3.79.34 \(\int \frac {18-42 x+32 x^2-8 x^3+(18 x-24 x^2+8 x^3) \log (\frac {x}{5})+(-60 x+72 x^2-24 x^3) \log ^2(\frac {x}{5})}{(9 x-12 x^2+4 x^3) \log ^2(\frac {x}{5})} \, dx\)

Optimal. Leaf size=34 \[ -2-x-5 (1+x)+\frac {2 x}{-3+2 x}+\frac {-2+2 x}{\log \left (\frac {x}{5}\right )} \]

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Rubi [A]  time = 0.44, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {1594, 27, 6688, 12, 683, 2330, 2297, 2298, 2302, 30} \begin {gather*} -6 x-\frac {3}{3-2 x}+\frac {2 x}{\log \left (\frac {x}{5}\right )}-\frac {2}{\log \left (\frac {x}{5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 - 42*x + 32*x^2 - 8*x^3 + (18*x - 24*x^2 + 8*x^3)*Log[x/5] + (-60*x + 72*x^2 - 24*x^3)*Log[x/5]^2)/((9
*x - 12*x^2 + 4*x^3)*Log[x/5]^2),x]

[Out]

-3/(3 - 2*x) - 6*x - 2/Log[x/5] + (2*x)/Log[x/5]

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

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} &=\int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{x \left (9-12 x+4 x^2\right ) \log ^2\left (\frac {x}{5}\right )} \, dx\\ &=\int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{x (-3+2 x)^2 \log ^2\left (\frac {x}{5}\right )} \, dx\\ &=\int 2 \left (-\frac {6 \left (5-6 x+2 x^2\right )}{(3-2 x)^2}+\frac {-1+\frac {1}{x}}{\log ^2\left (\frac {x}{5}\right )}+\frac {1}{\log \left (\frac {x}{5}\right )}\right ) \, dx\\ &=2 \int \left (-\frac {6 \left (5-6 x+2 x^2\right )}{(3-2 x)^2}+\frac {-1+\frac {1}{x}}{\log ^2\left (\frac {x}{5}\right )}+\frac {1}{\log \left (\frac {x}{5}\right )}\right ) \, dx\\ &=2 \int \frac {-1+\frac {1}{x}}{\log ^2\left (\frac {x}{5}\right )} \, dx+2 \int \frac {1}{\log \left (\frac {x}{5}\right )} \, dx-12 \int \frac {5-6 x+2 x^2}{(3-2 x)^2} \, dx\\ &=10 \text {li}\left (\frac {x}{5}\right )+2 \int \left (-\frac {1}{\log ^2\left (\frac {x}{5}\right )}+\frac {1}{x \log ^2\left (\frac {x}{5}\right )}\right ) \, dx-12 \int \left (\frac {1}{2}+\frac {1}{2 (-3+2 x)^2}\right ) \, dx\\ &=-\frac {3}{3-2 x}-6 x+10 \text {li}\left (\frac {x}{5}\right )-2 \int \frac {1}{\log ^2\left (\frac {x}{5}\right )} \, dx+2 \int \frac {1}{x \log ^2\left (\frac {x}{5}\right )} \, dx\\ &=-\frac {3}{3-2 x}-6 x+\frac {2 x}{\log \left (\frac {x}{5}\right )}+10 \text {li}\left (\frac {x}{5}\right )-2 \int \frac {1}{\log \left (\frac {x}{5}\right )} \, dx+2 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {x}{5}\right )\right )\\ &=-\frac {3}{3-2 x}-6 x-\frac {2}{\log \left (\frac {x}{5}\right )}+\frac {2 x}{\log \left (\frac {x}{5}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.13, size = 60, normalized size = 1.76 \begin {gather*} 2 \left (-\frac {3}{2 (3-2 x)}-\frac {3}{2} (-3+2 x)-5 \text {Ei}\left (\log \left (\frac {x}{5}\right )\right )-\frac {1}{\log \left (\frac {x}{5}\right )}+\frac {x}{\log \left (\frac {x}{5}\right )}+5 \text {li}\left (\frac {x}{5}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 42*x + 32*x^2 - 8*x^3 + (18*x - 24*x^2 + 8*x^3)*Log[x/5] + (-60*x + 72*x^2 - 24*x^3)*Log[x/5]^
2)/((9*x - 12*x^2 + 4*x^3)*Log[x/5]^2),x]

[Out]

2*(-3/(2*(3 - 2*x)) - (3*(-3 + 2*x))/2 - 5*ExpIntegralEi[Log[x/5]] - Log[x/5]^(-1) + x/Log[x/5] + 5*LogIntegra
l[x/5])

