\(\int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx\) [6564]

Optimal result

Integrand size = 34, antiderivative size = 27 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\log (10) \left (-5-\frac {3}{x}+\log (x) \left (-3+\left (-1+\frac {5}{3 x}\right ) \log (x)\right )\right ) \]

[Out]

(((5/3/x-1)*ln(x)-3)*ln(x)-3/x-5)*ln(10)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {12, 14, 45, 2372, 2338, 2342, 2341} \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {5 \log (10) \log ^2(x)}{3 x}-\log (10) \log ^2(x)-3 \log (10) \log (x)-\frac {3 \log (10)}{x} \]

[In]

Int[((9 - 9*x)*Log[10] + (10 - 6*x)*Log[10]*Log[x] - 5*Log[10]*Log[x]^2)/(3*x^2),x]

[Out]

(-3*Log[10])/x - 3*Log[10]*Log[x] - Log[10]*Log[x]^2 + (5*Log[10]*Log[x]^2)/(3*x)

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 45

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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, 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 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{x^2} \, dx \\ & = \frac {1}{3} \int \left (-\frac {9 (-1+x) \log (10)}{x^2}-\frac {2 (-5+3 x) \log (10) \log (x)}{x^2}-\frac {5 \log (10) \log ^2(x)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{3} (2 \log (10)) \int \frac {(-5+3 x) \log (x)}{x^2} \, dx\right )-\frac {1}{3} (5 \log (10)) \int \frac {\log ^2(x)}{x^2} \, dx-(3 \log (10)) \int \frac {-1+x}{x^2} \, dx \\ & = -\frac {10 \log (10) \log (x)}{3 x}-2 \log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x}+\frac {1}{3} (2 \log (10)) \int \frac {5+3 x \log (x)}{x^2} \, dx-(3 \log (10)) \int \left (-\frac {1}{x^2}+\frac {1}{x}\right ) \, dx-\frac {1}{3} (10 \log (10)) \int \frac {\log (x)}{x^2} \, dx \\ & = \frac {\log (10)}{3 x}-3 \log (10) \log (x)-2 \log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x}+\frac {1}{3} (2 \log (10)) \int \left (\frac {5}{x^2}+\frac {3 \log (x)}{x}\right ) \, dx \\ & = -\frac {3 \log (10)}{x}-3 \log (10) \log (x)-2 \log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x}+(2 \log (10)) \int \frac {\log (x)}{x} \, dx \\ & = -\frac {3 \log (10)}{x}-3 \log (10) \log (x)-\log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\frac {3 \log (10)}{x}-3 \log (10) \log (x)-\log (10) \log ^2(x)+\frac {5 \log (10) \log ^2(x)}{3 x} \]

[In]

Integrate[((9 - 9*x)*Log[10] + (10 - 6*x)*Log[10]*Log[x] - 5*Log[10]*Log[x]^2)/(3*x^2),x]

[Out]

(-3*Log[10])/x - 3*Log[10]*Log[x] - Log[10]*Log[x]^2 + (5*Log[10]*Log[x]^2)/(3*x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {-3 \ln \left (10\right ) x \ln \left (x \right )+\frac {5 \ln \left (10\right ) \ln \left (x \right )^{2}}{3}-\ln \left (10\right ) x \ln \left (x \right )^{2}-3 \ln \left (10\right )}{x}\)	\(34\)
risch	\(-\frac {\left (3 x \ln \left (5\right )+3 x \ln \left (2\right )-5 \ln \left (5\right )-5 \ln \left (2\right )\right ) \ln \left (x \right )^{2}}{3 x}-\frac {3 \left (x \ln \left (5\right ) \ln \left (x \right )+x \ln \left (2\right ) \ln \left (x \right )+\ln \left (5\right )+\ln \left (2\right )\right )}{x}\)	\(52\)
parts	\(-3 \ln \left (10\right ) \left (\ln \left (x \right )+\frac {1}{x}\right )-\frac {5 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )^{2}}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {2}{x}\right )}{3}-\frac {2 \ln \left (10\right ) \left (\frac {3 \ln \left (x \right )^{2}}{2}+\frac {5 \ln \left (x \right )}{x}+\frac {5}{x}\right )}{3}\)	\(61\)
default	\(-\frac {5 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )^{2}}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {2}{x}\right )}{3}-\ln \left (10\right ) \ln \left (x \right )^{2}+\frac {10 \ln \left (10\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{3}-3 \ln \left (10\right ) \ln \left (x \right )-\frac {3 \ln \left (10\right )}{x}\)	\(66\)

