3.75.30 \(\int \frac {-25+20 x+5 x^2-20 x^3+(50-30 x^2) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx\)

Optimal. Leaf size=22 \[ \frac {5 \left (\frac {4 x}{5}+\log (x)\right )}{4 x^2 \left (-5+x^2\right )^2} \]

________________________________________________________________________________________

Rubi [B]  time = 0.79, antiderivative size = 116, normalized size of antiderivative = 5.27, number of steps used = 28, number of rules used = 14, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6741, 12, 6742, 199, 207, 266, 44, 290, 325, 2357, 2304, 2338, 2335, 260} \begin {gather*} \frac {3 x}{40 \left (5-x^2\right )}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 \left (5-x^2\right ) x}-\frac {1}{4 \left (5-x^2\right )^2 x}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {\log (x)}{20 x^2}+\frac {3}{40 x}+\frac {\log (x)}{100} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + 20*x + 5*x^2 - 20*x^3 + (50 - 30*x^2)*Log[x])/(-500*x^3 + 300*x^5 - 60*x^7 + 4*x^9),x]

[Out]

3/(40*x) - 1/(4*x*(5 - x^2)^2) + x/(4*(5 - x^2)^2) - 1/(8*x*(5 - x^2)) + (3*x)/(40*(5 - x^2)) + Log[x]/100 + L
og[x]/(20*x^2) + Log[x]/(4*(5 - x^2)^2) + (x^2*Log[x])/(100*(5 - x^2))

Rule 12

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 2335

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

Rule 2338

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

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 6741

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

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 {25-20 x-5 x^2+20 x^3-\left (50-30 x^2\right ) \log (x)}{4 x^3 \left (5-x^2\right )^3} \, dx\\ &=\frac {1}{4} \int \frac {25-20 x-5 x^2+20 x^3-\left (50-30 x^2\right ) \log (x)}{x^3 \left (5-x^2\right )^3} \, dx\\ &=\frac {1}{4} \int \frac {5 \left (5-4 x-x^2+4 x^3-10 \log (x)+6 x^2 \log (x)\right )}{x^3 \left (5-x^2\right )^3} \, dx\\ &=\frac {5}{4} \int \frac {5-4 x-x^2+4 x^3-10 \log (x)+6 x^2 \log (x)}{x^3 \left (5-x^2\right )^3} \, dx\\ &=\frac {5}{4} \int \left (-\frac {4}{\left (-5+x^2\right )^3}-\frac {5}{x^3 \left (-5+x^2\right )^3}+\frac {4}{x^2 \left (-5+x^2\right )^3}+\frac {1}{x \left (-5+x^2\right )^3}-\frac {2 \left (-5+3 x^2\right ) \log (x)}{x^3 \left (-5+x^2\right )^3}\right ) \, dx\\ &=\frac {5}{4} \int \frac {1}{x \left (-5+x^2\right )^3} \, dx-\frac {5}{2} \int \frac {\left (-5+3 x^2\right ) \log (x)}{x^3 \left (-5+x^2\right )^3} \, dx-5 \int \frac {1}{\left (-5+x^2\right )^3} \, dx+5 \int \frac {1}{x^2 \left (-5+x^2\right )^3} \, dx-\frac {25}{4} \int \frac {1}{x^3 \left (-5+x^2\right )^3} \, dx\\ &=-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{(-5+x)^3 x} \, dx,x,x^2\right )+\frac {3}{4} \int \frac {1}{\left (-5+x^2\right )^2} \, dx-\frac {5}{4} \int \frac {1}{x^2 \left (-5+x^2\right )^2} \, dx-\frac {5}{2} \int \left (\frac {\log (x)}{25 x^3}+\frac {2 x \log (x)}{5 \left (-5+x^2\right )^3}-\frac {x \log (x)}{25 \left (-5+x^2\right )^2}\right ) \, dx-\frac {25}{8} \operatorname {Subst}\left (\int \frac {1}{(-5+x)^3 x^2} \, dx,x,x^2\right )\\ &=-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}-\frac {3}{40} \int \frac {1}{-5+x^2} \, dx-\frac {1}{10} \int \frac {\log (x)}{x^3} \, dx+\frac {1}{10} \int \frac {x \log (x)}{\left (-5+x^2\right )^2} \, dx+\frac {3}{8} \int \frac {1}{x^2 \left (-5+x^2\right )} \, dx+\frac {5}{8} \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x)^3}-\frac {1}{25 (-5+x)^2}+\frac {1}{125 (-5+x)}-\frac {1}{125 x}\right ) \, dx,x,x^2\right )-\frac {25}{8} \operatorname {Subst}\left (\int \left (\frac {1}{25 (-5+x)^3}-\frac {2}{125 (-5+x)^2}+\frac {3}{625 (-5+x)}-\frac {1}{125 x^2}-\frac {3}{625 x}\right ) \, dx,x,x^2\right )-\int \frac {x \log (x)}{\left (-5+x^2\right )^3} \, dx\\ &=\frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {3 \tanh ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{40 \sqrt {5}}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{100} \log \left (5-x^2\right )+\frac {1}{100} \int \frac {x}{-5+x^2} \, dx+\frac {3}{40} \int \frac {1}{-5+x^2} \, dx-\frac {1}{4} \int \frac {1}{x \left (-5+x^2\right )^2} \, dx\\ &=\frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{200} \log \left (5-x^2\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{(-5+x)^2 x} \, dx,x,x^2\right )\\ &=\frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{200} \log \left (5-x^2\right )-\frac {1}{8} \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx,x,x^2\right )\\ &=\frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{100}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 22, normalized size = 1.00 \begin {gather*} \frac {4 x+5 \log (x)}{4 x^2 \left (-5+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 20*x + 5*x^2 - 20*x^3 + (50 - 30*x^2)*Log[x])/(-500*x^3 + 300*x^5 - 60*x^7 + 4*x^9),x]

