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

Optimal. Leaf size=19 \[ 4 x \left (1+\log \left (3 (3+2 x) \left (\frac {1}{x}+\log (2)\right )\right )\right ) \]

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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 {gather*} -4 x-\frac {4 \left (3 \log ^2(2)-\log (4)\right ) \log (x \log (2)+1)}{\log ^2(2) (2-\log (8))}+4 x \log \left (\frac {3 (2 x+3) (x \log (2)+1)}{x}\right )+\frac {4 x \log (16)}{\log (4)}-\frac {4 \log (x \log (2)+1)}{\log (2)} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int 4 \left (\frac {x (2+\log (8)+x \log (16))}{(3+2 x) (1+x \log (2))}+\log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right )\right ) \, dx\\ &=4 \int \left (\frac {x (2+\log (8)+x \log (16))}{(3+2 x) (1+x \log (2))}+\log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right )\right ) \, dx\\ &=4 \int \frac {x (2+\log (8)+x \log (16))}{(3+2 x) (1+x \log (2))} \, dx+4 \int \log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right ) \, dx\\ &=4 x \log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right )-4 \int \frac {-3+x^2 \log (4)}{(3+2 x) (1+x \log (2))} \, dx+4 \int \left (\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+2 x) (-2+\log (8))}\right ) \, dx\\ &=\frac {4 x \log (16)}{\log (4)}-\frac {3 (4-\log (64)) \log (3+2 x)}{2-\log (8)}-\frac {4 \log (1+x \log (2))}{\log (2)}+4 x \log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right )-4 \int \left (1+\frac {-3 \log ^2(2)+\log (4)}{\log (2) (1+x \log (2)) (-2+\log (8))}-\frac {3 (-4+\log (64))}{2 (3+2 x) (-2+\log (8))}\right ) \, dx\\ &=-4 x+\frac {4 x \log (16)}{\log (4)}-\frac {4 \log (1+x \log (2))}{\log (2)}-\frac {4 \left (3 \log ^2(2)-\log (4)\right ) \log (1+x \log (2))}{\log ^2(2) (2-\log (8))}+4 x \log \left (\frac {3 (3+2 x) (1+x \log (2))}{x}\right )\\ \end {aligned} \end {gather*}

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

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 {gather*} 4 \, x \log \left (\frac {3 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \relax (2) + 2 \, x + 3\right )}}{x}\right ) + 4 \, x \end {gather*}

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 {gather*} 4 \, x \log \left (6 \, x^{2} \log \relax (2) + 9 \, x \log \relax (2) + 6 \, x + 9\right ) - 4 \, x \log \relax (x) + 4 \, x \end {gather*}

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	\(4 x +4 x \ln \left (\frac {\left (6 x^{2}+9 x \right ) \ln \relax (2)+6 x +9}{x}\right )\)	\(30\)
risch	\(4 x +4 x \ln \left (\frac {\left (6 x^{2}+9 x \right ) \ln \relax (2)+6 x +9}{x}\right )\)	\(30\)
default	\(4 x +4 x \ln \relax (3)+4 x \ln \left (\frac {2 x^{2} \ln \relax (2)+3 x \ln \relax (2)+2 x +3}{x}\right )\)	\(35\)

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 {gather*} 4 \, {\left (\frac {2 \, x}{\log \relax (2)} + \frac {4 \, \log \left (x \log \relax (2) + 1\right )}{3 \, \log \relax (2)^{3} - 2 \, \log \relax (2)^{2}} - \frac {9 \, \log \left (2 \, x + 3\right )}{3 \, \log \relax (2) - 2}\right )} \log \relax (2) - 6 \, {\left (\frac {2 \, \log \left (x \log \relax (2) + 1\right )}{3 \, \log \relax (2)^{2} - 2 \, \log \relax (2)} - \frac {3 \, \log \left (2 \, x + 3\right )}{3 \, \log \relax (2) - 2}\right )} \log \relax (2) + \frac {2 \, {\left (2 \, x {\left (\log \relax (3) - 1\right )} \log \relax (2) - 2 \, x \log \relax (2) \log \relax (x) + 2 \, {\left (x \log \relax (2) + 1\right )} \log \left (x \log \relax (2) + 1\right ) + {\left (2 \, x \log \relax (2) + 3 \, \log \relax (2)\right )} \log \left (2 \, x + 3\right )\right )}}{\log \relax (2)} - \frac {8 \, \log \left (x \log \relax (2) + 1\right )}{3 \, \log \relax (2)^{2} - 2 \, \log \relax (2)} + \frac {12 \, \log \left (2 \, x + 3\right )}{3 \, \log \relax (2) - 2} \end {gather*}

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 {gather*} 4\,x\,\left (\ln \left (\frac {6\,x+\ln \relax (2)\,\left (6\,x^2+9\,x\right )+9}{x}\right )+1\right ) \end {gather*}

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 {gather*} 4 x \log {\left (\frac {6 x + \left (6 x^{2} + 9 x\right ) \log {\relax (2 )} + 9}{x} \right )} + 4 x \end {gather*}

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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