3.6.39 \(\int \frac {(2+4 x) \log (\frac {1}{16} (x+x^2))}{x+x^2} \, dx\)

Optimal. Leaf size=11 \[ \log ^2\left (\frac {1}{16} x (1+x)\right ) \]

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Rubi [B]  time = 0.27, antiderivative size = 53, normalized size of antiderivative = 4.82, number of steps used = 17, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1593, 2528, 2524, 2357, 2301, 2317, 2391, 2418, 2390} \begin {gather*} 2 \log \left (\frac {1}{16} \left (x^2+x\right )\right ) \log (x)+2 \log (x+1) \log \left (\frac {1}{16} \left (x^2+x\right )\right )-\log ^2(x)-\log ^2(x+1)-2 \log (x+1) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + 4*x)*Log[(x + x^2)/16])/(x + x^2),x]

[Out]

-Log[x]^2 - 2*Log[x]*Log[1 + x] - Log[1 + x]^2 + 2*Log[x]*Log[(x + x^2)/16] + 2*Log[1 + x]*Log[(x + x^2)/16]

Rule 1593

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

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

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(2+4 x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )}{x (1+x)} \, dx\\ &=\int \left (\frac {2 \log \left (\frac {1}{16} \left (x+x^2\right )\right )}{x}+\frac {2 \log \left (\frac {1}{16} \left (x+x^2\right )\right )}{1+x}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {1}{16} \left (x+x^2\right )\right )}{x} \, dx+2 \int \frac {\log \left (\frac {1}{16} \left (x+x^2\right )\right )}{1+x} \, dx\\ &=2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )-2 \int \frac {(1+2 x) \log (x)}{x+x^2} \, dx-2 \int \frac {(1+2 x) \log (1+x)}{x+x^2} \, dx\\ &=2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )-2 \int \frac {(1+2 x) \log (x)}{x (1+x)} \, dx-2 \int \frac {(1+2 x) \log (1+x)}{x (1+x)} \, dx\\ &=2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )-2 \int \left (\frac {\log (x)}{x}+\frac {\log (x)}{1+x}\right ) \, dx-2 \int \left (\frac {\log (1+x)}{x}+\frac {\log (1+x)}{1+x}\right ) \, dx\\ &=2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )-2 \int \frac {\log (x)}{x} \, dx-2 \int \frac {\log (x)}{1+x} \, dx-2 \int \frac {\log (1+x)}{x} \, dx-2 \int \frac {\log (1+x)}{1+x} \, dx\\ &=-\log ^2(x)-2 \log (x) \log (1+x)+2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \text {Li}_2(-x)+2 \int \frac {\log (1+x)}{x} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x\right )\\ &=-\log ^2(x)-2 \log (x) \log (1+x)-\log ^2(1+x)+2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 53, normalized size = 4.82 \begin {gather*} -\log ^2(x)-2 \log (x) \log (1+x)-\log ^2(1+x)+2 \log (x) \log \left (\frac {1}{16} \left (x+x^2\right )\right )+2 \log (1+x) \log \left (\frac {1}{16} \left (x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + 4*x)*Log[(x + x^2)/16])/(x + x^2),x]

[Out]

-Log[x]^2 - 2*Log[x]*Log[1 + x] - Log[1 + x]^2 + 2*Log[x]*Log[(x + x^2)/16] + 2*Log[1 + x]*Log[(x + x^2)/16]

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fricas [A]  time = 0.60, size = 12, normalized size = 1.09 \begin {gather*} \log \left (\frac {1}{16} \, x^{2} + \frac {1}{16} \, x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)*log(1/16*x^2+1/16*x)/(x^2+x),x, algorithm="fricas")

[Out]

log(1/16*x^2 + 1/16*x)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (2 \, x + 1\right )} \log \left (\frac {1}{16} \, x^{2} + \frac {1}{16} \, x\right )}{x^{2} + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)*log(1/16*x^2+1/16*x)/(x^2+x),x, algorithm="giac")

[Out]

integrate(2*(2*x + 1)*log(1/16*x^2 + 1/16*x)/(x^2 + x), x)

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maple [A]  time = 0.23, size = 13, normalized size = 1.18

method	result	size
norman	\(\ln \left (\frac {1}{16} x^{2}+\frac {1}{16} x \right )^{2}\)	\(13\)
risch	\(\ln \left (\frac {1}{16} x^{2}+\frac {1}{16} x \right )^{2}\)	\(13\)
default	\(2 \ln \relax (x ) \ln \left (x^{2}+x \right )-\ln \relax (x )^{2}-2 \ln \relax (x ) \ln \left (x +1\right )+2 \ln \left (x +1\right ) \ln \left (x^{2}+x \right )-\ln \left (x +1\right )^{2}-8 \ln \relax (2) \ln \left (\left (x +1\right ) x \right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x+2)*ln(1/16*x^2+1/16*x)/(x^2+x),x,method=_RETURNVERBOSE)

[Out]

ln(1/16*x^2+1/16*x)^2

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maxima [B]  time = 0.45, size = 32, normalized size = 2.91 \begin {gather*} -2 \, {\left (4 \, \log \relax (2) - \log \relax (x)\right )} \log \left (x + 1\right ) + \log \left (x + 1\right )^{2} - 8 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)*log(1/16*x^2+1/16*x)/(x^2+x),x, algorithm="maxima")

[Out]

-2*(4*log(2) - log(x))*log(x + 1) + log(x + 1)^2 - 8*log(2)*log(x) + log(x)^2

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mupad [B]  time = 0.71, size = 13, normalized size = 1.18 \begin {gather*} {\left (\ln \left (x^2+x\right )-\ln \left (16\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/16 + x^2/16)*(4*x + 2))/(x + x^2),x)

[Out]

(log(x + x^2) - log(16))^2

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sympy [A]  time = 0.11, size = 10, normalized size = 0.91 \begin {gather*} \log {\left (\frac {x^{2}}{16} + \frac {x}{16} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)*ln(1/16*x**2+1/16*x)/(x**2+x),x)

[Out]

log(x**2/16 + x/16)**2

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