3.42.10 \(\int \frac {-19200+3840 x+(-1-32 x) \log (1+64 x+1024 x^2)}{(-5-159 x+32 x^2) \log (1+64 x+1024 x^2)} \, dx\)

Optimal. Leaf size=21 \[ -3-\log (-5+x)+60 \left (-5+\log \left (\log \left ((1+32 x)^2\right )\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6728, 2390, 2302, 29} \begin {gather*} 60 \log \left (\log \left ((32 x+1)^2\right )\right )-\log (5-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-19200 + 3840*x + (-1 - 32*x)*Log[1 + 64*x + 1024*x^2])/((-5 - 159*x + 32*x^2)*Log[1 + 64*x + 1024*x^2]),
x]

[Out]

-Log[5 - x] + 60*Log[Log[(1 + 32*x)^2]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{5-x}+\frac {3840}{(1+32 x) \log \left ((1+32 x)^2\right )}\right ) \, dx\\ &=-\log (5-x)+3840 \int \frac {1}{(1+32 x) \log \left ((1+32 x)^2\right )} \, dx\\ &=-\log (5-x)+120 \operatorname {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,1+32 x\right )\\ &=-\log (5-x)+60 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((1+32 x)^2\right )\right )\\ &=-\log (5-x)+60 \log \left (\log \left ((1+32 x)^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 20, normalized size = 0.95 \begin {gather*} -\log (160-32 x)+60 \log \left (\log \left ((1+32 x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-19200 + 3840*x + (-1 - 32*x)*Log[1 + 64*x + 1024*x^2])/((-5 - 159*x + 32*x^2)*Log[1 + 64*x + 1024*
x^2]),x]

[Out]

-Log[160 - 32*x] + 60*Log[Log[(1 + 32*x)^2]]

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fricas [A]  time = 0.57, size = 21, normalized size = 1.00 \begin {gather*} -\log \left (x - 5\right ) + 60 \, \log \left (\log \left (1024 \, x^{2} + 64 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-1)*log(1024*x^2+64*x+1)+3840*x-19200)/(32*x^2-159*x-5)/log(1024*x^2+64*x+1),x, algorithm="fr
icas")

[Out]

-log(x - 5) + 60*log(log(1024*x^2 + 64*x + 1))

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giac [A]  time = 0.29, size = 21, normalized size = 1.00 \begin {gather*} -\log \left (x - 5\right ) + 60 \, \log \left (\log \left (1024 \, x^{2} + 64 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-1)*log(1024*x^2+64*x+1)+3840*x-19200)/(32*x^2-159*x-5)/log(1024*x^2+64*x+1),x, algorithm="gi
ac")

[Out]

-log(x - 5) + 60*log(log(1024*x^2 + 64*x + 1))

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maple [A]  time = 0.10, size = 22, normalized size = 1.05

method	result	size
default	\(-\ln \left (x -5\right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\)	\(22\)
norman	\(-\ln \left (x -5\right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\)	\(22\)
risch	\(-\ln \left (x -5\right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x-1)*ln(1024*x^2+64*x+1)+3840*x-19200)/(32*x^2-159*x-5)/ln(1024*x^2+64*x+1),x,method=_RETURNVERBOSE)

[Out]

-ln(x-5)+60*ln(ln(1024*x^2+64*x+1))

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maxima [A]  time = 0.40, size = 16, normalized size = 0.76 \begin {gather*} -\log \left (x - 5\right ) + 60 \, \log \left (\log \left (32 \, x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-1)*log(1024*x^2+64*x+1)+3840*x-19200)/(32*x^2-159*x-5)/log(1024*x^2+64*x+1),x, algorithm="ma
xima")

[Out]

-log(x - 5) + 60*log(log(32*x + 1))

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mupad [B]  time = 0.29, size = 21, normalized size = 1.00 \begin {gather*} 60\,\ln \left (\ln \left (1024\,x^2+64\,x+1\right )\right )-\ln \left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(64*x + 1024*x^2 + 1)*(32*x + 1) - 3840*x + 19200)/(log(64*x + 1024*x^2 + 1)*(159*x - 32*x^2 + 5)),x)

[Out]

60*log(log(64*x + 1024*x^2 + 1)) - log(x - 5)

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sympy [A]  time = 0.13, size = 19, normalized size = 0.90 \begin {gather*} - \log {\left (x - 5 \right )} + 60 \log {\left (\log {\left (1024 x^{2} + 64 x + 1 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x-1)*ln(1024*x**2+64*x+1)+3840*x-19200)/(32*x**2-159*x-5)/ln(1024*x**2+64*x+1),x)

[Out]

-log(x - 5) + 60*log(log(1024*x**2 + 64*x + 1))

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