3.30.51 \(\int \frac {75-150 x+690 x^2+294 x^3+30 x^4+(150 x+60 x^2+6 x^3) \log (x)+(150 x+60 x^2+6 x^3) \log ^2(x)}{125+50 x+5 x^2} \, dx\)

Optimal. Leaf size=34 \[ x \left (x^2+x \left (x+\frac {3 \left (-x+\frac {5}{5+x}+x \log ^2(x)\right )}{5 x}\right )\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 31, normalized size of antiderivative = 0.91, number of steps used = 15, number of rules used = 6, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {27, 12, 6742, 43, 2304, 2305} \begin {gather*} 2 x^3-\frac {3 x^2}{5}+\frac {3}{5} x^2 \log ^2(x)-\frac {15}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 - 150*x + 690*x^2 + 294*x^3 + 30*x^4 + (150*x + 60*x^2 + 6*x^3)*Log[x] + (150*x + 60*x^2 + 6*x^3)*Log[
x]^2)/(125 + 50*x + 5*x^2),x]

[Out]

(-3*x^2)/5 + 2*x^3 - 15/(5 + x) + (3*x^2*Log[x]^2)/5

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 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 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[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 {75-150 x+690 x^2+294 x^3+30 x^4+\left (150 x+60 x^2+6 x^3\right ) \log (x)+\left (150 x+60 x^2+6 x^3\right ) \log ^2(x)}{5 (5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {75-150 x+690 x^2+294 x^3+30 x^4+\left (150 x+60 x^2+6 x^3\right ) \log (x)+\left (150 x+60 x^2+6 x^3\right ) \log ^2(x)}{(5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {75}{(5+x)^2}-\frac {150 x}{(5+x)^2}+\frac {690 x^2}{(5+x)^2}+\frac {294 x^3}{(5+x)^2}+\frac {30 x^4}{(5+x)^2}+6 x \log (x)+6 x \log ^2(x)\right ) \, dx\\ &=-\frac {15}{5+x}+\frac {6}{5} \int x \log (x) \, dx+\frac {6}{5} \int x \log ^2(x) \, dx+6 \int \frac {x^4}{(5+x)^2} \, dx-30 \int \frac {x}{(5+x)^2} \, dx+\frac {294}{5} \int \frac {x^3}{(5+x)^2} \, dx+138 \int \frac {x^2}{(5+x)^2} \, dx\\ &=-\frac {3 x^2}{10}-\frac {15}{5+x}+\frac {3}{5} x^2 \log (x)+\frac {3}{5} x^2 \log ^2(x)-\frac {6}{5} \int x \log (x) \, dx+6 \int \left (75-10 x+x^2+\frac {625}{(5+x)^2}-\frac {500}{5+x}\right ) \, dx-30 \int \left (-\frac {5}{(5+x)^2}+\frac {1}{5+x}\right ) \, dx+\frac {294}{5} \int \left (-10+x-\frac {125}{(5+x)^2}+\frac {75}{5+x}\right ) \, dx+138 \int \left (1+\frac {25}{(5+x)^2}-\frac {10}{5+x}\right ) \, dx\\ &=-\frac {3 x^2}{5}+2 x^3-\frac {15}{5+x}+\frac {3}{5} x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 0.94 \begin {gather*} \frac {3}{5} \left (-x^2+\frac {10 x^3}{3}-\frac {25}{5+x}+x^2 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 - 150*x + 690*x^2 + 294*x^3 + 30*x^4 + (150*x + 60*x^2 + 6*x^3)*Log[x] + (150*x + 60*x^2 + 6*x^3
)*Log[x]^2)/(125 + 50*x + 5*x^2),x]

[Out]

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

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fricas [A]  time = 0.88, size = 39, normalized size = 1.15 \begin {gather*} \frac {10 \, x^{4} + 47 \, x^{3} + 3 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (x)^{2} - 15 \, x^{2} - 75}{5 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3+60*x^2+150*x)*log(x)^2+(6*x^3+60*x^2+150*x)*log(x)+30*x^4+294*x^3+690*x^2-150*x+75)/(5*x^2+5
0*x+125),x, algorithm="fricas")

