3.37.31 \(\int \frac {1125-450 x+45 x^2+12095 x^4-4879 x^5+492 x^6+(-615 x^4+82 x^5) \log (x)}{225 x^2-90 x^3+9 x^4} \, dx\)

Optimal. Leaf size=23 \[ \frac {-5+\frac {41}{9} x^4 \left (4+\frac {\log (x)}{-5+x}\right )}{x} \]

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Rubi [A]  time = 0.44, antiderivative size = 43, normalized size of antiderivative = 1.87, number of steps used = 21, number of rules used = 11, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1594, 27, 12, 6742, 44, 43, 2357, 2295, 2314, 31, 2304} \begin {gather*} \frac {164 x^3}{9}+\frac {41}{9} x^2 \log (x)-\frac {5}{x}-\frac {1025 x \log (x)}{9 (5-x)}+\frac {205}{9} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1125 - 450*x + 45*x^2 + 12095*x^4 - 4879*x^5 + 492*x^6 + (-615*x^4 + 82*x^5)*Log[x])/(225*x^2 - 90*x^3 +
9*x^4),x]

[Out]

-5/x + (164*x^3)/9 + (205*x*Log[x])/9 - (1025*x*Log[x])/(9*(5 - x)) + (41*x^2*Log[x])/9

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 31

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

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 2295

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

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 2314

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 2357

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 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 {1125-450 x+45 x^2+12095 x^4-4879 x^5+492 x^6+\left (-615 x^4+82 x^5\right ) \log (x)}{x^2 \left (225-90 x+9 x^2\right )} \, dx\\ &=\int \frac {1125-450 x+45 x^2+12095 x^4-4879 x^5+492 x^6+\left (-615 x^4+82 x^5\right ) \log (x)}{9 (-5+x)^2 x^2} \, dx\\ &=\frac {1}{9} \int \frac {1125-450 x+45 x^2+12095 x^4-4879 x^5+492 x^6+\left (-615 x^4+82 x^5\right ) \log (x)}{(-5+x)^2 x^2} \, dx\\ &=\frac {1}{9} \int \left (\frac {45}{(-5+x)^2}+\frac {1125}{(-5+x)^2 x^2}-\frac {450}{(-5+x)^2 x}+\frac {12095 x^2}{(-5+x)^2}-\frac {4879 x^3}{(-5+x)^2}+\frac {492 x^4}{(-5+x)^2}+\frac {41 x^2 (-15+2 x) \log (x)}{(-5+x)^2}\right ) \, dx\\ &=\frac {5}{5-x}+\frac {41}{9} \int \frac {x^2 (-15+2 x) \log (x)}{(-5+x)^2} \, dx-50 \int \frac {1}{(-5+x)^2 x} \, dx+\frac {164}{3} \int \frac {x^4}{(-5+x)^2} \, dx+125 \int \frac {1}{(-5+x)^2 x^2} \, dx-\frac {4879}{9} \int \frac {x^3}{(-5+x)^2} \, dx+\frac {12095}{9} \int \frac {x^2}{(-5+x)^2} \, dx\\ &=\frac {5}{5-x}+\frac {41}{9} \int \left (5 \log (x)-\frac {125 \log (x)}{(-5+x)^2}+2 x \log (x)\right ) \, dx-50 \int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx+\frac {164}{3} \int \left (75+\frac {625}{(-5+x)^2}+\frac {500}{-5+x}+10 x+x^2\right ) \, dx+125 \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 {4879}{9} \int \left (10+\frac {125}{(-5+x)^2}+\frac {75}{-5+x}+x\right ) \, dx+\frac {12095}{9} \int \left (1+\frac {25}{(-5+x)^2}+\frac {10}{-5+x}\right ) \, dx\\ &=-\frac {5}{x}+\frac {205 x}{9}+\frac {41 x^2}{18}+\frac {164 x^3}{9}+\frac {1025}{9} \log (5-x)+\frac {82}{9} \int x \log (x) \, dx+\frac {205}{9} \int \log (x) \, dx-\frac {5125}{9} \int \frac {\log (x)}{(-5+x)^2} \, dx\\ &=-\frac {5}{x}+\frac {164 x^3}{9}+\frac {1025}{9} \log (5-x)+\frac {205}{9} x \log (x)-\frac {1025 x \log (x)}{9 (5-x)}+\frac {41}{9} x^2 \log (x)-\frac {1025}{9} \int \frac {1}{-5+x} \, dx\\ &=-\frac {5}{x}+\frac {164 x^3}{9}+\frac {205}{9} x \log (x)-\frac {1025 x \log (x)}{9 (5-x)}+\frac {41}{9} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 34, normalized size = 1.48 \begin {gather*} \frac {225-45 x-820 x^4+164 x^5+41 x^4 \log (x)}{9 (-5+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1125 - 450*x + 45*x^2 + 12095*x^4 - 4879*x^5 + 492*x^6 + (-615*x^4 + 82*x^5)*Log[x])/(225*x^2 - 90*
x^3 + 9*x^4),x]

[Out]

