\(\int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+(5775-1500 x+100 x^2) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx\) [9509]

Optimal result

Integrand size = 52, antiderivative size = 28 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=\frac {5 x^2+\log (x)}{\frac {2}{5 \left (2-\frac {15}{x}\right )}-x} \]

[Out]

(5*x^2+ln(x))/(2/(10-75/x)-x)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.173, Rules used = {1608, 27, 6874, 46, 45, 2404, 2341, 2351, 31} \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 x+\frac {77}{77-10 x}+\frac {200 x \log (x)}{5929 (77-10 x)}+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x} \]

[In]

Int[(-5775 + 1520*x - 28975*x^2 + 7700*x^3 - 500*x^4 + (5775 - 1500*x + 100*x^2)*Log[x])/(5929*x^2 - 1540*x^3
+ 100*x^4),x]

[Out]

77/(77 - 10*x) - 5*x + (20*Log[x])/5929 - (75*Log[x])/(77*x) + (200*x*Log[x])/(5929*(77 - 10*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 1608

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 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 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{x^2 \left (5929-1540 x+100 x^2\right )} \, dx \\ & = \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{x^2 (-77+10 x)^2} \, dx \\ & = \int \left (-\frac {28975}{(-77+10 x)^2}-\frac {5775}{x^2 (-77+10 x)^2}+\frac {1520}{x (-77+10 x)^2}+\frac {7700 x}{(-77+10 x)^2}-\frac {500 x^2}{(-77+10 x)^2}+\frac {25 \left (231-60 x+4 x^2\right ) \log (x)}{x^2 (-77+10 x)^2}\right ) \, dx \\ & = -\frac {5795}{2 (77-10 x)}+25 \int \frac {\left (231-60 x+4 x^2\right ) \log (x)}{x^2 (-77+10 x)^2} \, dx-500 \int \frac {x^2}{(-77+10 x)^2} \, dx+1520 \int \frac {1}{x (-77+10 x)^2} \, dx-5775 \int \frac {1}{x^2 (-77+10 x)^2} \, dx+7700 \int \frac {x}{(-77+10 x)^2} \, dx \\ & = -\frac {5795}{2 (77-10 x)}+25 \int \left (\frac {3 \log (x)}{77 x^2}+\frac {8 \log (x)}{77 (-77+10 x)^2}\right ) \, dx-500 \int \left (\frac {1}{100}+\frac {5929}{100 (-77+10 x)^2}+\frac {77}{50 (-77+10 x)}\right ) \, dx+1520 \int \left (\frac {1}{5929 x}+\frac {10}{77 (-77+10 x)^2}-\frac {10}{5929 (-77+10 x)}\right ) \, dx-5775 \int \left (\frac {1}{5929 x^2}+\frac {20}{456533 x}+\frac {100}{5929 (-77+10 x)^2}-\frac {200}{456533 (-77+10 x)}\right ) \, dx+7700 \int \left (\frac {77}{10 (-77+10 x)^2}+\frac {1}{10 (-77+10 x)}\right ) \, dx \\ & = \frac {77}{77-10 x}+\frac {75}{77 x}-5 x-\frac {20 \log (77-10 x)}{5929}+\frac {20 \log (x)}{5929}+\frac {75}{77} \int \frac {\log (x)}{x^2} \, dx+\frac {200}{77} \int \frac {\log (x)}{(-77+10 x)^2} \, dx \\ & = \frac {77}{77-10 x}-5 x-\frac {20 \log (77-10 x)}{5929}+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x}+\frac {200 x \log (x)}{5929 (77-10 x)}+\frac {200 \int \frac {1}{-77+10 x} \, dx}{5929} \\ & = \frac {77}{77-10 x}-5 x+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x}+\frac {200 x \log (x)}{5929 (77-10 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 \left (-\frac {77}{5 (77-10 x)}+x-\frac {4 \log (x)}{77 (77-10 x)}+\frac {15 \log (x)}{77 x}\right ) \]

[In]

Integrate[(-5775 + 1520*x - 28975*x^2 + 7700*x^3 - 500*x^4 + (5775 - 1500*x + 100*x^2)*Log[x])/(5929*x^2 - 154
0*x^3 + 100*x^4),x]

[Out]

-5*(-77/(5*(77 - 10*x)) + x - (4*Log[x])/(77*(77 - 10*x)) + (15*Log[x])/(77*x))