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fricas [A]  time = 0.89, size = 40, normalized size = 1.18 \begin {gather*} \frac {4 \, x^{2} - 3 \, {\left (4 \, x^{2} - 6 \, x - 1\right )} \log \left (\frac {1}{5} \, x\right ) - 10 \, x + 6}{{\left (2 \, x - 3\right )} \log \left (\frac {1}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5*x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x
^2+9*x)/log(1/5*x)^2,x, algorithm="fricas")

[Out]

(4*x^2 - 3*(4*x^2 - 6*x - 1)*log(1/5*x) - 10*x + 6)/((2*x - 3)*log(1/5*x))

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giac [A]  time = 0.16, size = 24, normalized size = 0.71 \begin {gather*} -6 \, x + \frac {2 \, {\left (x - 1\right )}}{\log \left (\frac {1}{5} \, x\right )} + \frac {3}{2 \, x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5*x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x
^2+9*x)/log(1/5*x)^2,x, algorithm="giac")

[Out]

-6*x + 2*(x - 1)/log(1/5*x) + 3/(2*x - 3)

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maple [A]  time = 0.05, size = 31, normalized size = 0.91

method	result	size
derivativedivides	\(-6 x +\frac {3}{2 x -3}+\frac {2 x}{\ln \left (\frac {x}{5}\right )}-\frac {2}{\ln \left (\frac {x}{5}\right )}\)	\(31\)
default	\(-6 x +\frac {3}{2 x -3}+\frac {2 x}{\ln \left (\frac {x}{5}\right )}-\frac {2}{\ln \left (\frac {x}{5}\right )}\)	\(31\)
risch	\(-\frac {3 \left (4 x^{2}-6 x -1\right )}{2 x -3}+\frac {2 x -2}{\ln \left (\frac {x}{5}\right )}\)	\(32\)
norman	\(\frac {6+30 \ln \left (\frac {x}{5}\right )-10 x +4 x^{2}-12 \ln \left (\frac {x}{5}\right ) x^{2}}{\left (2 x -3\right ) \ln \left (\frac {x}{5}\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-24*x^3+72*x^2-60*x)*ln(1/5*x)^2+(8*x^3-24*x^2+18*x)*ln(1/5*x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/
ln(1/5*x)^2,x,method=_RETURNVERBOSE)

[Out]

-6*x+3/(2*x-3)+2/ln(1/5*x)*x-2/ln(1/5*x)

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maxima [B]  time = 0.49, size = 63, normalized size = 1.85 \begin {gather*} -\frac {4 \, x^{2} {\left (3 \, \log \relax (5) + 1\right )} - 2 \, x {\left (9 \, \log \relax (5) + 5\right )} - 3 \, {\left (4 \, x^{2} - 6 \, x - 1\right )} \log \relax (x) - 3 \, \log \relax (5) + 6}{2 \, x \log \relax (5) - {\left (2 \, x - 3\right )} \log \relax (x) - 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5*x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x
^2+9*x)/log(1/5*x)^2,x, algorithm="maxima")

[Out]

-(4*x^2*(3*log(5) + 1) - 2*x*(9*log(5) + 5) - 3*(4*x^2 - 6*x - 1)*log(x) - 3*log(5) + 6)/(2*x*log(5) - (2*x -
3)*log(x) - 3*log(5))

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mupad [B]  time = 6.80, size = 42, normalized size = 1.24 \begin {gather*} \frac {20\,x-12\,x^2}{2\,x-3}+\frac {4\,x^2-10\,x+6}{\ln \left (\frac {x}{5}\right )\,\left (2\,x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(42*x - log(x/5)*(18*x - 24*x^2 + 8*x^3) + log(x/5)^2*(60*x - 72*x^2 + 24*x^3) - 32*x^2 + 8*x^3 - 18)/(lo
g(x/5)^2*(9*x - 12*x^2 + 4*x^3)),x)

[Out]

(20*x - 12*x^2)/(2*x - 3) + (4*x^2 - 10*x + 6)/(log(x/5)*(2*x - 3))

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sympy [A]  time = 0.18, size = 19, normalized size = 0.56 \begin {gather*} - 6 x + \frac {2 x - 2}{\log {\left (\frac {x}{5} \right )}} + \frac {3}{2 x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x**3+72*x**2-60*x)*ln(1/5*x)**2+(8*x**3-24*x**2+18*x)*ln(1/5*x)-8*x**3+32*x**2-42*x+18)/(4*x**
3-12*x**2+9*x)/ln(1/5*x)**2,x)

[Out]

-6*x + (2*x - 2)/log(x/5) + 3/(2*x - 3)

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