3.75.50 \(\int \frac {1375-3150 x+550 x^2-660 x^3+55 x^4-6 x^5-6000 x \log (x)}{1875+750 x^2+75 x^4} \, dx\)

Optimal. Leaf size=31 \[ -\left (1-\frac {x}{5}\right )^2+\frac {x}{3}-\frac {8 x \log (x)}{\frac {5}{x}+x} \]

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Rubi [A]  time = 0.21, antiderivative size = 53, normalized size of antiderivative = 1.71, number of steps used = 19, number of rules used = 11, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {28, 6742, 199, 203, 261, 288, 266, 43, 321, 2335, 260} \begin {gather*} -\frac {x^2}{25}-\frac {11 x}{6 \left (x^2+5\right )}-\frac {8 x^2 \log (x)}{x^2+5}-\frac {11 x^3}{30 \left (x^2+5\right )}+\frac {11 x}{10} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1375 - 3150*x + 550*x^2 - 660*x^3 + 55*x^4 - 6*x^5 - 6000*x*Log[x])/(1875 + 750*x^2 + 75*x^4),x]

[Out]

(11*x)/10 - x^2/25 - (11*x)/(6*(5 + x^2)) - (11*x^3)/(30*(5 + x^2)) - (8*x^2*Log[x])/(5 + x^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && 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 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -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 288

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

Rule 321

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

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=75 \int \frac {1375-3150 x+550 x^2-660 x^3+55 x^4-6 x^5-6000 x \log (x)}{\left (375+75 x^2\right )^2} \, dx\\ &=75 \int \left (\frac {11}{45 \left (5+x^2\right )^2}-\frac {14 x}{25 \left (5+x^2\right )^2}+\frac {22 x^2}{225 \left (5+x^2\right )^2}-\frac {44 x^3}{375 \left (5+x^2\right )^2}+\frac {11 x^4}{1125 \left (5+x^2\right )^2}-\frac {2 x^5}{1875 \left (5+x^2\right )^2}-\frac {16 x \log (x)}{15 \left (5+x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {2}{25} \int \frac {x^5}{\left (5+x^2\right )^2} \, dx\right )+\frac {11}{15} \int \frac {x^4}{\left (5+x^2\right )^2} \, dx+\frac {22}{3} \int \frac {x^2}{\left (5+x^2\right )^2} \, dx-\frac {44}{5} \int \frac {x^3}{\left (5+x^2\right )^2} \, dx+\frac {55}{3} \int \frac {1}{\left (5+x^2\right )^2} \, dx-42 \int \frac {x}{\left (5+x^2\right )^2} \, dx-80 \int \frac {x \log (x)}{\left (5+x^2\right )^2} \, dx\\ &=\frac {21}{5+x^2}-\frac {11 x}{6 \left (5+x^2\right )}-\frac {11 x^3}{30 \left (5+x^2\right )}-\frac {8 x^2 \log (x)}{5+x^2}-\frac {1}{25} \operatorname {Subst}\left (\int \frac {x^2}{(5+x)^2} \, dx,x,x^2\right )+\frac {11}{10} \int \frac {x^2}{5+x^2} \, dx+\frac {11}{6} \int \frac {1}{5+x^2} \, dx+\frac {11}{3} \int \frac {1}{5+x^2} \, dx-\frac {22}{5} \operatorname {Subst}\left (\int \frac {x}{(5+x)^2} \, dx,x,x^2\right )+8 \int \frac {x}{5+x^2} \, dx\\ &=\frac {11 x}{10}+\frac {21}{5+x^2}-\frac {11 x}{6 \left (5+x^2\right )}-\frac {11 x^3}{30 \left (5+x^2\right )}+\frac {11 \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{2 \sqrt {5}}-\frac {8 x^2 \log (x)}{5+x^2}+4 \log \left (5+x^2\right )-\frac {1}{25} \operatorname {Subst}\left (\int \left (1+\frac {25}{(5+x)^2}-\frac {10}{5+x}\right ) \, dx,x,x^2\right )-\frac {22}{5} \operatorname {Subst}\left (\int \left (-\frac {5}{(5+x)^2}+\frac {1}{5+x}\right ) \, dx,x,x^2\right )-\frac {11}{2} \int \frac {1}{5+x^2} \, dx\\ &=\frac {11 x}{10}-\frac {x^2}{25}-\frac {11 x}{6 \left (5+x^2\right )}-\frac {11 x^3}{30 \left (5+x^2\right )}-\frac {8 x^2 \log (x)}{5+x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 22, normalized size = 0.71 \begin {gather*} \frac {1}{75} x \left (55-3 x-\frac {600 x \log (x)}{5+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1375 - 3150*x + 550*x^2 - 660*x^3 + 55*x^4 - 6*x^5 - 6000*x*Log[x])/(1875 + 750*x^2 + 75*x^4),x]

