3.22.4 \(\int \frac {720 x+144 x^2-288 x^3+(360-648 x-432 x^2) \log (x)+(-360-144 x) \log ^2(x)}{125 x^3+75 x^4+15 x^5+x^6} \, dx\)

Optimal. Leaf size=20 \[ \frac {36 (-2 x-\log (x))^2}{x^2 (5+x)^2} \]

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Rubi [B]  time = 0.82, antiderivative size = 103, normalized size of antiderivative = 5.15, number of steps used = 36, number of rules used = 15, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.254, Rules used = {6741, 6742, 44, 2357, 2304, 2301, 2319, 2314, 31, 2317, 2391, 2305, 2347, 2344, 2318} \begin {gather*} \frac {36 \log ^2(x)}{25 x^2}+\frac {144}{(x+5)^2}-\frac {72 \log ^2(x)}{125 x}-\frac {72 x \log ^2(x)}{625 (x+5)}+\frac {36 \log ^2(x)}{25 (x+5)^2}+\frac {72 \log ^2(x)}{625}+\frac {144 \log (x)}{25 x}+\frac {144 x \log (x)}{125 (x+5)}-\frac {144 \log (x)}{5 (x+5)^2}-\frac {144 \log (x)}{125} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(720*x + 144*x^2 - 288*x^3 + (360 - 648*x - 432*x^2)*Log[x] + (-360 - 144*x)*Log[x]^2)/(125*x^3 + 75*x^4 +
15*x^5 + x^6),x]

[Out]

144/(5 + x)^2 - (144*Log[x])/125 + (144*Log[x])/(25*x) - (144*Log[x])/(5*(5 + x)^2) + (144*x*Log[x])/(125*(5 +
x)) + (72*Log[x]^2)/625 + (36*Log[x]^2)/(25*x^2) - (72*Log[x]^2)/(125*x) + (36*Log[x]^2)/(25*(5 + x)^2) - (72
*x*Log[x]^2)/(625*(5 + x))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, 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 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 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 2317

