3.26.58 \(\int \frac {e^2 (-2+2 x)-4 e^2 \log (64)+(2 e^2+4 e^2 \log (64)) \log (x)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac {x-2 e^2 (1-x+2 \log (64)) \log (x)}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {14, 43, 2304} \begin {gather*} 2 e^2 \log (x)-\frac {2 e^2 (1+\log (4096)) \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-2 + 2*x) - 4*E^2*Log[64] + (2*E^2 + 4*E^2*Log[64])*Log[x])/x^2,x]

[Out]

2*E^2*Log[x] - (2*E^2*(1 + Log[4096])*Log[x])/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} &=\int \left (\frac {2 e^2 (-1+x-2 \log (64))}{x^2}+\frac {2 e^2 (1+\log (4096)) \log (x)}{x^2}\right ) \, dx\\ &=\left (2 e^2\right ) \int \frac {-1+x-2 \log (64)}{x^2} \, dx+\left (2 e^2 (1+\log (4096))\right ) \int \frac {\log (x)}{x^2} \, dx\\ &=-\frac {2 e^2 (1+\log (4096))}{x}-\frac {2 e^2 (1+\log (4096)) \log (x)}{x}+\left (2 e^2\right ) \int \left (\frac {1}{x}+\frac {-1-\log (4096)}{x^2}\right ) \, dx\\ &=2 e^2 \log (x)-\frac {2 e^2 (1+\log (4096)) \log (x)}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 24, normalized size = 1.09 \begin {gather*} 2 e^2 \left (\log (x)-\frac {\log (x)}{x}-\frac {\log (4096) \log (x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-2 + 2*x) - 4*E^2*Log[64] + (2*E^2 + 4*E^2*Log[64])*Log[x])/x^2,x]

[Out]

2*E^2*(Log[x] - Log[x]/x - (Log[4096]*Log[x])/x)

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 20, normalized size = 0.91 \begin {gather*} \frac {2 \, {\left ({\left (x - 1\right )} e^{2} - 12 \, e^{2} \log \relax (2)\right )} \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*exp(2)*log(2)+2*exp(2))*log(x)-24*exp(2)*log(2)+(2*x-2)*exp(2))/x^2,x, algorithm="fricas")

[Out]

2*((x - 1)*e^2 - 12*e^2*log(2))*log(x)/x

________________________________________________________________________________________

giac [A]  time = 0.32, size = 26, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left (x e^{2} \log \relax (x) - 12 \, e^{2} \log \relax (2) \log \relax (x) - e^{2} \log \relax (x)\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*exp(2)*log(2)+2*exp(2))*log(x)-24*exp(2)*log(2)+(2*x-2)*exp(2))/x^2,x, algorithm="giac")

[Out]

2*(x*e^2*log(x) - 12*e^2*log(2)*log(x) - e^2*log(x))/x

________________________________________________________________________________________

maple [A]  time = 0.03, size = 23, normalized size = 1.05

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{2} \left (12 \ln \relax (2)+1\right ) \ln \relax (x )}{x}+2 \,{\mathrm e}^{2} \ln \relax (x )\)	\(23\)
norman	\(\frac {\left (-24 \,{\mathrm e}^{2} \ln \relax (2)-2 \,{\mathrm e}^{2}\right ) \ln \relax (x )+2 x \,{\mathrm e}^{2} \ln \relax (x )}{x}\)	\(27\)
default	\(24 \,{\mathrm e}^{2} \ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )+2 \,{\mathrm e}^{2} \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )+\frac {24 \,{\mathrm e}^{2} \ln \relax (2)}{x}+2 \,{\mathrm e}^{2} \ln \relax (x )+\frac {2 \,{\mathrm e}^{2}}{x}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*exp(2)*ln(2)+2*exp(2))*ln(x)-24*exp(2)*ln(2)+(2*x-2)*exp(2))/x^2,x,method=_RETURNVERBOSE)

[Out]

-2*exp(2)*(12*ln(2)+1)/x*ln(x)+2*exp(2)*ln(x)

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 53, normalized size = 2.41 \begin {gather*} -24 \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} e^{2} \log \relax (2) - 2 \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} e^{2} + 2 \, e^{2} \log \relax (x) + \frac {24 \, e^{2} \log \relax (2)}{x} + \frac {2 \, e^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*exp(2)*log(2)+2*exp(2))*log(x)-24*exp(2)*log(2)+(2*x-2)*exp(2))/x^2,x, algorithm="maxima")

[Out]

-24*(log(x)/x + 1/x)*e^2*log(2) - 2*(log(x)/x + 1/x)*e^2 + 2*e^2*log(x) + 24*e^2*log(2)/x + 2*e^2/x

________________________________________________________________________________________

mupad [B]  time = 1.42, size = 18, normalized size = 0.82 \begin {gather*} -\frac {2\,{\mathrm {e}}^2\,\ln \relax (x)\,\left (12\,\ln \relax (2)-x+1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*exp(2) + 24*exp(2)*log(2)) - 24*exp(2)*log(2) + exp(2)*(2*x - 2))/x^2,x)

[Out]

-(2*exp(2)*log(x)*(12*log(2) - x + 1))/x

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 27, normalized size = 1.23 \begin {gather*} 2 e^{2} \log {\relax (x )} + \frac {\left (- 24 e^{2} \log {\relax (2 )} - 2 e^{2}\right ) \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*exp(2)*ln(2)+2*exp(2))*ln(x)-24*exp(2)*ln(2)+(2*x-2)*exp(2))/x**2,x)

[Out]

2*exp(2)*log(x) + (-24*exp(2)*log(2) - 2*exp(2))*log(x)/x

________________________________________________________________________________________