3.98.43 \(\int \frac {6 e^{2 e^4}+6 x-3 e^{2 e^4} \log (x^2)}{x^2 \log (8)} \, dx\)

Optimal. Leaf size=22 \[ \frac {3 \left (e^{2 e^4}+x\right ) \log \left (x^2\right )}{x \log (8)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14, 43, 2304} \begin {gather*} \frac {3 e^{2 e^4} \log \left (x^2\right )}{x \log (8)}+\frac {6 \log (x)}{\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^(2*E^4) + 6*x - 3*E^(2*E^4)*Log[x^2])/(x^2*Log[8]),x]

[Out]

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

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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {6 e^{2 e^4}+6 x-3 e^{2 e^4} \log \left (x^2\right )}{x^2} \, dx}{\log (8)}\\ &=\frac {\int \left (\frac {6 \left (e^{2 e^4}+x\right )}{x^2}-\frac {3 e^{2 e^4} \log \left (x^2\right )}{x^2}\right ) \, dx}{\log (8)}\\ &=\frac {6 \int \frac {e^{2 e^4}+x}{x^2} \, dx}{\log (8)}-\frac {\left (3 e^{2 e^4}\right ) \int \frac {\log \left (x^2\right )}{x^2} \, dx}{\log (8)}\\ &=\frac {6 e^{2 e^4}}{x \log (8)}+\frac {3 e^{2 e^4} \log \left (x^2\right )}{x \log (8)}+\frac {6 \int \left (\frac {e^{2 e^4}}{x^2}+\frac {1}{x}\right ) \, dx}{\log (8)}\\ &=\frac {6 \log (x)}{\log (8)}+\frac {3 e^{2 e^4} \log \left (x^2\right )}{x \log (8)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 26, normalized size = 1.18 \begin {gather*} \frac {6 \log (x)+\frac {3 e^{2 e^4} \log \left (x^2\right )}{x}}{\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^(2*E^4) + 6*x - 3*E^(2*E^4)*Log[x^2])/(x^2*Log[8]),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.99, size = 19, normalized size = 0.86 \begin {gather*} \frac {{\left (x + e^{\left (2 \, e^{4}\right )}\right )} \log \left (x^{2}\right )}{x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*exp(exp(4))^2*log(x^2)+6*exp(exp(4))^2+6*x)/x^2/log(2),x, algorithm="fricas")

[Out]

(x + e^(2*e^4))*log(x^2)/(x*log(2))

________________________________________________________________________________________

giac [A]  time = 0.13, size = 24, normalized size = 1.09 \begin {gather*} \frac {e^{\left (2 \, e^{4}\right )} \log \left (x^{2}\right ) + 2 \, x \log \relax (x)}{x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*exp(exp(4))^2*log(x^2)+6*exp(exp(4))^2+6*x)/x^2/log(2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 27, normalized size = 1.23

method	result	size
norman	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}} \ln \left (x^{2}\right )}{\ln \relax (2) x}+\frac {2 \ln \relax (x )}{\ln \relax (2)}\)	\(27\)
risch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}} \ln \left (x^{2}\right )}{\ln \relax (2) x}+\frac {2 \ln \relax (x )}{\ln \relax (2)}\)	\(27\)
default	\(\frac {2 \ln \relax (x )+\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}} \ln \left (x^{2}\right )}{x}}{\ln \relax (2)}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-3*exp(exp(4))^2*ln(x^2)+6*exp(exp(4))^2+6*x)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.39, size = 40, normalized size = 1.82 \begin {gather*} \frac {{\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} e^{\left (2 \, e^{4}\right )} - \frac {2 \, e^{\left (2 \, e^{4}\right )}}{x} + 2 \, \log \relax (x)}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*exp(exp(4))^2*log(x^2)+6*exp(exp(4))^2+6*x)/x^2/log(2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.91, size = 19, normalized size = 0.86 \begin {gather*} \frac {\ln \left (x^2\right )\,\left (x+{\mathrm {e}}^{2\,{\mathrm {e}}^4}\right )}{x\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 2*exp(2*exp(4)) - log(x^2)*exp(2*exp(4)))/(x^2*log(2)),x)

[Out]

(log(x^2)*(x + exp(2*exp(4))))/(x*log(2))

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 24, normalized size = 1.09 \begin {gather*} \frac {2 \log {\relax (x )}}{\log {\relax (2 )}} + \frac {e^{2 e^{4}} \log {\left (x^{2} \right )}}{x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*exp(exp(4))**2*ln(x**2)+6*exp(exp(4))**2+6*x)/x**2/ln(2),x)

[Out]

2*log(x)/log(2) + exp(2*exp(4))*log(x**2)/(x*log(2))

________________________________________________________________________________________