3.95.59 \(\int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+(-175+70 x-7 x^2) \log (x)}{50 x^2-20 x^3+2 x^4} \, dx\)

Optimal. Leaf size=32 \[ 4+x-\frac {x}{5-x}+(14+2 x) \left (-x+\frac {\log (x)}{4 x}\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 33, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 7, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.123, Rules used = {1594, 27, 12, 6742, 44, 43, 2304} \begin {gather*} -2 x^2-13 x-\frac {5}{5-x}+\frac {\log (x)}{2}+\frac {7 \log (x)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(175 - 45*x - 663*x^2 + 61*x^3 + 54*x^4 - 8*x^5 + (-175 + 70*x - 7*x^2)*Log[x])/(50*x^2 - 20*x^3 + 2*x^4),
x]

[Out]

-5/(5 - x) - 13*x - 2*x^2 + Log[x]/2 + (7*Log[x])/(2*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 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 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 44

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] && L
tQ[m + n + 2, 0])

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 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
]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{x^2 \left (50-20 x+2 x^2\right )} \, dx\\ &=\int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{2 (-5+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \frac {175-45 x-663 x^2+61 x^3+54 x^4-8 x^5+\left (-175+70 x-7 x^2\right ) \log (x)}{(-5+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {663}{(-5+x)^2}+\frac {175}{(-5+x)^2 x^2}-\frac {45}{(-5+x)^2 x}+\frac {61 x}{(-5+x)^2}+\frac {54 x^2}{(-5+x)^2}-\frac {8 x^3}{(-5+x)^2}-\frac {7 \log (x)}{x^2}\right ) \, dx\\ &=-\frac {663}{2 (5-x)}-\frac {7}{2} \int \frac {\log (x)}{x^2} \, dx-4 \int \frac {x^3}{(-5+x)^2} \, dx-\frac {45}{2} \int \frac {1}{(-5+x)^2 x} \, dx+27 \int \frac {x^2}{(-5+x)^2} \, dx+\frac {61}{2} \int \frac {x}{(-5+x)^2} \, dx+\frac {175}{2} \int \frac {1}{(-5+x)^2 x^2} \, dx\\ &=-\frac {663}{2 (5-x)}+\frac {7}{2 x}+\frac {7 \log (x)}{2 x}-4 \int \left (10+\frac {125}{(-5+x)^2}+\frac {75}{-5+x}+x\right ) \, dx-\frac {45}{2} \int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx+27 \int \left (1+\frac {25}{(-5+x)^2}+\frac {10}{-5+x}\right ) \, dx+\frac {61}{2} \int \left (\frac {5}{(-5+x)^2}+\frac {1}{-5+x}\right ) \, dx+\frac {175}{2} \int \left (\frac {1}{25 (-5+x)^2}-\frac {2}{125 (-5+x)}+\frac {1}{25 x^2}+\frac {2}{125 x}\right ) \, dx\\ &=-\frac {5}{5-x}-13 x-2 x^2+\frac {\log (x)}{2}+\frac {7 \log (x)}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 31, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (-\frac {10}{5-x}-26 x-4 x^2+\log (x)+\frac {7 \log (x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(175 - 45*x - 663*x^2 + 61*x^3 + 54*x^4 - 8*x^5 + (-175 + 70*x - 7*x^2)*Log[x])/(50*x^2 - 20*x^3 + 2
*x^4),x]

[Out]

(-10/(5 - x) - 26*x - 4*x^2 + Log[x] + (7*Log[x])/x)/2

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fricas [A]  time = 0.70, size = 42, normalized size = 1.31 \begin {gather*} -\frac {4 \, x^{4} + 6 \, x^{3} - 130 \, x^{2} - {\left (x^{2} + 2 \, x - 35\right )} \log \relax (x) - 10 \, x}{2 \, {\left (x^{2} - 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-7*x^2+70*x-175)*log(x)-8*x^5+54*x^4+61*x^3-663*x^2-45*x+175)/(2*x^4-20*x^3+50*x^2),x, algorithm="
fricas")

