

Optimal. Leaf size=22 \[ -\frac {1}{x}+x-\frac {20 e^2 x}{3 \log (x \log (4))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {12, 6688, 2297, 2298} \begin {gather*} x-\frac {1}{x}-\frac {20 e^2 x}{3 \log (x \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*E^2*x^2 - 20*E^2*x^2*Log[x*Log[4]] + (3 + 3*x^2)*Log[x*Log[4]]^2)/(3*x^2*Log[x*Log[4]]^2),x]

[Out]

-x^(-1) + x - (20*E^2*x)/(3*Log[x*Log[4]])

Rule 12

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

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {20 e^2 x^2-20 e^2 x^2 \log (x \log (4))+\left (3+3 x^2\right ) \log ^2(x \log (4))}{x^2 \log ^2(x \log (4))} \, dx\\ &=\frac {1}{3} \int \left (3+\frac {3}{x^2}+\frac {20 e^2}{\log ^2(x \log (4))}-\frac {20 e^2}{\log (x \log (4))}\right ) \, dx\\ &=-\frac {1}{x}+x+\frac {1}{3} \left (20 e^2\right ) \int \frac {1}{\log ^2(x \log (4))} \, dx-\frac {1}{3} \left (20 e^2\right ) \int \frac {1}{\log (x \log (4))} \, dx\\ &=-\frac {1}{x}+x-\frac {20 e^2 x}{3 \log (x \log (4))}-\frac {20 e^2 \text {li}(x \log (4))}{3 \log (4)}+\frac {1}{3} \left (20 e^2\right ) \int \frac {1}{\log (x \log (4))} \, dx\\ &=-\frac {1}{x}+x-\frac {20 e^2 x}{3 \log (x \log (4))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} -\frac {1}{x}+x-\frac {20 e^2 x}{3 \log (x \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*E^2*x^2 - 20*E^2*x^2*Log[x*Log[4]] + (3 + 3*x^2)*Log[x*Log[4]]^2)/(3*x^2*Log[x*Log[4]]^2),x]

[Out]

-x^(-1) + x - (20*E^2*x)/(3*Log[x*Log[4]])

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 34, normalized size = 1.55 \begin {gather*} -\frac {20 \, x^{2} e^{2} - 3 \, {\left (x^{2} - 1\right )} \log \left (2 \, x \log \relax (2)\right )}{3 \, x \log \left (2 \, x \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+3)*log(2*x*log(2))^2-20*x^2*exp(2)*log(2*x*log(2))+20*x^2*exp(2))/x^2/log(2*x*log(2))^2,
x, algorithm="fricas")

[Out]

-1/3*(20*x^2*e^2 - 3*(x^2 - 1)*log(2*x*log(2)))/(x*log(2*x*log(2)))

________________________________________________________________________________________

giac [B]  time = 0.32, size = 52, normalized size = 2.36 \begin {gather*} -\frac {20 \, x^{2} e^{2} - 3 \, x^{2} \log \relax (2) - 3 \, x^{2} \log \left (x \log \relax (2)\right ) + 3 \, \log \relax (2) + 3 \, \log \left (x \log \relax (2)\right )}{3 \, {\left (x \log \relax (2) + x \log \left (x \log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+3)*log(2*x*log(2))^2-20*x^2*exp(2)*log(2*x*log(2))+20*x^2*exp(2))/x^2/log(2*x*log(2))^2,
x, algorithm="giac")

[Out]

-1/3*(20*x^2*e^2 - 3*x^2*log(2) - 3*x^2*log(x*log(2)) + 3*log(2) + 3*log(x*log(2)))/(x*log(2) + x*log(x*log(2)
))