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

Rule 2318

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])
^p)/(d*(d + e*x)), x] - Dist[(b*n*p)/d, Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 {720 x+144 x^2-288 x^3+\left (360-648 x-432 x^2\right ) \log (x)+(-360-144 x) \log ^2(x)}{x^3 (5+x)^3} \, dx\\ &=\int \left (-\frac {288}{(5+x)^3}+\frac {720}{x^2 (5+x)^3}+\frac {144}{x (5+x)^3}-\frac {72 \left (-5+9 x+6 x^2\right ) \log (x)}{x^3 (5+x)^3}-\frac {72 (5+2 x) \log ^2(x)}{x^3 (5+x)^3}\right ) \, dx\\ &=\frac {144}{(5+x)^2}-72 \int \frac {\left (-5+9 x+6 x^2\right ) \log (x)}{x^3 (5+x)^3} \, dx-72 \int \frac {(5+2 x) \log ^2(x)}{x^3 (5+x)^3} \, dx+144 \int \frac {1}{x (5+x)^3} \, dx+720 \int \frac {1}{x^2 (5+x)^3} \, dx\\ &=\frac {144}{(5+x)^2}-72 \int \left (-\frac {\log (x)}{25 x^3}+\frac {12 \log (x)}{125 x^2}-\frac {3 \log (x)}{625 x}-\frac {4 \log (x)}{5 (5+x)^3}-\frac {9 \log (x)}{125 (5+x)^2}+\frac {3 \log (x)}{625 (5+x)}\right ) \, dx-72 \int \left (\frac {\log ^2(x)}{25 x^3}-\frac {\log ^2(x)}{125 x^2}+\frac {\log ^2(x)}{25 (5+x)^3}+\frac {\log ^2(x)}{125 (5+x)^2}\right ) \, dx+144 \int \left (\frac {1}{125 x}-\frac {1}{5 (5+x)^3}-\frac {1}{25 (5+x)^2}-\frac {1}{125 (5+x)}\right ) \, dx+720 \int \left (\frac {1}{125 x^2}-\frac {3}{625 x}+\frac {1}{25 (5+x)^3}+\frac {2}{125 (5+x)^2}+\frac {3}{625 (5+x)}\right ) \, dx\\ &=-\frac {144}{25 x}+\frac {144}{(5+x)^2}-\frac {144}{25 (5+x)}-\frac {288 \log (x)}{125}+\frac {288}{125} \log (5+x)+\frac {216}{625} \int \frac {\log (x)}{x} \, dx-\frac {216}{625} \int \frac {\log (x)}{5+x} \, dx+\frac {72}{125} \int \frac {\log ^2(x)}{x^2} \, dx-\frac {72}{125} \int \frac {\log ^2(x)}{(5+x)^2} \, dx+\frac {72}{25} \int \frac {\log (x)}{x^3} \, dx-\frac {72}{25} \int \frac {\log ^2(x)}{x^3} \, dx-\frac {72}{25} \int \frac {\log ^2(x)}{(5+x)^3} \, dx+\frac {648}{125} \int \frac {\log (x)}{(5+x)^2} \, dx-\frac {864}{125} \int \frac {\log (x)}{x^2} \, dx+\frac {288}{5} \int \frac {\log (x)}{(5+x)^3} \, dx\\ &=-\frac {18}{25 x^2}+\frac {144}{125 x}+\frac {144}{(5+x)^2}-\frac {144}{25 (5+x)}-\frac {288 \log (x)}{125}-\frac {36 \log (x)}{25 x^2}+\frac {864 \log (x)}{125 x}-\frac {144 \log (x)}{5 (5+x)^2}+\frac {648 x \log (x)}{625 (5+x)}-\frac {216}{625} \log \left (1+\frac {x}{5}\right ) \log (x)+\frac {108 \log ^2(x)}{625}+\frac {36 \log ^2(x)}{25 x^2}-\frac {72 \log ^2(x)}{125 x}+\frac {36 \log ^2(x)}{25 (5+x)^2}-\frac {72 x \log ^2(x)}{625 (5+x)}+\frac {288}{125} \log (5+x)+\frac {144}{625} \int \frac {\log (x)}{5+x} \, dx+\frac {216}{625} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx-\frac {648}{625} \int \frac {1}{5+x} \, dx+\frac {144}{125} \int \frac {\log (x)}{x^2} \, dx-\frac {72}{25} \int \frac {\log (x)}{x^3} \, dx-\frac {72}{25} \int \frac {\log (x)}{x (5+x)^2} \, dx+\frac {144}{5} \int \frac {1}{x (5+x)^2} \, dx\\ &=\frac {144}{(5+x)^2}-\frac {144}{25 (5+x)}-\frac {288 \log (x)}{125}+\frac {144 \log (x)}{25 x}-\frac {144 \log (x)}{5 (5+x)^2}+\frac {648 x \log (x)}{625 (5+x)}-\frac {72}{625} \log \left (1+\frac {x}{5}\right ) \log (x)+\frac {108 \log ^2(x)}{625}+\frac {36 \log ^2(x)}{25 x^2}-\frac {72 \log ^2(x)}{125 x}+\frac {36 \log ^2(x)}{25 (5+x)^2}-\frac {72 x \log ^2(x)}{625 (5+x)}+\frac {792}{625} \log (5+x)-\frac {216 \text {Li}_2\left (-\frac {x}{5}\right )}{625}-\frac {144}{625} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx+\frac {72}{125} \int \frac {\log (x)}{(5+x)^2} \, dx-\frac {72}{125} \int \frac {\log (x)}{x (5+x)} \, dx+\frac {144}{5} \int \left (\frac {1}{25 x}-\frac {1}{5 (5+x)^2}-\frac {1}{25 (5+x)}\right ) \, dx\\ &=\frac {144}{(5+x)^2}-\frac {144 \log (x)}{125}+\frac {144 \log (x)}{25 x}-\frac {144 \log (x)}{5 (5+x)^2}+\frac {144 x \log (x)}{125 (5+x)}-\frac {72}{625} \log \left (1+\frac {x}{5}\right ) \log (x)+\frac {108 \log ^2(x)}{625}+\frac {36 \log ^2(x)}{25 x^2}-\frac {72 \log ^2(x)}{125 x}+\frac {36 \log ^2(x)}{25 (5+x)^2}-\frac {72 x \log ^2(x)}{625 (5+x)}+\frac {72}{625} \log (5+x)-\frac {72 \text {Li}_2\left (-\frac {x}{5}\right )}{625}-\frac {72}{625} \int \frac {1}{5+x} \, dx-\frac {72}{625} \int \frac {\log (x)}{x} \, dx+\frac {72}{625} \int \frac {\log (x)}{5+x} \, dx\\ &=\frac {144}{(5+x)^2}-\frac {144 \log (x)}{125}+\frac {144 \log (x)}{25 x}-\frac {144 \log (x)}{5 (5+x)^2}+\frac {144 x \log (x)}{125 (5+x)}+\frac {72 \log ^2(x)}{625}+\frac {36 \log ^2(x)}{25 x^2}-\frac {72 \log ^2(x)}{125 x}+\frac {36 \log ^2(x)}{25 (5+x)^2}-\frac {72 x \log ^2(x)}{625 (5+x)}-\frac {72 \text {Li}_2\left (-\frac {x}{5}\right )}{625}-\frac {72}{625} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx\\ &=\frac {144}{(5+x)^2}-\frac {144 \log (x)}{125}+\frac {144 \log (x)}{25 x}-\frac {144 \log (x)}{5 (5+x)^2}+\frac {144 x \log (x)}{125 (5+x)}+\frac {72 \log ^2(x)}{625}+\frac {36 \log ^2(x)}{25 x^2}-\frac {72 \log ^2(x)}{125 x}+\frac {36 \log ^2(x)}{25 (5+x)^2}-\frac {72 x \log ^2(x)}{625 (5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 43, normalized size = 2.15 \begin {gather*} -\frac {36 \left (4 x^2 \left (-125+x^2 \log (5)+x \log (9765625)+\log (298023223876953125)\right )-500 x \log (x)-125 \log ^2(x)\right )}{125 x^2 (5+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(720*x + 144*x^2 - 288*x^3 + (360 - 648*x - 432*x^2)*Log[x] + (-360 - 144*x)*Log[x]^2)/(125*x^3 + 75
*x^4 + 15*x^5 + x^6),x]