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07

method	result	size
norman	\(\frac {\frac {5775 x}{2}-10 x \ln \left (x \right )-50 x^{3}+75 \ln \left (x \right )}{x \left (10 x -77\right )}\)	\(30\)
parallelrisch	\(\frac {-500 x^{3}-100 x \ln \left (x \right )+28875 x +750 \ln \left (x \right )}{10 x \left (10 x -77\right )}\)	\(31\)
default	\(-5 x -\frac {77}{10 x -77}+\frac {20 \ln \left (x \right )}{5929}-\frac {200 \ln \left (x \right ) x}{5929 \left (10 x -77\right )}-\frac {75 \ln \left (x \right )}{77 x}\)	\(37\)
parts	\(-5 x -\frac {77}{10 x -77}+\frac {20 \ln \left (x \right )}{5929}-\frac {200 \ln \left (x \right ) x}{5929 \left (10 x -77\right )}-\frac {75 \ln \left (x \right )}{77 x}\)	\(37\)
risch	\(-\frac {5 \left (2 x -15\right ) \ln \left (x \right )}{x \left (10 x -77\right )}-\frac {50 x^{2}-385 x +77}{10 x -77}\)	\(40\)

[In]

int(((100*x^2-1500*x+5775)*ln(x)-500*x^4+7700*x^3-28975*x^2+1520*x-5775)/(100*x^4-1540*x^3+5929*x^2),x,method=
_RETURNVERBOSE)

[Out]

(5775/2*x-10*x*ln(x)-50*x^3+75*ln(x))/x/(10*x-77)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {50 \, x^{3} - 385 \, x^{2} + 5 \, {\left (2 \, x - 15\right )} \log \left (x\right ) + 77 \, x}{10 \, x^{2} - 77 \, x} \]

[In]

integrate(((100*x^2-1500*x+5775)*log(x)-500*x^4+7700*x^3-28975*x^2+1520*x-5775)/(100*x^4-1540*x^3+5929*x^2),x,
algorithm="fricas")

[Out]

-(50*x^3 - 385*x^2 + 5*(2*x - 15)*log(x) + 77*x)/(10*x^2 - 77*x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=- 5 x + \frac {\left (75 - 10 x\right ) \log {\left (x \right )}}{10 x^{2} - 77 x} - \frac {77}{10 x - 77} \]

[In]

integrate(((100*x**2-1500*x+5775)*ln(x)-500*x**4+7700*x**3-28975*x**2+1520*x-5775)/(100*x**4-1540*x**3+5929*x*
*2),x)

[Out]

-5*x + (75 - 10*x)*log(x)/(10*x**2 - 77*x) - 77/(10*x - 77)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).

Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 \, x - \frac {25 \, {\left ({\left (8 \, x^{2} + 2310 \, x - 17787\right )} \log \left (x\right ) + 2310 \, x - 17787\right )}}{5929 \, {\left (10 \, x^{2} - 77 \, x\right )}} + \frac {75 \, {\left (20 \, x - 77\right )}}{77 \, {\left (10 \, x^{2} - 77 \, x\right )}} - \frac {6679}{77 \, {\left (10 \, x - 77\right )}} + \frac {20}{5929} \, \log \left (x\right ) \]

[In]

integrate(((100*x^2-1500*x+5775)*log(x)-500*x^4+7700*x^3-28975*x^2+1520*x-5775)/(100*x^4-1540*x^3+5929*x^2),x,
algorithm="maxima")

[Out]

-5*x - 25/5929*((8*x^2 + 2310*x - 17787)*log(x) + 2310*x - 17787)/(10*x^2 - 77*x) + 75/77*(20*x - 77)/(10*x^2
- 77*x) - 6679/77/(10*x - 77) + 20/5929*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {5}{77} \, {\left (\frac {4}{10 \, x - 77} + \frac {15}{x}\right )} \log \left (x\right ) - 5 \, x - \frac {77}{10 \, x - 77} \]

[In]

integrate(((100*x^2-1500*x+5775)*log(x)-500*x^4+7700*x^3-28975*x^2+1520*x-5775)/(100*x^4-1540*x^3+5929*x^2),x,
algorithm="giac")

[Out]

-5/77*(4/(10*x - 77) + 15/x)*log(x) - 5*x - 77/(10*x - 77)

Mupad [B] (verification not implemented)

Time = 14.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {5\,\left (2\,x-15\right )\,\left (\ln \left (x\right )+5\,x^2\right )}{x\,\left (10\,x-77\right )} \]

[In]

int((1520*x + log(x)*(100*x^2 - 1500*x + 5775) - 28975*x^2 + 7700*x^3 - 500*x^4 - 5775)/(5929*x^2 - 1540*x^3 +
100*x^4),x)

[Out]

-(5*(2*x - 15)*(log(x) + 5*x^2))/(x*(10*x - 77))