3.88.9 \(\int \frac {-144+9 x^2+(240+60 x) \log (\frac {14 x}{3})-120 \log ^2(\frac {14 x}{3})+100 \log ^3(\frac {14 x}{3})-25 \log ^4(\frac {14 x}{3})}{9 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {\left (4+x+\frac {5}{3} \log ^2\left (\frac {14 x}{3}\right )\right )^2}{x} \]

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Rubi [B]  time = 0.19, antiderivative size = 78, normalized size of antiderivative = 3.71, number of steps used = 18, number of rules used = 7, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 43, 2334, 2301, 2305, 2304} \begin {gather*} x+\frac {16}{x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x}+\frac {40 \log ^2\left (\frac {14 x}{3}\right )}{3 x}-\frac {10 \log ^2(x)}{3}-\frac {20}{3} \left (\frac {4}{x}-\log (x)\right ) \log \left (\frac {14 x}{3}\right )+\frac {80 \log \left (\frac {14 x}{3}\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-144 + 9*x^2 + (240 + 60*x)*Log[(14*x)/3] - 120*Log[(14*x)/3]^2 + 100*Log[(14*x)/3]^3 - 25*Log[(14*x)/3]^
4)/(9*x^2),x]

[Out]

16/x + x - (10*Log[x]^2)/3 + (80*Log[(14*x)/3])/(3*x) - (20*(4/x - Log[x])*Log[(14*x)/3])/3 + (40*Log[(14*x)/3
]^2)/(3*x) + (25*Log[(14*x)/3]^4)/(9*x)

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {-144+9 x^2+(240+60 x) \log \left (\frac {14 x}{3}\right )-120 \log ^2\left (\frac {14 x}{3}\right )+100 \log ^3\left (\frac {14 x}{3}\right )-25 \log ^4\left (\frac {14 x}{3}\right )}{x^2} \, dx\\ &=\frac {1}{9} \int \left (\frac {9 \left (-16+x^2\right )}{x^2}+\frac {60 (4+x) \log \left (\frac {14 x}{3}\right )}{x^2}-\frac {120 \log ^2\left (\frac {14 x}{3}\right )}{x^2}+\frac {100 \log ^3\left (\frac {14 x}{3}\right )}{x^2}-\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {25}{9} \int \frac {\log ^4\left (\frac {14 x}{3}\right )}{x^2} \, dx\right )+\frac {20}{3} \int \frac {(4+x) \log \left (\frac {14 x}{3}\right )}{x^2} \, dx+\frac {100}{9} \int \frac {\log ^3\left (\frac {14 x}{3}\right )}{x^2} \, dx-\frac {40}{3} \int \frac {\log ^2\left (\frac {14 x}{3}\right )}{x^2} \, dx+\int \frac {-16+x^2}{x^2} \, dx\\ &=-\frac {20}{3} \left (\frac {4}{x}-\log (x)\right ) \log \left (\frac {14 x}{3}\right )+\frac {40 \log ^2\left (\frac {14 x}{3}\right )}{3 x}-\frac {100 \log ^3\left (\frac {14 x}{3}\right )}{9 x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x}-\frac {20}{3} \int \frac {-4+x \log (x)}{x^2} \, dx-\frac {100}{9} \int \frac {\log ^3\left (\frac {14 x}{3}\right )}{x^2} \, dx-\frac {80}{3} \int \frac {\log \left (\frac {14 x}{3}\right )}{x^2} \, dx+\frac {100}{3} \int \frac {\log ^2\left (\frac {14 x}{3}\right )}{x^2} \, dx+\int \left (1-\frac {16}{x^2}\right ) \, dx\\ &=\frac {128}{3 x}+x+\frac {80 \log \left (\frac {14 x}{3}\right )}{3 x}-\frac {20}{3} \left (\frac {4}{x}-\log (x)\right ) \log \left (\frac {14 x}{3}\right )-\frac {20 \log ^2\left (\frac {14 x}{3}\right )}{x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x}-\frac {20}{3} \int \left (-\frac {4}{x^2}+\frac {\log (x)}{x}\right ) \, dx-\frac {100}{3} \int \frac {\log ^2\left (\frac {14 x}{3}\right )}{x^2} \, dx+\frac {200}{3} \int \frac {\log \left (\frac {14 x}{3}\right )}{x^2} \, dx\\ &=-\frac {152}{3 x}+x-\frac {40 \log \left (\frac {14 x}{3}\right )}{x}-\frac {20}{3} \left (\frac {4}{x}-\log (x)\right ) \log \left (\frac {14 x}{3}\right )+\frac {40 \log ^2\left (\frac {14 x}{3}\right )}{3 x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x}-\frac {20}{3} \int \frac {\log (x)}{x} \, dx-\frac {200}{3} \int \frac {\log \left (\frac {14 x}{3}\right )}{x^2} \, dx\\ &=\frac {16}{x}+x-\frac {10 \log ^2(x)}{3}+\frac {80 \log \left (\frac {14 x}{3}\right )}{3 x}-\frac {20}{3} \left (\frac {4}{x}-\log (x)\right ) \log \left (\frac {14 x}{3}\right )+\frac {40 \log ^2\left (\frac {14 x}{3}\right )}{3 x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 49, normalized size = 2.33 \begin {gather*} \frac {16}{x}+x+\frac {10}{3} \log ^2\left (\frac {14 x}{3}\right )+\frac {40 \log ^2\left (\frac {14 x}{3}\right )}{3 x}+\frac {25 \log ^4\left (\frac {14 x}{3}\right )}{9 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-144 + 9*x^2 + (240 + 60*x)*Log[(14*x)/3] - 120*Log[(14*x)/3]^2 + 100*Log[(14*x)/3]^3 - 25*Log[(14*
x)/3]^4)/(9*x^2),x]

