3.73.49 $\int \frac{72+8x+(-8-8x)\log (1+x)}{(1+x){\log }^2(1+x)}dx$

Optimal. Leaf size=13 $$1-\frac{8(9+x)}{\log (1+x)}$$

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Rubi [A]  time = 0.27, antiderivative size = 20, normalized size of antiderivative = 1.54, number of steps used = 13, number of rules used = 10, integrand size = 27, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.370, Rules used = {6741, 12, 6742, 2411, 2353, 2297, 2298, 2302, 30, 2389} $$\begin{array}{c}-\frac{8(x+1)}{\log (x+1)}-\frac{64}{\log (x+1)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(72 + 8*x + (-8 - 8*x)*Log[1 + x])/((1 + x)*Log[1 + x]^2),x]

[Out]

-64/Log[1 + x] - (8*(1 + x))/Log[1 + 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 30

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

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 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 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 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 6741

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{8(9+x-\log (1+x)-x\log (1+x))}{(1+x){\log }^2(1+x)}dx\\ & =8\int \frac{9+x-\log (1+x)-x\log (1+x)}{(1+x){\log }^2(1+x)}dx\\ & =8\int (\frac{9+x}{(1+x){\log }^2(1+x)}-\frac{1}{\log (1+x)})dx\\ & =8\int \frac{9+x}{(1+x){\log }^2(1+x)}dx-8\int \frac{1}{\log (1+x)}dx\\ & =8\mathrm{Subst}(\int \frac{8+x}{x{\log }^2(x)}dx,x,1+x)-8\mathrm{Subst}(\int \frac{1}{\log (x)}dx,x,1+x)\\ & =-8\text{li}(1+x)+8\mathrm{Subst}(\int (\frac{1}{{\log }^2(x)}+\frac{8}{x{\log }^2(x)})dx,x,1+x)\\ & =-8\text{li}(1+x)+8\mathrm{Subst}(\int \frac{1}{{\log }^2(x)}dx,x,1+x)+64\mathrm{Subst}(\int \frac{1}{x{\log }^2(x)}dx,x,1+x)\\ & =-\frac{8(1+x)}{\log (1+x)}-8\text{li}(1+x)+8\mathrm{Subst}(\int \frac{1}{\log (x)}dx,x,1+x)+64\mathrm{Subst}(\int \frac{1}{x^2}dx,x,\log (1+x))\\ & =-\frac{64}{\log (1+x)}-\frac{8(1+x)}{\log (1+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 13, normalized size = 1.00 $$\begin{array}{c}\frac{8(-9-x)}{\log (1+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(72 + 8*x + (-8 - 8*x)*Log[1 + x])/((1 + x)*Log[1 + x]^2),x]

[Out]

(8*(-9 - x))/Log[1 + x]

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fricas [A]  time = 0.56, size = 11, normalized size = 0.85 $$\begin{array}{c}-\frac{8(x+9)}{\log (x+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x-8)*log(x+1)+8*x+72)/(x+1)/log(x+1)^2,x, algorithm="fricas")

[Out]

-8*(x + 9)/log(x + 1)

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giac [A]  time = 0.14, size = 11, normalized size = 0.85 $$\begin{array}{c}-\frac{8(x+9)}{\log (x+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x-8)*log(x+1)+8*x+72)/(x+1)/log(x+1)^2,x, algorithm="giac")

[Out]

-8*(x + 9)/log(x + 1)

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maple [A]  time = 0.12, size = 12, normalized size = 0.92

method	result	size
risch	$-\frac{8(x+9)}{\ln (x+1)}$	$12$
norman	$\frac{-8x-72}{\ln (x+1)}$	$13$
derivativedivides	$-\frac{8(x+1)}{\ln (x+1)}-\frac{64}{\ln (x+1)}$	$21$
default	$-\frac{8(x+1)}{\ln (x+1)}-\frac{64}{\ln (x+1)}$	$21$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x-8)*ln(x+1)+8*x+72)/(x+1)/ln(x+1)^2,x,method=_RETURNVERBOSE)

[Out]

-8*(x+9)/ln(x+1)

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maxima [A]  time = 0.37, size = 18, normalized size = 1.38 $$\begin{array}{c}-\frac{8x}{\log (x+1)}-\frac{72}{\log (x+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x-8)*log(x+1)+8*x+72)/(x+1)/log(x+1)^2,x, algorithm="maxima")

[Out]

-8*x/log(x + 1) - 72/log(x + 1)

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mupad [B]  time = 4.43, size = 11, normalized size = 0.85 $$\begin{array}{c}-\frac{8(x+9)}{\ln (x+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - log(x + 1)*(8*x + 8) + 72)/(log(x + 1)^2*(x + 1)),x)

[Out]

-(8*(x + 9))/log(x + 1)

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sympy [A]  time = 0.11, size = 10, normalized size = 0.77 $$\begin{array}{c}\frac{-8x-72}{\log (x+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x-8)*ln(x+1)+8*x+72)/(x+1)/ln(x+1)**2,x)

[Out]

(-8*x - 72)/log(x + 1)

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