[In]

int(1/3*(-5*ln(10)*ln(x)^2+(-6*x+10)*ln(10)*ln(x)+(-9*x+9)*ln(10))/x^2,x,method=_RETURNVERBOSE)

[Out]

(-3*ln(10)*x*ln(x)+5/3*ln(10)*ln(x)^2-ln(10)*x*ln(x)^2-3*ln(10))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\frac {{\left (3 \, x - 5\right )} \log \left (10\right ) \log \left (x\right )^{2} + 9 \, x \log \left (10\right ) \log \left (x\right ) + 9 \, \log \left (10\right )}{3 \, x} \]

[In]

integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(10))/x^2,x, algorithm="fricas")

[Out]

-1/3*((3*x - 5)*log(10)*log(x)^2 + 9*x*log(10)*log(x) + 9*log(10))/x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=- 3 \log {\left (10 \right )} \log {\left (x \right )} + \frac {\left (- 3 x \log {\left (10 \right )} + 5 \log {\left (10 \right )}\right ) \log {\left (x \right )}^{2}}{3 x} - \frac {3 \log {\left (10 \right )}}{x} \]

[In]

integrate(1/3*(-5*ln(10)*ln(x)**2+(-6*x+10)*ln(10)*ln(x)+(-9*x+9)*ln(10))/x**2,x)

[Out]

-3*log(10)*log(x) + (-3*x*log(10) + 5*log(10))*log(x)**2/(3*x) - 3*log(10)/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=-\log \left (10\right ) \log \left (x\right )^{2} - \frac {10}{3} \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (10\right ) - 3 \, \log \left (10\right ) \log \left (x\right ) + \frac {5 \, {\left (\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 2\right )} \log \left (10\right )}{3 \, x} - \frac {3 \, \log \left (10\right )}{x} \]

[In]

integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(10))/x^2,x, algorithm="maxima")

[Out]

-log(10)*log(x)^2 - 10/3*(log(x)/x + 1/x)*log(10) - 3*log(10)*log(x) + 5/3*(log(x)^2 + 2*log(x) + 2)*log(10)/x
- 3*log(10)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {1}{3} \, {\left (\frac {5 \, \log \left (10\right )}{x} - 3 \, \log \left (10\right )\right )} \log \left (x\right )^{2} - 3 \, \log \left (10\right ) \log \left (x\right ) - \frac {3 \, \log \left (10\right )}{x} \]

[In]

integrate(1/3*(-5*log(10)*log(x)^2+(-6*x+10)*log(10)*log(x)+(-9*x+9)*log(10))/x^2,x, algorithm="giac")

[Out]

1/3*(5*log(10)/x - 3*log(10))*log(x)^2 - 3*log(10)*log(x) - 3*log(10)/x

Mupad [B] (verification not implemented)

Time = 11.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {(9-9 x) \log (10)+(10-6 x) \log (10) \log (x)-5 \log (10) \log ^2(x)}{3 x^2} \, dx=\frac {\ln \left (10\right )\,\left (5\,{\ln \left (x\right )}^2-9\right )}{3\,x}-\frac {\ln \left (10\right )\,\left (3\,{\ln \left (x\right )}^2+9\,\ln \left (x\right )\right )}{3} \]

[In]

int(-((log(10)*(9*x - 9))/3 + (5*log(10)*log(x)^2)/3 + (log(10)*log(x)*(6*x - 10))/3)/x^2,x)

[Out]

(log(10)*(5*log(x)^2 - 9))/(3*x) - (log(10)*(9*log(x) + 3*log(x)^2))/3