[Out]

(x*(55 - 3*x - (600*x*Log[x])/(5 + x^2)))/75

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fricas [A]  time = 0.66, size = 35, normalized size = 1.13 \begin {gather*} -\frac {3 \, x^{4} - 55 \, x^{3} + 600 \, x^{2} \log \relax (x) + 15 \, x^{2} - 275 \, x}{75 \, {\left (x^{2} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6000*x*log(x)-6*x^5+55*x^4-660*x^3+550*x^2-3150*x+1375)/(75*x^4+750*x^2+1875),x, algorithm="fricas
")

[Out]

-1/75*(3*x^4 - 55*x^3 + 600*x^2*log(x) + 15*x^2 - 275*x)/(x^2 + 5)

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giac [A]  time = 0.23, size = 24, normalized size = 0.77 \begin {gather*} -\frac {1}{25} \, x^{2} + \frac {11}{15} \, x + \frac {40 \, \log \relax (x)}{x^{2} + 5} - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6000*x*log(x)-6*x^5+55*x^4-660*x^3+550*x^2-3150*x+1375)/(75*x^4+750*x^2+1875),x, algorithm="giac")

[Out]

-1/25*x^2 + 11/15*x + 40*log(x)/(x^2 + 5) - 8*log(x)

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maple [A]  time = 0.04, size = 24, normalized size = 0.77

method	result	size
default	\(-\frac {x^{2}}{25}+\frac {11 x}{15}-\frac {8 \ln \relax (x ) x^{2}}{x^{2}+5}\)	\(24\)
risch	\(\frac {40 \ln \relax (x )}{x^{2}+5}-\frac {x^{2}}{25}+\frac {11 x}{15}-8 \ln \relax (x )\)	\(25\)
norman	\(\frac {\frac {11 x}{3}+\frac {11 x^{3}}{15}-\frac {x^{4}}{25}-8 x^{2} \ln \relax (x )+1}{x^{2}+5}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6000*x*ln(x)-6*x^5+55*x^4-660*x^3+550*x^2-3150*x+1375)/(75*x^4+750*x^2+1875),x,method=_RETURNVERBOSE)

[Out]

-1/25*x^2+11/15*x-8*ln(x)*x^2/(x^2+5)

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maxima [A]  time = 0.45, size = 26, normalized size = 0.84 \begin {gather*} -\frac {1}{25} \, x^{2} + \frac {11}{15} \, x + \frac {40 \, \log \relax (x)}{x^{2} + 5} - 4 \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6000*x*log(x)-6*x^5+55*x^4-660*x^3+550*x^2-3150*x+1375)/(75*x^4+750*x^2+1875),x, algorithm="maxima
")

[Out]

-1/25*x^2 + 11/15*x + 40*log(x)/(x^2 + 5) - 4*log(x^2)

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mupad [B]  time = 5.95, size = 24, normalized size = 0.77 \begin {gather*} \frac {11\,x}{15}-8\,\ln \relax (x)-\frac {x^2}{25}+\frac {40\,\ln \relax (x)}{x^2+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3150*x + 6000*x*log(x) - 550*x^2 + 660*x^3 - 55*x^4 + 6*x^5 - 1375)/(750*x^2 + 75*x^4 + 1875),x)

[Out]

(11*x)/15 - 8*log(x) - x^2/25 + (40*log(x))/(x^2 + 5)

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sympy [A]  time = 0.15, size = 24, normalized size = 0.77 \begin {gather*} - \frac {x^{2}}{25} + \frac {11 x}{15} - 8 \log {\relax (x )} + \frac {40 \log {\relax (x )}}{x^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6000*x*ln(x)-6*x**5+55*x**4-660*x**3+550*x**2-3150*x+1375)/(75*x**4+750*x**2+1875),x)

[Out]

-x**2/25 + 11*x/15 - 8*log(x) + 40*log(x)/(x**2 + 5)

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