3.86.86 \(\int \frac {-4-4 e^9+12 x+4 x \log (\frac {81 x}{16})}{16 x+8 x \log (\frac {81 x}{16})+x \log ^2(\frac {81 x}{16})} \, dx\)

Optimal. Leaf size=18 \[ \frac {4 \left (1+e^9+x\right )}{4+\log \left (\frac {81 x}{16}\right )} \]

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Rubi [A]  time = 0.32, antiderivative size = 31, normalized size of antiderivative = 1.72, number of steps used = 13, number of rules used = 9, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6688, 12, 6742, 2353, 2297, 2299, 2178, 2302, 30} \begin {gather*} \frac {4 x}{\log \left (\frac {81 x}{16}\right )+4}+\frac {4 \left (1+e^9\right )}{\log \left (\frac {81 x}{16}\right )+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 4*E^9 + 12*x + 4*x*Log[(81*x)/16])/(16*x + 8*x*Log[(81*x)/16] + x*Log[(81*x)/16]^2),x]

[Out]

(4*(1 + E^9))/(4 + Log[(81*x)/16]) + (4*x)/(4 + Log[(81*x)/16])

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

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

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

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

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6688

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

Rule 6742

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 \left (-1-e^9+3 x+x \log \left (\frac {81 x}{16}\right )\right )}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2} \, dx\\ &=4 \int \frac {-1-e^9+3 x+x \log \left (\frac {81 x}{16}\right )}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2} \, dx\\ &=4 \int \left (\frac {-1-e^9-x}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2}+\frac {1}{4+\log \left (\frac {81 x}{16}\right )}\right ) \, dx\\ &=4 \int \frac {-1-e^9-x}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2} \, dx+4 \int \frac {1}{4+\log \left (\frac {81 x}{16}\right )} \, dx\\ &=\frac {64}{81} \operatorname {Subst}\left (\int \frac {e^x}{4+x} \, dx,x,\log \left (\frac {81 x}{16}\right )\right )+4 \int \left (-\frac {1}{\left (4+\log \left (\frac {81 x}{16}\right )\right )^2}+\frac {-1-e^9}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2}\right ) \, dx\\ &=\frac {64 \text {Ei}\left (4+\log \left (\frac {81 x}{16}\right )\right )}{81 e^4}-4 \int \frac {1}{\left (4+\log \left (\frac {81 x}{16}\right )\right )^2} \, dx-\left (4 \left (1+e^9\right )\right ) \int \frac {1}{x \left (4+\log \left (\frac {81 x}{16}\right )\right )^2} \, dx\\ &=\frac {64 \text {Ei}\left (4+\log \left (\frac {81 x}{16}\right )\right )}{81 e^4}+\frac {4 x}{4+\log \left (\frac {81 x}{16}\right )}-4 \int \frac {1}{4+\log \left (\frac {81 x}{16}\right )} \, dx-\left (4 \left (1+e^9\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,4+\log \left (\frac {81 x}{16}\right )\right )\\ &=\frac {64 \text {Ei}\left (4+\log \left (\frac {81 x}{16}\right )\right )}{81 e^4}+\frac {4 \left (1+e^9\right )}{4+\log \left (\frac {81 x}{16}\right )}+\frac {4 x}{4+\log \left (\frac {81 x}{16}\right )}-\frac {64}{81} \operatorname {Subst}\left (\int \frac {e^x}{4+x} \, dx,x,\log \left (\frac {81 x}{16}\right )\right )\\ &=\frac {4 \left (1+e^9\right )}{4+\log \left (\frac {81 x}{16}\right )}+\frac {4 x}{4+\log \left (\frac {81 x}{16}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 18, normalized size = 1.00 \begin {gather*} \frac {4 \left (1+e^9+x\right )}{4+\log \left (\frac {81 x}{16}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 4*E^9 + 12*x + 4*x*Log[(81*x)/16])/(16*x + 8*x*Log[(81*x)/16] + x*Log[(81*x)/16]^2),x]

[Out]

(4*(1 + E^9 + x))/(4 + Log[(81*x)/16])

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fricas [A]  time = 1.10, size = 15, normalized size = 0.83 \begin {gather*} \frac {4 \, {\left (x + e^{9} + 1\right )}}{\log \left (\frac {81}{16} \, x\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(81/16*x)-4*exp(9)+12*x-4)/(x*log(81/16*x)^2+8*x*log(81/16*x)+16*x),x, algorithm="fricas")

[Out]

4*(x + e^9 + 1)/(log(81/16*x) + 4)

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giac [A]  time = 0.19, size = 15, normalized size = 0.83 \begin {gather*} \frac {4 \, {\left (x + e^{9} + 1\right )}}{\log \left (\frac {81}{16} \, x\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(81/16*x)-4*exp(9)+12*x-4)/(x*log(81/16*x)^2+8*x*log(81/16*x)+16*x),x, algorithm="giac")

[Out]

4*(x + e^9 + 1)/(log(81/16*x) + 4)

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maple [A]  time = 0.05, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {4 x +4+4 \,{\mathrm e}^{9}}{\ln \left (\frac {81 x}{16}\right )+4}\)	\(16\)
norman	\(\frac {4 x +4+4 \,{\mathrm e}^{9}}{\ln \left (\frac {81 x}{16}\right )+4}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(81/16*x)-4*exp(9)+12*x-4)/(x*ln(81/16*x)^2+8*x*ln(81/16*x)+16*x),x,method=_RETURNVERBOSE)

[Out]

4*(x+1+exp(9))/(ln(81/16*x)+4)

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maxima [B]  time = 0.47, size = 52, normalized size = 2.89 \begin {gather*} \frac {4 \, x}{4 \, \log \relax (3) - 4 \, \log \relax (2) + \log \relax (x) + 4} + \frac {4 \, e^{9}}{4 \, \log \relax (3) - 4 \, \log \relax (2) + \log \relax (x) + 4} + \frac {4}{4 \, \log \relax (3) - 4 \, \log \relax (2) + \log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(81/16*x)-4*exp(9)+12*x-4)/(x*log(81/16*x)^2+8*x*log(81/16*x)+16*x),x, algorithm="maxima")

[Out]

4*x/(4*log(3) - 4*log(2) + log(x) + 4) + 4*e^9/(4*log(3) - 4*log(2) + log(x) + 4) + 4/(4*log(3) - 4*log(2) + l
og(x) + 4)

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mupad [B]  time = 5.93, size = 18, normalized size = 1.00 \begin {gather*} \frac {4\,x+4\,{\mathrm {e}}^9+4}{\ln \left (\frac {81\,x}{16}\right )+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x - 4*exp(9) + 4*x*log((81*x)/16) - 4)/(16*x + 8*x*log((81*x)/16) + x*log((81*x)/16)^2),x)

[Out]

(4*x + 4*exp(9) + 4)/(log((81*x)/16) + 4)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.94 \begin {gather*} \frac {4 x + 4 + 4 e^{9}}{\log {\left (\frac {81 x}{16} \right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(81/16*x)-4*exp(9)+12*x-4)/(x*ln(81/16*x)**2+8*x*ln(81/16*x)+16*x),x)

[Out]

(4*x + 4 + 4*exp(9))/(log(81*x/16) + 4)

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