[Out]

1/5*(10*x^4 + 47*x^3 + 3*(x^3 + 5*x^2)*log(x)^2 - 15*x^2 - 75)/(x + 5)

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giac [A]  time = 0.20, size = 27, normalized size = 0.79 \begin {gather*} \frac {3}{5} \, x^{2} \log \relax (x)^{2} + 2 \, x^{3} - \frac {3}{5} \, x^{2} - \frac {15}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3+60*x^2+150*x)*log(x)^2+(6*x^3+60*x^2+150*x)*log(x)+30*x^4+294*x^3+690*x^2-150*x+75)/(5*x^2+5
0*x+125),x, algorithm="giac")

[Out]

3/5*x^2*log(x)^2 + 2*x^3 - 3/5*x^2 - 15/(x + 5)

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maple [A]  time = 0.53, size = 28, normalized size = 0.82

method	result	size
default	\(\frac {3 x^{2} \ln \relax (x )^{2}}{5}+2 x^{3}-\frac {3 x^{2}}{5}-\frac {15}{5+x}\)	\(28\)
risch	\(\frac {3 x^{2} \ln \relax (x )^{2}}{5}+\frac {10 x^{4}+47 x^{3}-15 x^{2}-75}{25+5 x}\)	\(35\)
norman	\(\frac {-3 x^{2}+\frac {47 x^{3}}{5}+2 x^{4}+3 x^{2} \ln \relax (x )^{2}+\frac {3 x^{3} \ln \relax (x )^{2}}{5}-15}{5+x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^3+60*x^2+150*x)*ln(x)^2+(6*x^3+60*x^2+150*x)*ln(x)+30*x^4+294*x^3+690*x^2-150*x+75)/(5*x^2+50*x+125)
,x,method=_RETURNVERBOSE)

[Out]

3/5*x^2*ln(x)^2+2*x^3-3/5*x^2-15/(5+x)

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maxima [A]  time = 0.40, size = 27, normalized size = 0.79 \begin {gather*} \frac {3}{5} \, x^{2} \log \relax (x)^{2} + 2 \, x^{3} - \frac {3}{5} \, x^{2} - \frac {15}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3+60*x^2+150*x)*log(x)^2+(6*x^3+60*x^2+150*x)*log(x)+30*x^4+294*x^3+690*x^2-150*x+75)/(5*x^2+5
0*x+125),x, algorithm="maxima")

[Out]

3/5*x^2*log(x)^2 + 2*x^3 - 3/5*x^2 - 15/(x + 5)

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mupad [B]  time = 1.87, size = 32, normalized size = 0.94 \begin {gather*} x^2\,\left (\frac {3\,{\ln \relax (x)}^2}{5}-\frac {3}{5}\right )-\frac {75\,x}{5\,x^2+25\,x}+2\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(150*x + 60*x^2 + 6*x^3) - 150*x + 690*x^2 + 294*x^3 + 30*x^4 + log(x)*(150*x + 60*x^2 + 6*x^3)
+ 75)/(50*x + 5*x^2 + 125),x)

[Out]

x^2*((3*log(x)^2)/5 - 3/5) - (75*x)/(25*x + 5*x^2) + 2*x^3

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sympy [A]  time = 0.15, size = 27, normalized size = 0.79 \begin {gather*} 2 x^{3} + \frac {3 x^{2} \log {\relax (x )}^{2}}{5} - \frac {3 x^{2}}{5} - \frac {15}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**3+60*x**2+150*x)*ln(x)**2+(6*x**3+60*x**2+150*x)*ln(x)+30*x**4+294*x**3+690*x**2-150*x+75)/(5
*x**2+50*x+125),x)

[Out]

2*x**3 + 3*x**2*log(x)**2/5 - 3*x**2/5 - 15/(x + 5)

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