[Out]

(4*x + 5*Log[x])/(4*x^2*(-5 + x^2)^2)

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 26, normalized size = 1.18 \begin {gather*} \frac {4 \, x + 5 \, \log \relax (x)}{4 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x^2+50)*log(x)-20*x^3+5*x^2+20*x-25)/(4*x^9-60*x^7+300*x^5-500*x^3),x, algorithm="fricas")

[Out]

1/4*(4*x + 5*log(x))/(x^6 - 10*x^4 + 25*x^2)

________________________________________________________________________________________

giac [B]  time = 0.19, size = 55, normalized size = 2.50 \begin {gather*} -\frac {1}{20} \, {\left (\frac {x^{2} - 10}{x^{4} - 10 \, x^{2} + 25} - \frac {1}{x^{2}}\right )} \log \relax (x) - \frac {x^{3} - 10 \, x}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {1}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x^2+50)*log(x)-20*x^3+5*x^2+20*x-25)/(4*x^9-60*x^7+300*x^5-500*x^3),x, algorithm="giac")

[Out]

-1/20*((x^2 - 10)/(x^4 - 10*x^2 + 25) - 1/x^2)*log(x) - 1/25*(x^3 - 10*x)/(x^4 - 10*x^2 + 25) + 1/25/x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 18, normalized size = 0.82

method	result	size
norman	\(\frac {x +\frac {5 \ln \relax (x )}{4}}{\left (x^{2}-5\right )^{2} x^{2}}\)	\(18\)
risch	\(\frac {5 \ln \relax (x )}{4 x^{2} \left (x^{4}-10 x^{2}+25\right )}+\frac {1}{x \left (x^{4}-10 x^{2}+25\right )}\)	\(37\)
default	\(\frac {1}{25 x}+\frac {\ln \relax (x )}{50}-\frac {4 x^{3}+\frac {5}{2} x^{2}-40 x -\frac {25}{2}}{100 \left (x^{2}-5\right )^{2}}+\frac {1}{40 x^{2}-200}-\frac {\ln \relax (x ) x^{2} \left (x^{2}-10\right )}{100 \left (x^{2}-5\right )^{2}}+\frac {\ln \relax (x )}{20 x^{2}}-\frac {\ln \relax (x ) x^{2}}{100 \left (x^{2}-5\right )}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-30*x^2+50)*ln(x)-20*x^3+5*x^2+20*x-25)/(4*x^9-60*x^7+300*x^5-500*x^3),x,method=_RETURNVERBOSE)

[Out]

(x+5/4*ln(x))/(x^2-5)^2/x^2

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 158, normalized size = 7.18 \begin {gather*} \frac {10 \, x^{4} - 75 \, x^{2} - 2 \, {\left (2 \, x^{6} - 20 \, x^{4} + 50 \, x^{2} - 125\right )} \log \relax (x) + 125}{200 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} - \frac {6 \, x^{4} - 45 \, x^{2} + 50}{80 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} + \frac {3 \, x^{4} - 25 \, x^{2} + 40}{40 \, {\left (x^{5} - 10 \, x^{3} + 25 \, x\right )}} - \frac {3 \, x^{3} - 25 \, x}{40 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {2 \, x^{2} - 15}{80 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {1}{50} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x^2+50)*log(x)-20*x^3+5*x^2+20*x-25)/(4*x^9-60*x^7+300*x^5-500*x^3),x, algorithm="maxima")

[Out]

1/200*(10*x^4 - 75*x^2 - 2*(2*x^6 - 20*x^4 + 50*x^2 - 125)*log(x) + 125)/(x^6 - 10*x^4 + 25*x^2) - 1/80*(6*x^4
- 45*x^2 + 50)/(x^6 - 10*x^4 + 25*x^2) + 1/40*(3*x^4 - 25*x^2 + 40)/(x^5 - 10*x^3 + 25*x) - 1/40*(3*x^3 - 25*
x)/(x^4 - 10*x^2 + 25) + 1/80*(2*x^2 - 15)/(x^4 - 10*x^2 + 25) + 1/50*log(x)

________________________________________________________________________________________

mupad [B]  time = 4.62, size = 20, normalized size = 0.91 \begin {gather*} \frac {4\,x+5\,\ln \relax (x)}{4\,x^2\,{\left (x^2-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x^3 - 5*x^2 - 20*x + log(x)*(30*x^2 - 50) + 25)/(500*x^3 - 300*x^5 + 60*x^7 - 4*x^9),x)

[Out]

(4*x + 5*log(x))/(4*x^2*(x^2 - 5)^2)

________________________________________________________________________________________

sympy [A]  time = 0.23, size = 32, normalized size = 1.45 \begin {gather*} \frac {5 \log {\relax (x )}}{4 x^{6} - 40 x^{4} + 100 x^{2}} + \frac {1}{x^{5} - 10 x^{3} + 25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x**2+50)*ln(x)-20*x**3+5*x**2+20*x-25)/(4*x**9-60*x**7+300*x**5-500*x**3),x)

[Out]

5*log(x)/(4*x**6 - 40*x**4 + 100*x**2) + 1/(x**5 - 10*x**3 + 25*x)

________________________________________________________________________________________