3.6.96 \(\int \frac {-x+3 x^2+(-12+33 x+9 x^2) \log (4+x)+(-12 x-3 x^2) \log (4+x) \log (\frac {3}{5 x^3 \log (4+x)})}{(4 x+x^2) \log (4+x)} \, dx\)

Optimal. Leaf size=20 \[ (1-3 x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right ) \]

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Rubi [A]  time = 0.42, antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 17, number of rules used = 8, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1593, 6688, 2411, 2353, 2298, 2302, 29, 2549} \begin {gather*} -3 x \log \left (\frac {3}{5 x^3 \log (x+4)}\right )-3 \log (x)-\log (\log (x+4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + 3*x^2 + (-12 + 33*x + 9*x^2)*Log[4 + x] + (-12*x - 3*x^2)*Log[4 + x]*Log[3/(5*x^3*Log[4 + x])])/((4*
x + x^2)*Log[4 + x]),x]

[Out]

-3*Log[x] - 3*x*Log[3/(5*x^3*Log[4 + x])] - Log[Log[4 + x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{x (4+x) \log (4+x)} \, dx\\ &=\int \left (9-\frac {3}{x}+\frac {-1+3 x}{(4+x) \log (4+x)}-3 \log \left (\frac {3}{5 x^3 \log (4+x)}\right )\right ) \, dx\\ &=9 x-3 \log (x)-3 \int \log \left (\frac {3}{5 x^3 \log (4+x)}\right ) \, dx+\int \frac {-1+3 x}{(4+x) \log (4+x)} \, dx\\ &=9 x-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )+3 \int \left (-3-\frac {x}{(4+x) \log (4+x)}\right ) \, dx+\operatorname {Subst}\left (\int \frac {-13+3 x}{x \log (x)} \, dx,x,4+x\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-3 \int \frac {x}{(4+x) \log (4+x)} \, dx+\operatorname {Subst}\left (\int \left (\frac {3}{\log (x)}-\frac {13}{x \log (x)}\right ) \, dx,x,4+x\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )+3 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4+x\right )-3 \operatorname {Subst}\left (\int \frac {-4+x}{x \log (x)} \, dx,x,4+x\right )-13 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4+x\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )+3 \text {li}(4+x)-3 \operatorname {Subst}\left (\int \left (\frac {1}{\log (x)}-\frac {4}{x \log (x)}\right ) \, dx,x,4+x\right )-13 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (4+x)\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-13 \log (\log (4+x))+3 \text {li}(4+x)-3 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4+x\right )+12 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4+x\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-13 \log (\log (4+x))+12 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (4+x)\right )\\ &=-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-\log (\log (4+x))\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.06, size = 42, normalized size = 2.10 \begin {gather*} -3 \text {Ei}(\log (4+x))-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-\log (\log (4+x))+3 \text {li}(4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + 3*x^2 + (-12 + 33*x + 9*x^2)*Log[4 + x] + (-12*x - 3*x^2)*Log[4 + x]*Log[3/(5*x^3*Log[4 + x])]
)/((4*x + x^2)*Log[4 + x]),x]

[Out]

-3*ExpIntegralEi[Log[4 + x]] - 3*Log[x] - 3*x*Log[3/(5*x^3*Log[4 + x])] - Log[Log[4 + x]] + 3*LogIntegral[4 +
x]

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fricas [A]  time = 0.80, size = 19, normalized size = 0.95 \begin {gather*} -{\left (3 \, x - 1\right )} \log \left (\frac {3}{5 \, x^{3} \log \left (x + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*log(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x
, algorithm="fricas")

[Out]

-(3*x - 1)*log(3/5/(x^3*log(x + 4)))

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giac [A]  time = 0.40, size = 30, normalized size = 1.50 \begin {gather*} -3 \, x \log \relax (3) + 3 \, x \log \left (5 \, x^{3} \log \left (x + 4\right )\right ) - 3 \, \log \relax (x) - \log \left (\log \left (x + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*log(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x
, algorithm="giac")

[Out]

-3*x*log(3) + 3*x*log(5*x^3*log(x + 4)) - 3*log(x) - log(log(x + 4))

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maple [A]  time = 0.44, size = 28, normalized size = 1.40

method	result	size
norman	\(-3 x \ln \left (\frac {3}{5 x^{3} \ln \left (4+x \right )}\right )-3 \ln \relax (x )-\ln \left (\ln \left (4+x \right )\right )\)	\(28\)
default	\(-\ln \left (\ln \left (4+x \right )\right )-3 \ln \relax (x )-3 x \ln \relax (3)+3 x \ln \relax (5)-3 \ln \left (\frac {1}{x^{3} \ln \left (4+x \right )}\right ) x -36\)	\(38\)
risch	\(3 x \ln \left (\ln \left (4+x \right )\right )+9 x \ln \relax (x )-\frac {3 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{2}}{2}-\frac {3 i \pi x \,\mathrm {csgn}\left (\frac {i}{\ln \left (4+x \right )}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{2}}{2}-\frac {3 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-\frac {3 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )}{2}+\frac {3 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (4+x \right )}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )}{2}+\frac {3 i \pi x \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{3}}{2}-\frac {3 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+3 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {3 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}-\frac {3 i \pi x \mathrm {csgn}\left (i x^{3}\right )^{3}}{2}+\frac {3 i \pi x \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}+3 x \ln \relax (5)-3 x \ln \relax (3)-3 \ln \relax (x )-\ln \left (\ln \left (4+x \right )\right )\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2-12*x)*ln(4+x)*ln(3/5/x^3/ln(4+x))+(9*x^2+33*x-12)*ln(4+x)+3*x^2-x)/(x^2+4*x)/ln(4+x),x,method=_RE
TURNVERBOSE)

[Out]

-3*x*ln(3/5/x^3/ln(4+x))-3*ln(x)-ln(ln(4+x))

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maxima [A]  time = 0.57, size = 35, normalized size = 1.75 \begin {gather*} 3 \, x {\left (\log \relax (5) - \log \relax (3)\right )} + 3 \, {\left (3 \, x - 1\right )} \log \relax (x) + 3 \, x \log \left (\log \left (x + 4\right )\right ) - \log \left (\log \left (x + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*log(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x
, algorithm="maxima")

[Out]

3*x*(log(5) - log(3)) + 3*(3*x - 1)*log(x) + 3*x*log(log(x + 4)) - log(log(x + 4))

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mupad [B]  time = 0.93, size = 27, normalized size = 1.35 \begin {gather*} -\ln \left (\ln \left (x+4\right )\right )-3\,\ln \relax (x)-3\,x\,\ln \left (\frac {3}{5\,x^3\,\ln \left (x+4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - log(x + 4)*(33*x + 9*x^2 - 12) - 3*x^2 + log(x + 4)*log(3/(5*x^3*log(x + 4)))*(12*x + 3*x^2))/(log(x
+ 4)*(4*x + x^2)),x)

[Out]

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

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sympy [A]  time = 0.57, size = 32, normalized size = 1.60 \begin {gather*} \left (- 3 x - 2\right ) \log {\left (\frac {3}{5 x^{3} \log {\left (x + 4 \right )}} \right )} - 9 \log {\relax (x )} - 3 \log {\left (\log {\left (x + 4 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2-12*x)*ln(4+x)*ln(3/5/x**3/ln(4+x))+(9*x**2+33*x-12)*ln(4+x)+3*x**2-x)/(x**2+4*x)/ln(4+x),x
)

[Out]

(-3*x - 2)*log(3/(5*x**3*log(x + 4))) - 9*log(x) - 3*log(log(x + 4))

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