3.67.24 $\int \frac{1260x^2+16x^3+(-252x^2-3x^3)\log (84+x)}{2100+25x+(-840-10x)\log (84+x)+(84+x){\log }^2(84+x)}dx$

Optimal. Leaf size=16 $$\log (4)-\frac{x^3}{-5+\log (84+x)}$$

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Rubi [B]  time = 0.76, antiderivative size = 61, normalized size of antiderivative = 3.81, number of steps used = 28, number of rules used = 14, integrand size = 55, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.254, Rules used = {6688, 6742, 2411, 2353, 2297, 2299, 2178, 2302, 30, 2306, 2309, 2399, 2389, 2390} $$\begin{array}{c}\frac{(x+84{)}^3}{5-\log (x+84)}-\frac{252(x+84{)}^2}{5-\log (x+84)}+\frac{21168(x+84)}{5-\log (x+84)}-\frac{592704}{5-\log (x+84)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(1260*x^2 + 16*x^3 + (-252*x^2 - 3*x^3)*Log[84 + x])/(2100 + 25*x + (-840 - 10*x)*Log[84 + x] + (84 + x)*L
og[84 + x]^2),x]

[Out]

-592704/(5 - Log[84 + x]) + (21168*(84 + x))/(5 - Log[84 + x]) - (252*(84 + x)^2)/(5 - Log[84 + x]) + (84 + x)
^3/(5 - Log[84 + 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 2306

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 2309

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 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 2389

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

Rule 2390

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

Rule 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

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 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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{x^2(4(315+4x)-3(84+x)\log (84+x))}{(84+x)(5-\log (84+x){)}^2}dx\\ & =\int (\frac{x^3}{(84+x)(-5+\log (84+x){)}^2}-\frac{3x^2}{-5+\log (84+x)})dx\\ & =-(3\int \frac{x^2}{-5+\log (84+x)}dx)+\int \frac{x^3}{(84+x)(-5+\log (84+x){)}^2}dx\\ & =-(3\int (\frac{7056}{-5+\log (84+x)}-\frac{168(84+x)}{-5+\log (84+x)}+\frac{(84+x{)}^2}{-5+\log (84+x)})dx)+\mathrm{Subst}(\int \frac{(-84+x{)}^3}{x(-5+\log (x){)}^2}dx,x,84+x)\\ & =-(3\int \frac{(84+x{)}^2}{-5+\log (84+x)}dx)+504\int \frac{84+x}{-5+\log (84+x)}dx-21168\int \frac{1}{-5+\log (84+x)}dx+\mathrm{Subst}(\int (\frac{21168}{(-5+\log (x){)}^2}-\frac{592704}{x(-5+\log (x){)}^2}-\frac{252x}{(-5+\log (x){)}^2}+\frac{x^2}{(-5+\log (x){)}^2})dx,x,84+x)\\ & =-\left(3\mathrm{Subst}(\int \frac{x^2}{-5+\log (x)}dx,x,84+x)\right)-252\mathrm{Subst}(\int \frac{x}{(-5+\log (x){)}^2}dx,x,84+x)+504\mathrm{Subst}(\int \frac{x}{-5+\log (x)}dx,x,84+x)+21168\mathrm{Subst}(\int \frac{1}{(-5+\log (x){)}^2}dx,x,84+x)-21168\mathrm{Subst}(\int \frac{1}{-5+\log (x)}dx,x,84+x)-592704\mathrm{Subst}(\int \frac{1}{x(-5+\log (x){)}^2}dx,x,84+x)+\mathrm{Subst}(\int \frac{x^2}{(-5+\log (x){)}^2}dx,x,84+x)\\ & =\frac{21168(84+x)}{5-\log (84+x)}-\frac{252(84+x{)}^2}{5-\log (84+x)}+\frac{(84+x{)}^3}{5-\log (84+x)}-3\mathrm{Subst}(\int \frac{e^{3x}}{-5+x}dx,x,\log (84+x))+3\mathrm{Subst}(\int \frac{x^2}{-5+\log (x)}dx,x,84+x)+504\mathrm{Subst}(\int \frac{e^{2x}}{-5+x}dx,x,\log (84+x))-504\mathrm{Subst}(\int \frac{x}{-5+\log (x)}dx,x,84+x)-21168\mathrm{Subst}(\int \frac{e^x}{-5+x}dx,x,\log (84+x))+21168\mathrm{Subst}(\int \frac{1}{-5+\log (x)}dx,x,84+x)-592704\mathrm{Subst}(\int \frac{1}{x^2}dx,x,-5+\log (84+x))\\ & =-3e^{15}\text{Ei}(-3(5-\log (84+x)))+504e^{10}\text{Ei}(-2(5-\log (84+x)))-21168e^5\text{Ei}(-5+\log (84+x))-\frac{592704}{5-\log (84+x)}+\frac{21168(84+x)}{5-\log (84+x)}-\frac{252(84+x{)}^2}{5-\log (84+x)}+\frac{(84+x{)}^3}{5-\log (84+x)}+3\mathrm{Subst}(\int \frac{e^{3x}}{-5+x}dx,x,\log (84+x))-504\mathrm{Subst}(\int \frac{e^{2x}}{-5+x}dx,x,\log (84+x))+21168\mathrm{Subst}(\int \frac{e^x}{-5+x}dx,x,\log (84+x))\\ & =-\frac{592704}{5-\log (84+x)}+\frac{21168(84+x)}{5-\log (84+x)}-\frac{252(84+x{)}^2}{5-\log (84+x)}+\frac{(84+x{)}^3}{5-\log (84+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.14, size = 13, normalized size = 0.81 $$\begin{array}{c}-\frac{x^3}{-5+\log (84+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(1260*x^2 + 16*x^3 + (-252*x^2 - 3*x^3)*Log[84 + x])/(2100 + 25*x + (-840 - 10*x)*Log[84 + x] + (84
+ x)*Log[84 + x]^2),x]

