\(\int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+(-121+x^2+22 \log (2)-\log ^2(2)) \log (x)}{x^2} \, dx\) [7787]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 44, antiderivative size = 22 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=5+x+x \left (2+\frac {(11+x-\log (2))^2 \log (x)}{x^2}\right ) \]

[Out]

5+(ln(x)/x^2*(11-ln(2)+x)^2+2)*x+x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14, 2404, 2332, 2341} \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=3 x+x \log (x)+(22-\log (4)) \log (x)+\frac {(11-\log (2))^2 \log (x)}{x} \]

[In]

Int[(121 + 22*x + 4*x^2 + (-22 - 2*x)*Log[2] + Log[2]^2 + (-121 + x^2 + 22*Log[2] - Log[2]^2)*Log[x])/x^2,x]

[Out]

3*x + x*Log[x] + ((11 - Log[2])^2*Log[x])/x + (22 - Log[4])*Log[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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2341

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
]

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4 x^2+(-11+\log (2))^2+x (22-\log (4))}{x^2}+\frac {(11+x-\log (2)) (-11+x+\log (2)) \log (x)}{x^2}\right ) \, dx \\ & = \int \frac {4 x^2+(-11+\log (2))^2+x (22-\log (4))}{x^2} \, dx+\int \frac {(11+x-\log (2)) (-11+x+\log (2)) \log (x)}{x^2} \, dx \\ & = \int \left (4+\frac {(-11+\log (2))^2}{x^2}+\frac {22-\log (4)}{x}\right ) \, dx+\int \left (\log (x)-\frac {(-11+\log (2))^2 \log (x)}{x^2}\right ) \, dx \\ & = 4 x-\frac {(11-\log (2))^2}{x}+(22-\log (4)) \log (x)-(-11+\log (2))^2 \int \frac {\log (x)}{x^2} \, dx+\int \log (x) \, dx \\ & = 3 x+x \log (x)+\frac {(11-\log (2))^2 \log (x)}{x}+(22-\log (4)) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=3 x+22 \log (x)+\frac {121 \log (x)}{x}+x \log (x)-2 \log (2) \log (x)-\frac {22 \log (2) \log (x)}{x}+\frac {\log ^2(2) \log (x)}{x} \]

[In]

Integrate[(121 + 22*x + 4*x^2 + (-22 - 2*x)*Log[2] + Log[2]^2 + (-121 + x^2 + 22*Log[2] - Log[2]^2)*Log[x])/x^
2,x]

[Out]

3*x + 22*Log[x] + (121*Log[x])/x + x*Log[x] - 2*Log[2]*Log[x] - (22*Log[2]*Log[x])/x + (Log[2]^2*Log[x])/x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55

method result size
risch \(\frac {\left (\ln \left (2\right )^{2}+x^{2}-22 \ln \left (2\right )+121\right ) \ln \left (x \right )}{x}-2 \ln \left (2\right ) \ln \left (x \right )+22 \ln \left (x \right )+3 x\) \(34\)
norman \(\frac {x^{2} \ln \left (x \right )+\left (\ln \left (2\right )^{2}-22 \ln \left (2\right )+121\right ) \ln \left (x \right )+\left (-2 \ln \left (2\right )+22\right ) x \ln \left (x \right )+3 x^{2}}{x}\) \(40\)
parallelrisch \(\frac {\ln \left (2\right )^{2} \ln \left (x \right )-2 x \ln \left (2\right ) \ln \left (x \right )+x^{2} \ln \left (x \right )-22 \ln \left (2\right ) \ln \left (x \right )+3 x^{2}+22 x \ln \left (x \right )+121 \ln \left (x \right )}{x}\) \(46\)
default \(-\ln \left (2\right )^{2} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x \ln \left (x \right )+3 x +22 \ln \left (2\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )-\frac {\ln \left (2\right )^{2}}{x}-2 \ln \left (2\right ) \ln \left (x \right )+\frac {121 \ln \left (x \right )}{x}+\frac {22 \ln \left (2\right )}{x}+22 \ln \left (x \right )\) \(78\)
parts \(3 x +\left (-2 \ln \left (2\right )+22\right ) \ln \left (x \right )-\frac {\ln \left (2\right )^{2}-22 \ln \left (2\right )+121}{x}-\ln \left (2\right )^{2} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x \ln \left (x \right )+22 \ln \left (2\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+\frac {121 \ln \left (x \right )}{x}+\frac {121}{x}\) \(81\)

