3.41.72 $\int \frac{8x+(12x+16x^2)\log (2)+(12+8x+(12x+8x^2)\log (2))\log (\frac{9+6x+(9x+6x^2)\log (2)}{x})}{3+2x+(3x+2x^2)\log (2)}dx$

Optimal. Leaf size=19 $$4x(1+\log (3(3+2x)(\frac{1}{x}+\log (2))))$$

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Rubi [B]  time = 0.22, antiderivative size = 78, normalized size of antiderivative = 4.11, number of steps used = 8, number of rules used = 5, integrand size = 76, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.066, Rules used = {6688, 12, 142, 2523, 894} $$\begin{array}{c}-4x-\frac{4(3{\log }^2(2)-\log (4))\log (x\log (2)+1)}{{\log }^2(2)(2-\log (8))}+4x\log \left(\frac{3(2x+3)(x\log (2)+1)}{x}\right)+\frac{4x\log (16)}{\log (4)}-\frac{4\log (x\log (2)+1)}{\log (2)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(8*x + (12*x + 16*x^2)*Log[2] + (12 + 8*x + (12*x + 8*x^2)*Log[2])*Log[(9 + 6*x + (9*x + 6*x^2)*Log[2])/x]
)/(3 + 2*x + (3*x + 2*x^2)*Log[2]),x]

[Out]

-4*x + (4*x*Log[16])/Log[4] - (4*Log[1 + x*Log[2]])/Log[2] - (4*(3*Log[2]^2 - Log[4])*Log[1 + x*Log[2]])/(Log[
2]^2*(2 - Log[8])) + 4*x*Log[(3*(3 + 2*x)*(1 + x*Log[2]))/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 142

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h},
x] && (IGtQ[m, 0] || IntegersQ[m, n])

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6688

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int 4(\frac{x(2+\log (8)+x\log (16))}{(3+2x)(1+x\log (2))}+\log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right))dx\\ & =4\int (\frac{x(2+\log (8)+x\log (16))}{(3+2x)(1+x\log (2))}+\log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right))dx\\ & =4\int \frac{x(2+\log (8)+x\log (16))}{(3+2x)(1+x\log (2))}dx+4\int \log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right)dx\\ & =4x\log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right)-4\int \frac{-3+x^2\log (4)}{(3+2x)(1+x\log (2))}dx+4\int (\frac{\log (4)-\log (2)\log (8)}{\log (2)(1+x\log (2))(-2+\log (8))}+\frac{\log (16)}{\log (4)}-\frac{3(-4+\log (64))}{2(3+2x)(-2+\log (8))})dx\\ & =\frac{4x\log (16)}{\log (4)}-\frac{3(4-\log (64))\log (3+2x)}{2-\log (8)}-\frac{4\log (1+x\log (2))}{\log (2)}+4x\log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right)-4\int (1+\frac{-3{\log }^2(2)+\log (4)}{\log (2)(1+x\log (2))(-2+\log (8))}-\frac{3(-4+\log (64))}{2(3+2x)(-2+\log (8))})dx\\ & =-4x+\frac{4x\log (16)}{\log (4)}-\frac{4\log (1+x\log (2))}{\log (2)}-\frac{4(3{\log }^2(2)-\log (4))\log (1+x\log (2))}{{\log }^2(2)(2-\log (8))}+4x\log \left(\frac{3(3+2x)(1+x\log (2))}{x}\right)\end{array}\end{array}$$

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Mathematica [B]  time = 0.12, size = 44, normalized size = 2.32 $$\begin{array}{c}\frac{4x(-\log (4)+\log (2)\log (8)+\log (2)(-2+\log (8))\log (6+\frac{9}{x}+x\log (64)+\log (512)))}{\log (2)(-2+\log (8))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(8*x + (12*x + 16*x^2)*Log[2] + (12 + 8*x + (12*x + 8*x^2)*Log[2])*Log[(9 + 6*x + (9*x + 6*x^2)*Log[
2])/x])/(3 + 2*x + (3*x + 2*x^2)*Log[2]),x]

[Out]

(4*x*(-Log[4] + Log[2]*Log[8] + Log[2]*(-2 + Log[8])*Log[6 + 9/x + x*Log[64] + Log[512]]))/(Log[2]*(-2 + Log[8
]))

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fricas [A]  time = 0.54, size = 30, normalized size = 1.58 $$\begin{array}{c}4x\log \left(\frac{3((2x^2+3x)\log (2)+2x+3)}{x}\right)+4x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+(16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*
x)*log(2)+2*x+3),x, algorithm="fricas")