[Out]

-(x^3/(-5 + Log[84 + x]))

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fricas [A]  time = 0.45, size = 13, normalized size = 0.81 $$\begin{array}{c}-\frac{x^3}{\log (x+84)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3-252*x^2)*log(x+84)+16*x^3+1260*x^2)/((x+84)*log(x+84)^2+(-10*x-840)*log(x+84)+25*x+2100),x,
algorithm="fricas")

[Out]

-x^3/(log(x + 84) - 5)

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giac [A]  time = 3.51, size = 13, normalized size = 0.81 $$\begin{array}{c}-\frac{x^3}{\log (x+84)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3-252*x^2)*log(x+84)+16*x^3+1260*x^2)/((x+84)*log(x+84)^2+(-10*x-840)*log(x+84)+25*x+2100),x,
algorithm="giac")

[Out]

-x^3/(log(x + 84) - 5)

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maple [A]  time = 0.08, size = 14, normalized size = 0.88

method	result	size
norman	$-\frac{x^3}{\ln (x+84)-5}$	$14$
risch	$-\frac{x^3}{\ln (x+84)-5}$	$14$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^3-252*x^2)*ln(x+84)+16*x^3+1260*x^2)/((x+84)*ln(x+84)^2+(-10*x-840)*ln(x+84)+25*x+2100),x,method=_R
ETURNVERBOSE)

[Out]

-x^3/(ln(x+84)-5)

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maxima [A]  time = 0.41, size = 13, normalized size = 0.81 $$\begin{array}{c}-\frac{x^3}{\log (x+84)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3-252*x^2)*log(x+84)+16*x^3+1260*x^2)/((x+84)*log(x+84)^2+(-10*x-840)*log(x+84)+25*x+2100),x,
algorithm="maxima")

[Out]

-x^3/(log(x + 84) - 5)

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mupad [B]  time = 0.20, size = 13, normalized size = 0.81 $$\begin{array}{c}-\frac{x^3}{\ln (x+84)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1260*x^2 - log(x + 84)*(252*x^2 + 3*x^3) + 16*x^3)/(25*x - log(x + 84)*(10*x + 840) + log(x + 84)^2*(x +
84) + 2100),x)

[Out]

-x^3/(log(x + 84) - 5)

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sympy [A]  time = 0.11, size = 10, normalized size = 0.62 $$\begin{array}{c}-\frac{x^3}{\log (x+84)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**3-252*x**2)*ln(x+84)+16*x**3+1260*x**2)/((x+84)*ln(x+84)**2+(-10*x-840)*ln(x+84)+25*x+2100),
x)

[Out]

-x**3/(log(x + 84) - 5)

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