[In]

int(((-ln(2)^2+22*ln(2)+x^2-121)*ln(x)+ln(2)^2+(-2*x-22)*ln(2)+4*x^2+22*x+121)/x^2,x,method=_RETURNVERBOSE)

[Out]

(ln(2)^2+x^2-22*ln(2)+121)/x*ln(x)-2*ln(2)*ln(x)+22*ln(x)+3*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=\frac {3 \, x^{2} + {\left (x^{2} - 2 \, {\left (x + 11\right )} \log \left (2\right ) + \log \left (2\right )^{2} + 22 \, x + 121\right )} \log \left (x\right )}{x} \]

[In]

integrate(((-log(2)^2+22*log(2)+x^2-121)*log(x)+log(2)^2+(-2*x-22)*log(2)+4*x^2+22*x+121)/x^2,x, algorithm="fr
icas")

[Out]

(3*x^2 + (x^2 - 2*(x + 11)*log(2) + log(2)^2 + 22*x + 121)*log(x))/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=3 x - 2 \left (-11 + \log {\left (2 \right )}\right ) \log {\left (x \right )} + \frac {\left (x^{2} - 22 \log {\left (2 \right )} + \log {\left (2 \right )}^{2} + 121\right ) \log {\left (x \right )}}{x} \]

[In]

integrate(((-ln(2)**2+22*ln(2)+x**2-121)*ln(x)+ln(2)**2+(-2*x-22)*ln(2)+4*x**2+22*x+121)/x**2,x)

[Out]

3*x - 2*(-11 + log(2))*log(x) + (x**2 - 22*log(2) + log(2)**2 + 121)*log(x)/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (22) = 44\).

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.18 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx={\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (2\right )^{2} - 22 \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (2\right ) + x \log \left (x\right ) - 2 \, \log \left (2\right ) \log \left (x\right ) + 3 \, x - \frac {\log \left (2\right )^{2}}{x} + \frac {22 \, \log \left (2\right )}{x} + \frac {121 \, \log \left (x\right )}{x} + 22 \, \log \left (x\right ) \]

[In]

integrate(((-log(2)^2+22*log(2)+x^2-121)*log(x)+log(2)^2+(-2*x-22)*log(2)+4*x^2+22*x+121)/x^2,x, algorithm="ma
xima")

[Out]

(log(x)/x + 1/x)*log(2)^2 - 22*(log(x)/x + 1/x)*log(2) + x*log(x) - 2*log(2)*log(x) + 3*x - log(2)^2/x + 22*lo
g(2)/x + 121*log(x)/x + 22*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx={\left (x + \frac {\log \left (2\right )^{2} - 22 \, \log \left (2\right ) + 121}{x}\right )} \log \left (x\right ) - 2 \, {\left (\log \left (2\right ) - 11\right )} \log \left (x\right ) + 3 \, x \]

[In]

integrate(((-log(2)^2+22*log(2)+x^2-121)*log(x)+log(2)^2+(-2*x-22)*log(2)+4*x^2+22*x+121)/x^2,x, algorithm="gi
ac")

[Out]

(x + (log(2)^2 - 22*log(2) + 121)/x)*log(x) - 2*(log(2) - 11)*log(x) + 3*x

Mupad [B] (verification not implemented)

Time = 14.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {121+22 x+4 x^2+(-22-2 x) \log (2)+\log ^2(2)+\left (-121+x^2+22 \log (2)-\log ^2(2)\right ) \log (x)}{x^2} \, dx=x\,\left (\ln \left (x\right )+3\right )-\ln \left (x\right )\,\left (\ln \left (4\right )-22\right )+\frac {\ln \left (x\right )\,{\left (\ln \left (2\right )-11\right )}^2}{x} \]

[In]

int((22*x - log(2)*(2*x + 22) + log(x)*(22*log(2) - log(2)^2 + x^2 - 121) + log(2)^2 + 4*x^2 + 121)/x^2,x)

[Out]

x*(log(x) + 3) - log(x)*(log(4) - 22) + (log(x)*(log(2) - 11)^2)/x