(225 - 45*x - 820*x^4 + 164*x^5 + 41*x^4*Log[x])/(9*(-5 + x)*x)

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fricas [A]  time = 0.61, size = 33, normalized size = 1.43 \begin {gather*} \frac {164 \, x^{5} + 41 \, x^{4} \log \relax (x) - 820 \, x^{4} - 45 \, x + 225}{9 \, {\left (x^{2} - 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((82*x^5-615*x^4)*log(x)+492*x^6-4879*x^5+12095*x^4+45*x^2-450*x+1125)/(9*x^4-90*x^3+225*x^2),x, alg
orithm="fricas")

[Out]

1/9*(164*x^5 + 41*x^4*log(x) - 820*x^4 - 45*x + 225)/(x^2 - 5*x)

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giac [A]  time = 0.18, size = 33, normalized size = 1.43 \begin {gather*} \frac {164}{9} \, x^{3} + \frac {41}{9} \, {\left (x^{2} + 5 \, x + \frac {125}{x - 5}\right )} \log \relax (x) - \frac {5}{x} + \frac {1025}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((82*x^5-615*x^4)*log(x)+492*x^6-4879*x^5+12095*x^4+45*x^2-450*x+1125)/(9*x^4-90*x^3+225*x^2),x, alg
orithm="giac")

[Out]

164/9*x^3 + 41/9*(x^2 + 5*x + 125/(x - 5))*log(x) - 5/x + 1025/9*log(x)

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maple [A]  time = 0.05, size = 32, normalized size = 1.39

method	result	size
norman	\(\frac {25-5 x -\frac {820 x^{4}}{9}+\frac {164 x^{5}}{9}+\frac {41 x^{4} \ln \relax (x )}{9}}{\left (x -5\right ) x}\)	\(32\)
default	\(\frac {164 x^{3}}{9}-\frac {5}{x}+\frac {41 x^{2} \ln \relax (x )}{9}+\frac {205 x \ln \relax (x )}{9}+\frac {1025 \ln \relax (x ) x}{9 \left (x -5\right )}\)	\(34\)
risch	\(\frac {41 \left (x^{3}-25 x +125\right ) \ln \relax (x )}{9 \left (x -5\right )}+\frac {164 x^{4}+1025 x \ln \relax (x )-45}{9 x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((82*x^5-615*x^4)*ln(x)+492*x^6-4879*x^5+12095*x^4+45*x^2-450*x+1125)/(9*x^4-90*x^3+225*x^2),x,method=_RET
URNVERBOSE)

[Out]

(25-5*x-820/9*x^4+164/9*x^5+41/9*x^4*ln(x))/(x-5)/x

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maxima [B]  time = 0.51, size = 72, normalized size = 3.13 \begin {gather*} \frac {164}{9} \, x^{3} + \frac {41}{18} \, x^{2} + \frac {205}{9} \, x - \frac {5 \, {\left (2 \, x - 5\right )}}{x^{2} - 5 \, x} - \frac {41 \, {\left (x^{3} + 5 \, x^{2} - 2 \, {\left (x^{3} - 25 \, x + 125\right )} \log \relax (x) - 50 \, x\right )}}{18 \, {\left (x - 5\right )}} + \frac {5}{x - 5} + \frac {1025}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((82*x^5-615*x^4)*log(x)+492*x^6-4879*x^5+12095*x^4+45*x^2-450*x+1125)/(9*x^4-90*x^3+225*x^2),x, alg
orithm="maxima")

[Out]

164/9*x^3 + 41/18*x^2 + 205/9*x - 5*(2*x - 5)/(x^2 - 5*x) - 41/18*(x^3 + 5*x^2 - 2*(x^3 - 25*x + 125)*log(x) -
50*x)/(x - 5) + 5/(x - 5) + 1025/9*log(x)

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mupad [B]  time = 2.20, size = 37, normalized size = 1.61 \begin {gather*} -\frac {41\,x^4\,\ln \relax (x)-9\,x^2-820\,x^4+164\,x^5+225}{45\,x-9\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*x^2 - log(x)*(615*x^4 - 82*x^5) - 450*x + 12095*x^4 - 4879*x^5 + 492*x^6 + 1125)/(225*x^2 - 90*x^3 + 9
*x^4),x)

[Out]

-(41*x^4*log(x) - 9*x^2 - 820*x^4 + 164*x^5 + 225)/(45*x - 9*x^2)

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sympy [A]  time = 0.21, size = 34, normalized size = 1.48 \begin {gather*} \frac {164 x^{3}}{9} + \frac {1025 \log {\relax (x )}}{9} + \frac {\left (41 x^{3} - 1025 x + 5125\right ) \log {\relax (x )}}{9 x - 45} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((82*x**5-615*x**4)*ln(x)+492*x**6-4879*x**5+12095*x**4+45*x**2-450*x+1125)/(9*x**4-90*x**3+225*x**2
),x)

[Out]

164*x**3/9 + 1025*log(x)/9 + (41*x**3 - 1025*x + 5125)*log(x)/(9*x - 45) - 5/x

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