[Out]

16/x + x + (10*Log[(14*x)/3]^2)/3 + (40*Log[(14*x)/3]^2)/(3*x) + (25*Log[(14*x)/3]^4)/(9*x)

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fricas [A]  time = 0.65, size = 31, normalized size = 1.48 \begin {gather*} \frac {25 \, \log \left (\frac {14}{3} \, x\right )^{4} + 30 \, {\left (x + 4\right )} \log \left (\frac {14}{3} \, x\right )^{2} + 9 \, x^{2} + 144}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-25*log(14/3*x)^4+100*log(14/3*x)^3-120*log(14/3*x)^2+(60*x+240)*log(14/3*x)+9*x^2-144)/x^2,x,
algorithm="fricas")

[Out]

1/9*(25*log(14/3*x)^4 + 30*(x + 4)*log(14/3*x)^2 + 9*x^2 + 144)/x

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giac [A]  time = 0.15, size = 33, normalized size = 1.57 \begin {gather*} \frac {10}{3} \, {\left (\frac {4}{x} + 1\right )} \log \left (\frac {14}{3} \, x\right )^{2} + \frac {25 \, \log \left (\frac {14}{3} \, x\right )^{4}}{9 \, x} + x + \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-25*log(14/3*x)^4+100*log(14/3*x)^3-120*log(14/3*x)^2+(60*x+240)*log(14/3*x)+9*x^2-144)/x^2,x,
algorithm="giac")

[Out]

10/3*(4/x + 1)*log(14/3*x)^2 + 25/9*log(14/3*x)^4/x + x + 16/x

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maple [A]  time = 0.05, size = 35, normalized size = 1.67

method	result	size
norman	\(\frac {16+x^{2}+\frac {40 \ln \left (\frac {14 x}{3}\right )^{2}}{3}+\frac {25 \ln \left (\frac {14 x}{3}\right )^{4}}{9}+\frac {10 x \ln \left (\frac {14 x}{3}\right )^{2}}{3}}{x}\)	\(35\)
risch	\(\frac {25 \ln \left (\frac {14 x}{3}\right )^{4}}{9 x}+\frac {10 \left (4+x \right ) \ln \left (\frac {14 x}{3}\right )^{2}}{3 x}+\frac {x^{2}+16}{x}\)	\(36\)
derivativedivides	\(\frac {25 \ln \left (\frac {14 x}{3}\right )^{4}}{9 x}+\frac {40 \ln \left (\frac {14 x}{3}\right )^{2}}{3 x}+\frac {16}{x}+\frac {10 \ln \left (\frac {14 x}{3}\right )^{2}}{3}+x\)	\(38\)
default	\(\frac {25 \ln \left (\frac {14 x}{3}\right )^{4}}{9 x}+\frac {40 \ln \left (\frac {14 x}{3}\right )^{2}}{3 x}+\frac {16}{x}+\frac {10 \ln \left (\frac {14 x}{3}\right )^{2}}{3}+x\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-25*ln(14/3*x)^4+100*ln(14/3*x)^3-120*ln(14/3*x)^2+(60*x+240)*ln(14/3*x)+9*x^2-144)/x^2,x,method=_RET
URNVERBOSE)

