3.33.92 \(\int \frac {-92 e-56 e \log (\frac {x}{15})-8 e \log ^2(\frac {x}{15})}{16 x^3+8 x^3 \log (\frac {x}{15})+x^3 \log ^2(\frac {x}{15})} \, dx\) [3292]

Optimal. Leaf size=19 \[ \frac {e \left (4-\frac {4}{4+\log \left (\frac {x}{15}\right )}\right )}{x^2} \]

[Out]

1/x^2*(4-4/(4+ln(1/15*x)))*exp(1)

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Rubi [A]
time = 0.38, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 6, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6820, 12, 6874, 2343, 2346, 2209} \begin {gather*} \frac {4 e}{x^2}-\frac {4 e}{x^2 \left (\log \left (\frac {x}{15}\right )+4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-92*E - 56*E*Log[x/15] - 8*E*Log[x/15]^2)/(16*x^3 + 8*x^3*Log[x/15] + x^3*Log[x/15]^2),x]

[Out]

(4*E)/x^2 - (4*E)/(x^2*(4 + Log[x/15]))

Rule 12

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log
[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e \left (-23-14 \log \left (\frac {x}{15}\right )-2 \log ^2\left (\frac {x}{15}\right )\right )}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx\\ &=(4 e) \int \frac {-23-14 \log \left (\frac {x}{15}\right )-2 \log ^2\left (\frac {x}{15}\right )}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx\\ &=(4 e) \int \left (-\frac {2}{x^3}+\frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2}+\frac {2}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )}\right ) \, dx\\ &=\frac {4 e}{x^2}+(4 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )^2} \, dx+(8 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )} \, dx\\ &=\frac {4 e}{x^2}-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )}+\frac {1}{225} (8 e) \text {Subst}\left (\int \frac {e^{-2 x}}{4+x} \, dx,x,\log \left (\frac {x}{15}\right )\right )-(8 e) \int \frac {1}{x^3 \left (4+\log \left (\frac {x}{15}\right )\right )} \, dx\\ &=\frac {4 e}{x^2}+\frac {8}{225} e^9 \text {Ei}\left (-2 \left (4+\log \left (\frac {x}{15}\right )\right )\right )-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )}-\frac {1}{225} (8 e) \text {Subst}\left (\int \frac {e^{-2 x}}{4+x} \, dx,x,\log \left (\frac {x}{15}\right )\right )\\ &=\frac {4 e}{x^2}-\frac {4 e}{x^2 \left (4+\log \left (\frac {x}{15}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.05, size = 18, normalized size = 0.95 \begin {gather*} -\frac {4 e \left (-1+\frac {1}{4+\log \left (\frac {x}{15}\right )}\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-92*E - 56*E*Log[x/15] - 8*E*Log[x/15]^2)/(16*x^3 + 8*x^3*Log[x/15] + x^3*Log[x/15]^2),x]

[Out]

(-4*E*(-1 + (4 + Log[x/15])^(-1)))/x^2

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Maple [A]
time = 1.35, size = 22, normalized size = 1.16

method	result	size
derivativedivides	\(\frac {4 \,{\mathrm e} \left (\ln \left (\frac {x}{15}\right )+3\right )}{\left (4+\ln \left (\frac {x}{15}\right )\right ) x^{2}}\)	\(22\)
default	\(\frac {4 \,{\mathrm e} \left (\ln \left (\frac {x}{15}\right )+3\right )}{\left (4+\ln \left (\frac {x}{15}\right )\right ) x^{2}}\)	\(22\)
risch	\(\frac {4 \,{\mathrm e}}{x^{2}}-\frac {4 \,{\mathrm e}}{x^{2} \left (4+\ln \left (\frac {x}{15}\right )\right )}\)	\(24\)
norman	\(\frac {4 \,{\mathrm e} \ln \left (\frac {x}{15}\right )+12 \,{\mathrm e}}{x^{2} \left (4+\ln \left (\frac {x}{15}\right )\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(1)*ln(1/15*x)^2-56*exp(1)*ln(1/15*x)-92*exp(1))/(x^3*ln(1/15*x)^2+8*x^3*ln(1/15*x)+16*x^3),x,metho
d=_RETURNVERBOSE)

[Out]

4*exp(1)*(ln(1/15*x)+3)/(4+ln(1/15*x))/x^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).
time = 0.50, size = 38, normalized size = 2.00 \begin {gather*} \frac {4 \, {\left ({\left (\log \left (5\right ) + \log \left (3\right ) - 3\right )} e - e \log \left (x\right )\right )}}{x^{2} {\left (\log \left (5\right ) + \log \left (3\right ) - 4\right )} - x^{2} \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="maxima")

[Out]

4*((log(5) + log(3) - 3)*e - e*log(x))/(x^2*(log(5) + log(3) - 4) - x^2*log(x))

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Fricas [A]
time = 0.40, size = 30, normalized size = 1.58 \begin {gather*} \frac {4 \, {\left (e \log \left (\frac {1}{15} \, x\right ) + 3 \, e\right )}}{x^{2} \log \left (\frac {1}{15} \, x\right ) + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="fricas")

[Out]

4*(e*log(1/15*x) + 3*e)/(x^2*log(1/15*x) + 4*x^2)

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Sympy [A]
time = 0.05, size = 26, normalized size = 1.37 \begin {gather*} - \frac {4 e}{x^{2} \log {\left (\frac {x}{15} \right )} + 4 x^{2}} + \frac {4 e}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*ln(1/15*x)**2-56*exp(1)*ln(1/15*x)-92*exp(1))/(x**3*ln(1/15*x)**2+8*x**3*ln(1/15*x)+16*x*
*3),x)

[Out]

-4*E/(x**2*log(x/15) + 4*x**2) + 4*E/x**2

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Giac [A]
time = 0.40, size = 30, normalized size = 1.58 \begin {gather*} \frac {4 \, {\left (e \log \left (\frac {1}{15} \, x\right ) + 3 \, e\right )}}{x^{2} \log \left (\frac {1}{15} \, x\right ) + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="giac")

[Out]

4*(e*log(1/15*x) + 3*e)/(x^2*log(1/15*x) + 4*x^2)

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Mupad [B]
time = 2.18, size = 21, normalized size = 1.11 \begin {gather*} \frac {4\,\mathrm {e}\,\left (\ln \left (\frac {x}{15}\right )+3\right )}{x^2\,\left (\ln \left (\frac {x}{15}\right )+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(92*exp(1) + 56*log(x/15)*exp(1) + 8*log(x/15)^2*exp(1))/(8*x^3*log(x/15) + 16*x^3 + x^3*log(x/15)^2),x)

[Out]

(4*exp(1)*(log(x/15) + 3))/(x^2*(log(x/15) + 4))

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