[Out]

(-36*(4*x^2*(-125 + x^2*Log[5] + x*Log[9765625] + Log[298023223876953125]) - 500*x*Log[x] - 125*Log[x]^2))/(12
5*x^2*(5 + x)^2)

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fricas [A]  time = 0.95, size = 33, normalized size = 1.65 \begin {gather*} \frac {36 \, {\left (4 \, x^{2} + 4 \, x \log \relax (x) + \log \relax (x)^{2}\right )}}{x^{4} + 10 \, x^{3} + 25 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-360)*log(x)^2+(-432*x^2-648*x+360)*log(x)-288*x^3+144*x^2+720*x)/(x^6+15*x^5+75*x^4+125*x^3
),x, algorithm="fricas")

[Out]

36*(4*x^2 + 4*x*log(x) + log(x)^2)/(x^4 + 10*x^3 + 25*x^2)

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giac [B]  time = 0.20, size = 70, normalized size = 3.50 \begin {gather*} \frac {36}{125} \, {\left (\frac {2 \, x + 15}{x^{2} + 10 \, x + 25} - \frac {2 \, x - 5}{x^{2}}\right )} \log \relax (x)^{2} - \frac {144}{25} \, {\left (\frac {x + 10}{x^{2} + 10 \, x + 25} - \frac {1}{x}\right )} \log \relax (x) + \frac {144}{x^{2} + 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-360)*log(x)^2+(-432*x^2-648*x+360)*log(x)-288*x^3+144*x^2+720*x)/(x^6+15*x^5+75*x^4+125*x^3
),x, algorithm="giac")

