3.58.64 \(\int \frac {(-8 x-4 x^2) \log (\frac {2+x}{x})+(6+8 x+(-6-11 x-4 x^2) \log (\frac {2+x}{x})) \log (3+4 x)}{(6 x+11 x^2+4 x^3) \log (\frac {2+x}{x}) \log (3+4 x)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {\log (2)}{x \log \left (1+\frac {2}{x}\right ) \log (3+4 x)}\right ) \]

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Rubi [A]  time = 0.79, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1594, 6728, 6685, 2390, 2302, 29} \begin {gather*} -\log \left (x \log \left (\frac {2}{x}+1\right )\right )-\log (\log (4 x+3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-8*x - 4*x^2)*Log[(2 + x)/x] + (6 + 8*x + (-6 - 11*x - 4*x^2)*Log[(2 + x)/x])*Log[3 + 4*x])/((6*x + 11*x
^2 + 4*x^3)*Log[(2 + x)/x]*Log[3 + 4*x]),x]

[Out]

-Log[x*Log[1 + 2/x]] - Log[Log[3 + 4*x]]

Rule 29

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

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 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 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 \frac {\left (-8 x-4 x^2\right ) \log \left (\frac {2+x}{x}\right )+\left (6+8 x+\left (-6-11 x-4 x^2\right ) \log \left (\frac {2+x}{x}\right )\right ) \log (3+4 x)}{x \left (6+11 x+4 x^2\right ) \log \left (\frac {2+x}{x}\right ) \log (3+4 x)} \, dx\\ &=\int \left (\frac {2-2 \log \left (\frac {2+x}{x}\right )-x \log \left (\frac {2+x}{x}\right )}{x (2+x) \log \left (1+\frac {2}{x}\right )}-\frac {4}{(3+4 x) \log (3+4 x)}\right ) \, dx\\ &=-\left (4 \int \frac {1}{(3+4 x) \log (3+4 x)} \, dx\right )+\int \frac {2-2 \log \left (\frac {2+x}{x}\right )-x \log \left (\frac {2+x}{x}\right )}{x (2+x) \log \left (1+\frac {2}{x}\right )} \, dx\\ &=-\log \left (x \log \left (1+\frac {2}{x}\right )\right )-\operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,3+4 x\right )\\ &=-\log \left (x \log \left (1+\frac {2}{x}\right )\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (3+4 x)\right )\\ &=-\log \left (x \log \left (1+\frac {2}{x}\right )\right )-\log (\log (3+4 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.05, size = 25, normalized size = 1.00 \begin {gather*} -\log (x)-\log \left (\log \left (\frac {2+x}{x}\right )\right )-\log (\log (3+4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8*x - 4*x^2)*Log[(2 + x)/x] + (6 + 8*x + (-6 - 11*x - 4*x^2)*Log[(2 + x)/x])*Log[3 + 4*x])/((6*x
+ 11*x^2 + 4*x^3)*Log[(2 + x)/x]*Log[3 + 4*x]),x]

[Out]

-Log[x] - Log[Log[(2 + x)/x]] - Log[Log[3 + 4*x]]

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fricas [A]  time = 0.60, size = 25, normalized size = 1.00 \begin {gather*} -\log \relax (x) - \log \left (\log \left (4 \, x + 3\right )\right ) - \log \left (\log \left (\frac {x + 2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-11*x-6)*log((2+x)/x)+8*x+6)*log(3+4*x)+(-4*x^2-8*x)*log((2+x)/x))/(4*x^3+11*x^2+6*x)/log((
2+x)/x)/log(3+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 26, normalized size = 1.04 \begin {gather*} -\log \relax (x) - \log \left (\log \left (x + 2\right ) - \log \relax (x)\right ) - \log \left (\log \left (4 \, x + 3\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-11*x-6)*log((2+x)/x)+8*x+6)*log(3+4*x)+(-4*x^2-8*x)*log((2+x)/x))/(4*x^3+11*x^2+6*x)/log((
2+x)/x)/log(3+4*x),x, algorithm="giac")

[Out]

-log(x) - log(log(x + 2) - log(x)) - log(log(4*x + 3))

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maple [A]  time = 0.12, size = 26, normalized size = 1.04

method	result	size
default	\(-\ln \relax (x )-\ln \left (\ln \left (3+4 x \right )\right )-\ln \left (\ln \left (\frac {2}{x}+1\right )\right )\)	\(26\)
risch	\(-\ln \relax (x )-\ln \left (\ln \left (2+x \right )-\frac {i \left (\pi \mathrm {csgn}\left (\frac {i \left (2+x \right )}{x}\right )^{3}-\pi \mathrm {csgn}\left (\frac {i \left (2+x \right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )-\pi \mathrm {csgn}\left (\frac {i \left (2+x \right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )\right )+\pi \,\mathrm {csgn}\left (\frac {i \left (2+x \right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (2+x \right )\right )-2 i \ln \relax (x )\right )}{2}\right )-\ln \left (\ln \left (3+4 x \right )\right )\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^2-11*x-6)*ln((2+x)/x)+8*x+6)*ln(3+4*x)+(-4*x^2-8*x)*ln((2+x)/x))/(4*x^3+11*x^2+6*x)/ln((2+x)/x)/ln
(3+4*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)-ln(ln(3+4*x))-ln(ln(2/x+1))

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maxima [A]  time = 0.40, size = 26, normalized size = 1.04 \begin {gather*} -\log \relax (x) - \log \left (\log \left (x + 2\right ) - \log \relax (x)\right ) - \log \left (\log \left (4 \, x + 3\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-11*x-6)*log((2+x)/x)+8*x+6)*log(3+4*x)+(-4*x^2-8*x)*log((2+x)/x))/(4*x^3+11*x^2+6*x)/log((
2+x)/x)/log(3+4*x),x, algorithm="maxima")

[Out]

-log(x) - log(log(x + 2) - log(x)) - log(log(4*x + 3))

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mupad [B]  time = 4.13, size = 25, normalized size = 1.00 \begin {gather*} -\ln \left (\ln \left (4\,x+3\right )\right )-\ln \relax (x)-\ln \left (\ln \left (\frac {x+2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x + 2)/x)*(8*x + 4*x^2) - log(4*x + 3)*(8*x - log((x + 2)/x)*(11*x + 4*x^2 + 6) + 6))/(log((x + 2)/
x)*log(4*x + 3)*(6*x + 11*x^2 + 4*x^3)),x)

[Out]

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

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sympy [A]  time = 0.36, size = 20, normalized size = 0.80 \begin {gather*} - \log {\relax (x )} - \log {\left (\log {\left (\frac {x + 2}{x} \right )} \right )} - \log {\left (\log {\left (4 x + 3 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**2-11*x-6)*ln((2+x)/x)+8*x+6)*ln(3+4*x)+(-4*x**2-8*x)*ln((2+x)/x))/(4*x**3+11*x**2+6*x)/ln((
2+x)/x)/ln(3+4*x),x)

[Out]

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

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