3.69.16 $\int \frac{-54x+(162+54x)\log (\frac{25x}{4e^4})-162{\log }^2(\frac{25x}{4e^4})}{x^3}dx$

Optimal. Leaf size=29 $$9{(x-\frac{-x+x^2+3\log \left(\frac{25x}{4e^4}\right)}{x})}^2$$

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Rubi [B]  time = 0.11, antiderivative size = 66, normalized size of antiderivative = 2.28, number of steps used = 8, number of rules used = 7, integrand size = 36, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.194, Rules used = {14, 2305, 2304, 37, 2334, 12, 43} $$\begin{array}{c}\frac{9(x+3{)}^2(-\log (x)+4-\log \left(\frac{25}{4}\right))}{x^2}+\frac{81{(2(2-\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^2}-\frac{81(-\log (x)+4-\log \left(\frac{25}{4}\right))}{x^2}+9\log (x)\end{array}$$

Antiderivative was successfully verified.

[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 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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{54}{x^2}-\frac{162{(4(1-\frac{1}{2}\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^3}+\frac{54(3+x)(-4(1-\frac{1}{2}\log \left(\frac{5}{2}\right))+\log (x))}{x^3})dx\\ & =\frac{54}{x}+54\int \frac{(3+x)(-4(1-\frac{1}{2}\log \left(\frac{5}{2}\right))+\log (x))}{x^3}dx-162\int \frac{{(4(1-\frac{1}{2}\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^3}dx\\ & =\frac{54}{x}+\frac{81{(2(2-\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^2}+\frac{9(3+x{)}^2(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}-54\int -\frac{(3+x{)}^2}{6x^3}dx+162\int \frac{4(1-\frac{1}{2}\log \left(\frac{5}{2}\right))-\log (x)}{x^3}dx\\ & =\frac{81}{2x^2}+\frac{54}{x}+\frac{81{(2(2-\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^2}-\frac{81(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+\frac{9(3+x{)}^2(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+9\int \frac{(3+x{)}^2}{x^3}dx\\ & =\frac{81}{2x^2}+\frac{54}{x}+\frac{81{(2(2-\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^2}-\frac{81(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+\frac{9(3+x{)}^2(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+9\int (\frac{9}{x^3}+\frac{6}{x^2}+\frac{1}{x})dx\\ & =\frac{81{(2(2-\log \left(\frac{5}{2}\right))-\log (x))}^2}{x^2}-\frac{81(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+\frac{9(3+x{)}^2(4-\log \left(\frac{25}{4}\right)-\log (x))}{x^2}+9\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.10, size = 55, normalized size = 1.90 $$\begin{array}{c}\frac{27(96+16x-21\log \left(\frac{25}{4}\right)-4x\log \left(\frac{25}{4}\right)+3(-9+\log \left(\frac{625}{16}\right))\log \left(\frac{25x}{4}\right)+\log (x)(-21-4x+6\log \left(\frac{25x}{4}\right)))}{2x^2}\end{array}$$

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.57, size = 25, normalized size = 0.86 $$\begin{array}{c}-\frac{27(2x\log \left(\frac{25}{4}x{e}^{(-4)}\right)-3\log {\left(\frac{25}{4}x{e}^{(-4)}\right)}^2)}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.14, size = 32, normalized size = 1.10 $$\begin{array}{c}-\frac{54(x+12)\log \left(\frac{25}{4}x\right)}{x^2}+\frac{81\log {\left(\frac{25}{4}x\right)}^2}{x^2}+\frac{216(x+6)}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 26, normalized size = 0.90

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [B]  time = 0.35, size = 53, normalized size = 1.83 $$\begin{array}{c}-\frac{54\log \left(\frac{25}{4}x{e}^{(-4)}\right)}{x}+\frac{81(2\log {\left(\frac{25}{4}x{e}^{(-4)}\right)}^2+2\log \left(\frac{25}{4}x{e}^{(-4)}\right)+1)}{2x^2}-\frac{81\log \left(\frac{25}{4}x{e}^{(-4)}\right)}{x^2}-\frac{81}{2x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.22, size = 23, normalized size = 0.79 $$\begin{array}{c}-\frac{27\ln \left(\frac{25x{\mathrm{e}}^{-4}}{4}\right)(2x-3\ln \left(\frac{25x{\mathrm{e}}^{-4}}{4}\right))}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.15, size = 29, normalized size = 1.00 $$\begin{array}{c}-\frac{54\log \left(\frac{25x}{4e^4}\right)}{x}+\frac{81\log {\left(\frac{25x}{4e^4}\right)}^2}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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