[Out]

(16+x^2+40/3*ln(14/3*x)^2+25/9*ln(14/3*x)^4+10/3*x*ln(14/3*x)^2)/x

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maxima [B]  time = 0.35, size = 105, normalized size = 5.00 \begin {gather*} \frac {10}{3} \, \log \left (\frac {14}{3} \, x\right )^{2} + x + \frac {25 \, {\left (\log \left (\frac {14}{3} \, x\right )^{4} + 4 \, \log \left (\frac {14}{3} \, x\right )^{3} + 12 \, \log \left (\frac {14}{3} \, x\right )^{2} + 24 \, \log \left (\frac {14}{3} \, x\right ) + 24\right )}}{9 \, x} - \frac {100 \, {\left (\log \left (\frac {14}{3} \, x\right )^{3} + 3 \, \log \left (\frac {14}{3} \, x\right )^{2} + 6 \, \log \left (\frac {14}{3} \, x\right ) + 6\right )}}{9 \, x} + \frac {40 \, {\left (\log \left (\frac {14}{3} \, x\right )^{2} + 2 \, \log \left (\frac {14}{3} \, x\right ) + 2\right )}}{3 \, x} - \frac {80 \, \log \left (\frac {14}{3} \, x\right )}{3 \, x} - \frac {32}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-25*log(14/3*x)^4+100*log(14/3*x)^3-120*log(14/3*x)^2+(60*x+240)*log(14/3*x)+9*x^2-144)/x^2,x,
algorithm="maxima")

[Out]

10/3*log(14/3*x)^2 + x + 25/9*(log(14/3*x)^4 + 4*log(14/3*x)^3 + 12*log(14/3*x)^2 + 24*log(14/3*x) + 24)/x - 1
00/9*(log(14/3*x)^3 + 3*log(14/3*x)^2 + 6*log(14/3*x) + 6)/x + 40/3*(log(14/3*x)^2 + 2*log(14/3*x) + 2)/x - 80
/3*log(14/3*x)/x - 32/3/x

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mupad [B]  time = 5.43, size = 32, normalized size = 1.52 \begin {gather*} x+\frac {\frac {25\,{\ln \left (\frac {14\,x}{3}\right )}^4}{9}+\frac {40\,{\ln \left (\frac {14\,x}{3}\right )}^2}{3}+16}{x}+\frac {10\,{\ln \left (\frac {14\,x}{3}\right )}^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((40*log((14*x)/3)^2)/3 - (100*log((14*x)/3)^3)/9 + (25*log((14*x)/3)^4)/9 - x^2 - (log((14*x)/3)*(60*x +
240))/9 + 16)/x^2,x)

[Out]

x + ((40*log((14*x)/3)^2)/3 + (25*log((14*x)/3)^4)/9 + 16)/x + (10*log((14*x)/3)^2)/3

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sympy [A]  time = 0.19, size = 34, normalized size = 1.62 \begin {gather*} x + \frac {\left (10 x + 40\right ) \log {\left (\frac {14 x}{3} \right )}^{2}}{3 x} + \frac {25 \log {\left (\frac {14 x}{3} \right )}^{4}}{9 x} + \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-25*ln(14/3*x)**4+100*ln(14/3*x)**3-120*ln(14/3*x)**2+(60*x+240)*ln(14/3*x)+9*x**2-144)/x**2,x)

[Out]

x + (10*x + 40)*log(14*x/3)**2/(3*x) + 25*log(14*x/3)**4/(9*x) + 16/x

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