[Out]

-1/2*(4*x^4 + 6*x^3 - 130*x^2 - (x^2 + 2*x - 35)*log(x) - 10*x)/(x^2 - 5*x)

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giac [A]  time = 0.14, size = 27, normalized size = 0.84 \begin {gather*} -2 \, x^{2} - 13 \, x + \frac {7 \, \log \relax (x)}{2 \, x} + \frac {5}{x - 5} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-7*x^2+70*x-175)*log(x)-8*x^5+54*x^4+61*x^3-663*x^2-45*x+175)/(2*x^4-20*x^3+50*x^2),x, algorithm="
giac")

[Out]

-2*x^2 - 13*x + 7/2*log(x)/x + 5/(x - 5) + 1/2*log(x)

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maple [A]  time = 0.05, size = 28, normalized size = 0.88

method	result	size
default	\(\frac {7 \ln \relax (x )}{2 x}-2 x^{2}-13 x +\frac {\ln \relax (x )}{2}+\frac {5}{x -5}\)	\(28\)
norman	\(\frac {330 x +x \ln \relax (x )+\frac {x^{2} \ln \relax (x )}{2}-3 x^{3}-2 x^{4}-\frac {35 \ln \relax (x )}{2}}{\left (x -5\right ) x}\)	\(39\)
risch	\(\frac {7 \ln \relax (x )}{2 x}+\frac {-4 x^{3}+x \ln \relax (x )-6 x^{2}-5 \ln \relax (x )+130 x +10}{2 x -10}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-7*x^2+70*x-175)*ln(x)-8*x^5+54*x^4+61*x^3-663*x^2-45*x+175)/(2*x^4-20*x^3+50*x^2),x,method=_RETURNVERBO
SE)

[Out]

7/2*ln(x)/x-2*x^2-13*x+1/2*ln(x)+5/(x-5)

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maxima [A]  time = 0.39, size = 45, normalized size = 1.41 \begin {gather*} -2 \, x^{2} - 13 \, x - \frac {7 \, {\left (2 \, x - 5\right )}}{2 \, {\left (x^{2} - 5 \, x\right )}} + \frac {7 \, {\left (\log \relax (x) + 1\right )}}{2 \, x} + \frac {17}{2 \, {\left (x - 5\right )}} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-7*x^2+70*x-175)*log(x)-8*x^5+54*x^4+61*x^3-663*x^2-45*x+175)/(2*x^4-20*x^3+50*x^2),x, algorithm="
maxima")

[Out]

-2*x^2 - 13*x - 7/2*(2*x - 5)/(x^2 - 5*x) + 7/2*(log(x) + 1)/x + 17/2/(x - 5) + 1/2*log(x)

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mupad [B]  time = 5.71, size = 37, normalized size = 1.16 \begin {gather*} \frac {\ln \relax (x)}{2}-13\,x-2\,x^2-\frac {\frac {35\,\ln \relax (x)}{2}-x\,\left (\frac {7\,\ln \relax (x)}{2}+5\right )}{x\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(45*x + log(x)*(7*x^2 - 70*x + 175) + 663*x^2 - 61*x^3 - 54*x^4 + 8*x^5 - 175)/(50*x^2 - 20*x^3 + 2*x^4),
x)

[Out]

log(x)/2 - 13*x - 2*x^2 - ((35*log(x))/2 - x*((7*log(x))/2 + 5))/(x*(x - 5))

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sympy [A]  time = 0.19, size = 26, normalized size = 0.81 \begin {gather*} - 2 x^{2} - 13 x + \frac {\log {\relax (x )}}{2} + \frac {5}{x - 5} + \frac {7 \log {\relax (x )}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-7*x**2+70*x-175)*ln(x)-8*x**5+54*x**4+61*x**3-663*x**2-45*x+175)/(2*x**4-20*x**3+50*x**2),x)

[Out]

-2*x**2 - 13*x + log(x)/2 + 5/(x - 5) + 7*log(x)/(2*x)

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