\(\int \frac {-54 x+(162+54 x) \log (\frac {25 x}{4 e^4})-162 \log ^2(\frac {25 x}{4 e^4})}{x^3} \, dx\) [6816]

Optimal result

Integrand size = 36, antiderivative size = 29 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=9 \left (x-\frac {-x+x^2+3 \log \left (\frac {25 x}{4 e^4}\right )}{x}\right )^2 \]

[Out]

3*(x-(3*ln(25/4*x/exp(2)^2)+x^2-x)/x)*(3*x-3*(3*ln(25/4*x/exp(2)^2)+x^2-x)/x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(29)=58\).

Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {14, 2342, 2341, 37, 2372, 12, 45} \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=\frac {9 (x+3)^2 \left (-\log (x)+4-\log \left (\frac {25}{4}\right )\right )}{x^2}+\frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}-\frac {81 \left (-\log (x)+4-\log \left (\frac {25}{4}\right )\right )}{x^2}+9 \log (x) \]

[In]

Int[(-54*x + (162 + 54*x)*Log[(25*x)/(4*E^4)] - 162*Log[(25*x)/(4*E^4)]^2)/x^3,x]

[Out]

(81*(2*(2 - Log[5/2]) - Log[x])^2)/x^2 - (81*(4 - Log[25/4] - Log[x]))/x^2 + (9*(3 + x)^2*(4 - Log[25/4] - Log
[x]))/x^2 + 9*Log[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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

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 2341

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 2342

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 2372

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]}, Dist[a + b*Log[c*x^n], u, 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{align*} \text {integral}& = \int \left (-\frac {54}{x^2}-\frac {162 \left (4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^3}+\frac {54 (3+x) \left (-4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )+\log (x)\right )}{x^3}\right ) \, dx \\ & = \frac {54}{x}+54 \int \frac {(3+x) \left (-4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )+\log (x)\right )}{x^3} \, dx-162 \int \frac {\left (4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^3} \, dx \\ & = \frac {54}{x}+\frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}+\frac {9 (3+x)^2 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}-54 \int -\frac {(3+x)^2}{6 x^3} \, dx+162 \int \frac {4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )-\log (x)}{x^3} \, dx \\ & = \frac {81}{2 x^2}+\frac {54}{x}+\frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}-\frac {81 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+\frac {9 (3+x)^2 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+9 \int \frac {(3+x)^2}{x^3} \, dx \\ & = \frac {81}{2 x^2}+\frac {54}{x}+\frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}-\frac {81 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+\frac {9 (3+x)^2 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+9 \int \left (\frac {9}{x^3}+\frac {6}{x^2}+\frac {1}{x}\right ) \, dx \\ & = \frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}-\frac {81 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+\frac {9 (3+x)^2 \left (4-\log \left (\frac {25}{4}\right )-\log (x)\right )}{x^2}+9 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=\frac {27 \left (96+16 x-21 \log \left (\frac {25}{4}\right )-4 x \log \left (\frac {25}{4}\right )+3 \left (-9+\log \left (\frac {625}{16}\right )\right ) \log \left (\frac {25 x}{4}\right )+\log (x) \left (-21-4 x+6 \log \left (\frac {25 x}{4}\right )\right )\right )}{2 x^2} \]

[In]

Integrate[(-54*x + (162 + 54*x)*Log[(25*x)/(4*E^4)] - 162*Log[(25*x)/(4*E^4)]^2)/x^3,x]

[Out]

(27*(96 + 16*x - 21*Log[25/4] - 4*x*Log[25/4] + 3*(-9 + Log[625/16])*Log[(25*x)/4] + Log[x]*(-21 - 4*x + 6*Log
[(25*x)/4])))/(2*x^2)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{x^{2}}-\frac {54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}\)	\(26\)
norman	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}-54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right ) x}{x^{2}}\)	\(29\)
parallelrisch	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}-54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right ) x}{x^{2}}\)	\(29\)
parts	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{x^{2}}-\frac {54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}\)	\(30\)
derivativedivides	\(\frac {25 \,{\mathrm e}^{-8} \left (216 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{4} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{25 x}-\frac {4 \,{\mathrm e}^{4}}{25 x}\right )+\frac {864 \,{\mathrm e}^{8}}{25 x}+\frac {1296 \,{\mathrm e}^{8} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{25 x^{2}}\right )}{16}\)	\(69\)
default	\(\frac {25 \,{\mathrm e}^{-8} \left (216 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{4} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{25 x}-\frac {4 \,{\mathrm e}^{4}}{25 x}\right )+\frac {864 \,{\mathrm e}^{8}}{25 x}+\frac {1296 \,{\mathrm e}^{8} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{25 x^{2}}\right )}{16}\)	\(69\)

[In]

int((-162*ln(25/4*x/exp(2)^2)^2+(54*x+162)*ln(25/4*x/exp(2)^2)-54*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

81/x^2*ln(25/4*x*exp(-4))^2-54*ln(25/4*x*exp(-4))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {27 \, {\left (2 \, x \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right ) - 3 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )^{2}\right )}}{x^{2}} \]

[In]

integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54*x)/x^3,x, algorithm="fricas")

[Out]

-27*(2*x*log(25/4*x*e^(-4)) - 3*log(25/4*x*e^(-4))^2)/x^2

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=- \frac {54 \log {\left (\frac {25 x}{4 e^{4}} \right )}}{x} + \frac {81 \log {\left (\frac {25 x}{4 e^{4}} \right )}^{2}}{x^{2}} \]

[In]

integrate((-162*ln(25/4*x/exp(2)**2)**2+(54*x+162)*ln(25/4*x/exp(2)**2)-54*x)/x**3,x)

[Out]

-54*log(25*x*exp(-4)/4)/x + 81*log(25*x*exp(-4)/4)**2/x**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {54 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )}{x} + \frac {81 \, {\left (2 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )^{2} + 2 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right ) + 1\right )}}{2 \, x^{2}} - \frac {81 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )}{x^{2}} - \frac {81}{2 \, x^{2}} \]

[In]

integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54*x)/x^3,x, algorithm="maxima")

[Out]

-54*log(25/4*x*e^(-4))/x + 81/2*(2*log(25/4*x*e^(-4))^2 + 2*log(25/4*x*e^(-4)) + 1)/x^2 - 81*log(25/4*x*e^(-4)
)/x^2 - 81/2/x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {54 \, {\left (x + 12\right )} \log \left (\frac {25}{4} \, x\right )}{x^{2}} + \frac {81 \, \log \left (\frac {25}{4} \, x\right )^{2}}{x^{2}} + \frac {216 \, {\left (x + 6\right )}}{x^{2}} \]

[In]

integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54*x)/x^3,x, algorithm="giac")

[Out]

-54*(x + 12)*log(25/4*x)/x^2 + 81*log(25/4*x)^2/x^2 + 216*(x + 6)/x^2

Mupad [B] (verification not implemented)

Time = 12.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {27\,\ln \left (\frac {25\,x\,{\mathrm {e}}^{-4}}{4}\right )\,\left (2\,x-3\,\ln \left (\frac {25\,x\,{\mathrm {e}}^{-4}}{4}\right )\right )}{x^2} \]

[In]

int(-(54*x + 162*log((25*x*exp(-4))/4)^2 - log((25*x*exp(-4))/4)*(54*x + 162))/x^3,x)

[Out]

-(27*log((25*x*exp(-4))/4)*(2*x - 3*log((25*x*exp(-4))/4)))/x^2