[Out]

4*x*log(3*((2*x^2 + 3*x)*log(2) + 2*x + 3)/x) + 4*x

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giac [A]  time = 0.24, size = 30, normalized size = 1.58 $$\begin{array}{c}4x\log (6x^2\log (2)+9x\log (2)+6x+9)-4x\log (x)+4x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+(16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*
x)*log(2)+2*x+3),x, algorithm="giac")

[Out]

4*x*log(6*x^2*log(2) + 9*x*log(2) + 6*x + 9) - 4*x*log(x) + 4*x

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maple [A]  time = 0.14, size = 30, normalized size = 1.58

method	result	size
norman	$4x+4x\ln \left(\frac{(6x^2+9x)\ln (2)+6x+9}{x}\right)$	$30$
risch	$4x+4x\ln \left(\frac{(6x^2+9x)\ln (2)+6x+9}{x}\right)$	$30$
default	$4x+4x\ln (3)+4x\ln \left(\frac{2x^2\ln (2)+3x\ln (2)+2x+3}{x}\right)$	$35$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2+12*x)*ln(2)+8*x+12)*ln(((6*x^2+9*x)*ln(2)+6*x+9)/x)+(16*x^2+12*x)*ln(2)+8*x)/((2*x^2+3*x)*ln(2)+2
*x+3),x,method=_RETURNVERBOSE)

[Out]

4*x+4*x*ln(((6*x^2+9*x)*ln(2)+6*x+9)/x)

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maxima [B]  time = 0.51, size = 189, normalized size = 9.95 $$\begin{array}{c}4(\frac{2x}{\log (2)}+\frac{4\log (x\log (2)+1)}{3\log (2{)}^3-2\log (2{)}^2}-\frac{9\log (2x+3)}{3\log (2)-2})\log (2)-6(\frac{2\log (x\log (2)+1)}{3\log (2{)}^2-2\log (2)}-\frac{3\log (2x+3)}{3\log (2)-2})\log (2)+\frac{2(2x(\log (3)-1)\log (2)-2x\log (2)\log (x)+2(x\log (2)+1)\log (x\log (2)+1)+(2x\log (2)+3\log (2))\log (2x+3))}{\log (2)}-\frac{8\log (x\log (2)+1)}{3\log (2{)}^2-2\log (2)}+\frac{12\log (2x+3)}{3\log (2)-2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+(16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*
x)*log(2)+2*x+3),x, algorithm="maxima")

[Out]

4*(2*x/log(2) + 4*log(x*log(2) + 1)/(3*log(2)^3 - 2*log(2)^2) - 9*log(2*x + 3)/(3*log(2) - 2))*log(2) - 6*(2*l
og(x*log(2) + 1)/(3*log(2)^2 - 2*log(2)) - 3*log(2*x + 3)/(3*log(2) - 2))*log(2) + 2*(2*x*(log(3) - 1)*log(2)
- 2*x*log(2)*log(x) + 2*(x*log(2) + 1)*log(x*log(2) + 1) + (2*x*log(2) + 3*log(2))*log(2*x + 3))/log(2) - 8*lo
g(x*log(2) + 1)/(3*log(2)^2 - 2*log(2)) + 12*log(2*x + 3)/(3*log(2) - 2)

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mupad [B]  time = 3.47, size = 27, normalized size = 1.42 $$\begin{array}{c}4x(\ln \left(\frac{6x+\ln (2)(6x^2+9x)+9}{x}\right)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(2)*(12*x + 16*x^2) + log((6*x + log(2)*(9*x + 6*x^2) + 9)/x)*(8*x + log(2)*(12*x + 8*x^2) + 12)
)/(2*x + log(2)*(3*x + 2*x^2) + 3),x)

[Out]

4*x*(log((6*x + log(2)*(9*x + 6*x^2) + 9)/x) + 1)

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sympy [A]  time = 0.25, size = 26, normalized size = 1.37 $$\begin{array}{c}4x\log \left(\frac{6x+(6x^2+9x)\log (2)+9}{x}\right)+4x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2+12*x)*ln(2)+8*x+12)*ln(((6*x**2+9*x)*ln(2)+6*x+9)/x)+(16*x**2+12*x)*ln(2)+8*x)/((2*x**2+3*
x)*ln(2)+2*x+3),x)

[Out]

4*x*log((6*x + (6*x**2 + 9*x)*log(2) + 9)/x) + 4*x

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