3.79.11 \(\int \frac {9+576 x+(x+72 x^2) \log (4)+(144 x+16 x^2 \log (4)) \log (9+x \log (4))}{9 x+288 x^2+(x^2+32 x^3) \log (4)+(72 x^2+8 x^3 \log (4)) \log (9+x \log (4))} \, dx\)

Optimal. Leaf size=17 \[ \log \left (x+8 x^2 (4+\log (9+x \log (4)))\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6741, 6685} \begin {gather*} \log (x (32 x+8 x \log (x \log (4)+9)+1)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 + 576*x + (x + 72*x^2)*Log[4] + (144*x + 16*x^2*Log[4])*Log[9 + x*Log[4]])/(9*x + 288*x^2 + (x^2 + 32*x
^3)*Log[4] + (72*x^2 + 8*x^3*Log[4])*Log[9 + x*Log[4]]),x]

[Out]

Log[x*(1 + 32*x + 8*x*Log[9 + x*Log[4]])]

Rule 6685

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[q]]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9+576 x+\left (x+72 x^2\right ) \log (4)+\left (144 x+16 x^2 \log (4)\right ) \log (9+x \log (4))}{x (9+x \log (4)) (1+32 x+8 x \log (9+x \log (4)))} \, dx\\ &=\log (x (1+32 x+8 x \log (9+x \log (4))))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.49, size = 19, normalized size = 1.12 \begin {gather*} \log (x)+\log (1+32 x+8 x \log (9+x \log (4))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + 576*x + (x + 72*x^2)*Log[4] + (144*x + 16*x^2*Log[4])*Log[9 + x*Log[4]])/(9*x + 288*x^2 + (x^2
+ 32*x^3)*Log[4] + (72*x^2 + 8*x^3*Log[4])*Log[9 + x*Log[4]]),x]

[Out]

Log[x] + Log[1 + 32*x + 8*x*Log[9 + x*Log[4]]]

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fricas [A]  time = 0.84, size = 26, normalized size = 1.53 \begin {gather*} 2 \, \log \relax (x) + \log \left (\frac {8 \, x \log \left (2 \, x \log \relax (2) + 9\right ) + 32 \, x + 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2*log(2)+144*x)*log(2*x*log(2)+9)+2*(72*x^2+x)*log(2)+576*x+9)/((16*x^3*log(2)+72*x^2)*log(2*
x*log(2)+9)+2*(32*x^3+x^2)*log(2)+288*x^2+9*x),x, algorithm="fricas")

[Out]

2*log(x) + log((8*x*log(2*x*log(2) + 9) + 32*x + 1)/x)

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giac [A]  time = 0.16, size = 20, normalized size = 1.18 \begin {gather*} \log \left (8 \, x \log \left (2 \, x \log \relax (2) + 9\right ) + 32 \, x + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2*log(2)+144*x)*log(2*x*log(2)+9)+2*(72*x^2+x)*log(2)+576*x+9)/((16*x^3*log(2)+72*x^2)*log(2*
x*log(2)+9)+2*(32*x^3+x^2)*log(2)+288*x^2+9*x),x, algorithm="giac")

[Out]

log(8*x*log(2*x*log(2) + 9) + 32*x + 1) + log(x)

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maple [A]  time = 0.10, size = 21, normalized size = 1.24

method	result	size
norman	\(\ln \relax (x )+\ln \left (8 x \ln \left (2 x \ln \relax (2)+9\right )+32 x +1\right )\)	\(21\)
risch	\(2 \ln \relax (x )+\ln \left (\ln \left (2 x \ln \relax (2)+9\right )+\frac {32 x +1}{8 x}\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^2*ln(2)+144*x)*ln(2*x*ln(2)+9)+2*(72*x^2+x)*ln(2)+576*x+9)/((16*x^3*ln(2)+72*x^2)*ln(2*x*ln(2)+9)+2
*(32*x^3+x^2)*ln(2)+288*x^2+9*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(8*x*ln(2*x*ln(2)+9)+32*x+1)

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maxima [A]  time = 0.49, size = 27, normalized size = 1.59 \begin {gather*} 2 \, \log \relax (x) + \log \left (\frac {8 \, x \log \left (2 \, x \log \relax (2) + 9\right ) + 32 \, x + 1}{8 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2*log(2)+144*x)*log(2*x*log(2)+9)+2*(72*x^2+x)*log(2)+576*x+9)/((16*x^3*log(2)+72*x^2)*log(2*
x*log(2)+9)+2*(32*x^3+x^2)*log(2)+288*x^2+9*x),x, algorithm="maxima")

[Out]

2*log(x) + log(1/8*(8*x*log(2*x*log(2) + 9) + 32*x + 1)/x)

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mupad [B]  time = 7.26, size = 26, normalized size = 1.53 \begin {gather*} \ln \left (\frac {32\,x+8\,x\,\ln \left (2\,x\,\ln \relax (2)+9\right )+1}{x}\right )+2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((576*x + log(2*x*log(2) + 9)*(144*x + 32*x^2*log(2)) + 2*log(2)*(x + 72*x^2) + 9)/(9*x + 2*log(2)*(x^2 + 3
2*x^3) + log(2*x*log(2) + 9)*(16*x^3*log(2) + 72*x^2) + 288*x^2),x)

[Out]

log((32*x + 8*x*log(2*x*log(2) + 9) + 1)/x) + 2*log(x)

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sympy [A]  time = 0.31, size = 24, normalized size = 1.41 \begin {gather*} 2 \log {\relax (x )} + \log {\left (\log {\left (2 x \log {\relax (2 )} + 9 \right )} + \frac {32 x + 1}{8 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**2*ln(2)+144*x)*ln(2*x*ln(2)+9)+2*(72*x**2+x)*ln(2)+576*x+9)/((16*x**3*ln(2)+72*x**2)*ln(2*x*
ln(2)+9)+2*(32*x**3+x**2)*ln(2)+288*x**2+9*x),x)

[Out]

2*log(x) + log(log(2*x*log(2) + 9) + (32*x + 1)/(8*x))

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