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

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

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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 {gather*} \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) \end {gather*}

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 55, normalized size = 1.90 \begin {gather*} \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} \end {gather*}

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 {gather*} -\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}} \end {gather*}

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 {gather*} -\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}} \end {gather*}

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	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{x^{2}}-\frac {54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}\)	\(26\)
norman	\(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}-54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right ) x}{x^{2}}\)	\(29\)
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	\(\frac {25 \,{\mathrm e}^{-8} \left (216 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{4} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{25 x}-\frac {4 \,{\mathrm e}^{4}}{25 x}\right )+\frac {864 \,{\mathrm e}^{8}}{25 x}+\frac {1296 \,{\mathrm e}^{8} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{25 x^{2}}\right )}{16}\)	\(69\)

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 {gather*} -\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}} \end {gather*}

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 {gather*} -\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} \end {gather*}

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 {gather*} - \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}} \end {gather*}

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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