________________________________________________________________________________________

maple [A]  time = 0.07, size = 24, normalized size = 1.09

method	result	size
risch	\(\frac {x^{2}-1}{x}-\frac {20 x \,{\mathrm e}^{2}}{3 \ln \left (2 x \ln \relax (2)\right )}\)	\(24\)
norman	\(\frac {x^{2} \ln \left (2 x \ln \relax (2)\right )-\frac {20 x^{2} {\mathrm e}^{2}}{3}-\ln \left (2 x \ln \relax (2)\right )}{x \ln \left (2 x \ln \relax (2)\right )}\)	\(39\)
derivativedivides	\(\frac {2 \ln \relax (2) \left (\frac {5 \,{\mathrm e}^{2} \expIntegralEi \left (1, -\ln \left (2 x \ln \relax (2)\right )\right )}{\ln \relax (2)^{2}}-\frac {3}{2 x \ln \relax (2)}+\frac {3 x}{2 \ln \relax (2)}+\frac {5 \,{\mathrm e}^{2} \left (-\frac {2 x \ln \relax (2)}{\ln \left (2 x \ln \relax (2)\right )}-\expIntegralEi \left (1, -\ln \left (2 x \ln \relax (2)\right )\right )\right )}{\ln \relax (2)^{2}}\right )}{3}\)	\(74\)
default	\(\frac {2 \ln \relax (2) \left (\frac {5 \,{\mathrm e}^{2} \expIntegralEi \left (1, -\ln \left (2 x \ln \relax (2)\right )\right )}{\ln \relax (2)^{2}}-\frac {3}{2 x \ln \relax (2)}+\frac {3 x}{2 \ln \relax (2)}+\frac {5 \,{\mathrm e}^{2} \left (-\frac {2 x \ln \relax (2)}{\ln \left (2 x \ln \relax (2)\right )}-\expIntegralEi \left (1, -\ln \left (2 x \ln \relax (2)\right )\right )\right )}{\ln \relax (2)^{2}}\right )}{3}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((3*x^2+3)*ln(2*x*ln(2))^2-20*x^2*exp(2)*ln(2*x*ln(2))+20*x^2*exp(2))/x^2/ln(2*x*ln(2))^2,x,method=_RE
TURNVERBOSE)

[Out]

(x^2-1)/x-20/3*x/ln(2*x*ln(2))*exp(2)

________________________________________________________________________________________

maxima [C]  time = 0.61, size = 40, normalized size = 1.82 \begin {gather*} x - \frac {10 \, {\rm Ei}\left (\log \left (2 \, x \log \relax (2)\right )\right ) e^{2}}{3 \, \log \relax (2)} + \frac {10 \, e^{2} \Gamma \left (-1, -\log \left (2 \, x \log \relax (2)\right )\right )}{3 \, \log \relax (2)} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+3)*log(2*x*log(2))^2-20*x^2*exp(2)*log(2*x*log(2))+20*x^2*exp(2))/x^2/log(2*x*log(2))^2,
x, algorithm="maxima")

[Out]

x - 10/3*Ei(log(2*x*log(2)))*e^2/log(2) + 10/3*e^2*gamma(-1, -log(2*x*log(2)))/log(2) - 1/x

________________________________________________________________________________________

mupad [B]  time = 1.18, size = 20, normalized size = 0.91 \begin {gather*} x-\frac {1}{x}-\frac {20\,x\,{\mathrm {e}}^2}{3\,\ln \left (2\,x\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2*x*log(2))^2*(3*x^2 + 3))/3 + (20*x^2*exp(2))/3 - (20*x^2*exp(2)*log(2*x*log(2)))/3)/(x^2*log(2*x*l
og(2))^2),x)

[Out]

x - 1/x - (20*x*exp(2))/(3*log(2*x*log(2)))

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 20, normalized size = 0.91 \begin {gather*} x - \frac {20 x e^{2}}{3 \log {\left (2 x \log {\relax (2 )} \right )}} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x**2+3)*ln(2*x*ln(2))**2-20*x**2*exp(2)*ln(2*x*ln(2))+20*x**2*exp(2))/x**2/ln(2*x*ln(2))**2,
x)

[Out]

x - 20*x*exp(2)/(3*log(2*x*log(2))) - 1/x

________________________________________________________________________________________