[Out]

36/125*((2*x + 15)/(x^2 + 10*x + 25) - (2*x - 5)/x^2)*log(x)^2 - 144/25*((x + 10)/(x^2 + 10*x + 25) - 1/x)*log
(x) + 144/(x^2 + 10*x + 25)

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maple [A]  time = 0.08, size = 27, normalized size = 1.35

method	result	size
norman	\(\frac {144 x^{2}+36 \ln \relax (x )^{2}+144 x \ln \relax (x )}{\left (5+x \right )^{2} x^{2}}\)	\(27\)
risch	\(\frac {36 \ln \relax (x )^{2}}{x^{2} \left (x^{2}+10 x +25\right )}+\frac {144 \ln \relax (x )}{x \left (x^{2}+10 x +25\right )}+\frac {144}{x^{2}+10 x +25}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-144*x-360)*ln(x)^2+(-432*x^2-648*x+360)*ln(x)-288*x^3+144*x^2+720*x)/(x^6+15*x^5+75*x^4+125*x^3),x,meth
od=_RETURNVERBOSE)

[Out]

(144*x^2+36*ln(x)^2+144*x*ln(x))/(5+x)^2/x^2

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maxima [B]  time = 0.55, size = 99, normalized size = 4.95 \begin {gather*} \frac {36 \, {\left (8 \, x^{3} + 60 \, x^{2} + 100 \, x \log \relax (x) + 25 \, \log \relax (x)^{2} + 100 \, x\right )}}{25 \, {\left (x^{4} + 10 \, x^{3} + 25 \, x^{2}\right )}} - \frac {72 \, {\left (6 \, x^{2} + 45 \, x + 50\right )}}{25 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )}} + \frac {72 \, {\left (2 \, x + 15\right )}}{25 \, {\left (x^{2} + 10 \, x + 25\right )}} + \frac {144}{x^{2} + 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-360)*log(x)^2+(-432*x^2-648*x+360)*log(x)-288*x^3+144*x^2+720*x)/(x^6+15*x^5+75*x^4+125*x^3
),x, algorithm="maxima")

[Out]

36/25*(8*x^3 + 60*x^2 + 100*x*log(x) + 25*log(x)^2 + 100*x)/(x^4 + 10*x^3 + 25*x^2) - 72/25*(6*x^2 + 45*x + 50
)/(x^3 + 10*x^2 + 25*x) + 72/25*(2*x + 15)/(x^2 + 10*x + 25) + 144/(x^2 + 10*x + 25)

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mupad [B]  time = 1.27, size = 26, normalized size = 1.30 \begin {gather*} \frac {144\,x^2+144\,x\,\ln \relax (x)+36\,{\ln \relax (x)}^2}{x^2\,{\left (x+5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(648*x + 432*x^2 - 360) - 720*x - 144*x^2 + 288*x^3 + log(x)^2*(144*x + 360))/(125*x^3 + 75*x^4 +
15*x^5 + x^6),x)

[Out]

(36*log(x)^2 + 144*x*log(x) + 144*x^2)/(x^2*(x + 5)^2)

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sympy [B]  time = 0.24, size = 48, normalized size = 2.40 \begin {gather*} \frac {36 \log {\relax (x )}^{2}}{x^{4} + 10 x^{3} + 25 x^{2}} + \frac {144 \log {\relax (x )}}{x^{3} + 10 x^{2} + 25 x} + \frac {288}{2 x^{2} + 20 x + 50} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-360)*ln(x)**2+(-432*x**2-648*x+360)*ln(x)-288*x**3+144*x**2+720*x)/(x**6+15*x**5+75*x**4+12
5*x**3),x)

[Out]

36*log(x)**2/(x**4 + 10*x**3 + 25*x**2) + 144*log(x)/(x**3 + 10*x**2 + 25*x) + 288/(2*x**2 + 20*x + 50)

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