3.64.31 \(\int \frac {-1-4 x-2 x^2+4 x^3+(-x-4 x^2-4 x^3) \log (x)+(-x+(-x-4 x^2-4 x^3) \log (x)) \log (2 x)}{x+4 x^2+4 x^3} \, dx\)

Optimal. Leaf size=26 \[ 2-\left (1-x+\frac {x}{1+2 x}+x \log (x)\right ) \log (2 x) \]

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Rubi [A]  time = 2.18, antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps used = 50, number of rules used = 17, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1594, 27, 6688, 6742, 44, 43, 2295, 2314, 31, 32, 2557, 2344, 2301, 2317, 2391, 2351, 2361} \begin {gather*} -x \log (2 x) \log (x)-\log (x)+x \log (2 x)-\frac {x \log (2 x)}{2 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 4*x - 2*x^2 + 4*x^3 + (-x - 4*x^2 - 4*x^3)*Log[x] + (-x + (-x - 4*x^2 - 4*x^3)*Log[x])*Log[2*x])/(x
+ 4*x^2 + 4*x^3),x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2314

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

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 2344

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

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 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1-4 x-2 x^2+4 x^3+\left (-x-4 x^2-4 x^3\right ) \log (x)+\left (-x+\left (-x-4 x^2-4 x^3\right ) \log (x)\right ) \log (2 x)}{x \left (1+4 x+4 x^2\right )} \, dx\\ &=\int \frac {-1-4 x-2 x^2+4 x^3+\left (-x-4 x^2-4 x^3\right ) \log (x)+\left (-x+\left (-x-4 x^2-4 x^3\right ) \log (x)\right ) \log (2 x)}{x (1+2 x)^2} \, dx\\ &=\int \frac {-1-4 x-2 x^2+4 x^3-x \log (2 x)-x (1+2 x)^2 \log (x) (1+\log (2 x))}{x (1+2 x)^2} \, dx\\ &=\int \left (-\frac {4}{(1+2 x)^2}-\frac {1}{x (1+2 x)^2}-\frac {2 x}{(1+2 x)^2}+\frac {4 x^2}{(1+2 x)^2}-\log (x)-\frac {\left (1+\log (x)+4 x \log (x)+4 x^2 \log (x)\right ) \log (2 x)}{(1+2 x)^2}\right ) \, dx\\ &=\frac {2}{1+2 x}-2 \int \frac {x}{(1+2 x)^2} \, dx+4 \int \frac {x^2}{(1+2 x)^2} \, dx-\int \frac {1}{x (1+2 x)^2} \, dx-\int \log (x) \, dx-\int \frac {\left (1+\log (x)+4 x \log (x)+4 x^2 \log (x)\right ) \log (2 x)}{(1+2 x)^2} \, dx\\ &=x+\frac {2}{1+2 x}-x \log (x)-2 \int \left (-\frac {1}{2 (1+2 x)^2}+\frac {1}{2 (1+2 x)}\right ) \, dx+4 \int \left (\frac {1}{4}+\frac {1}{4 (1+2 x)^2}-\frac {1}{2 (1+2 x)}\right ) \, dx-\int \left (\frac {1}{x}-\frac {2}{(1+2 x)^2}-\frac {2}{1+2 x}\right ) \, dx-\int \left (\frac {\log (2 x)}{(1+2 x)^2}+\frac {\log (x) \log (2 x)}{(1+2 x)^2}+\frac {4 x \log (x) \log (2 x)}{(1+2 x)^2}+\frac {4 x^2 \log (x) \log (2 x)}{(1+2 x)^2}\right ) \, dx\\ &=2 x-\log (x)-x \log (x)-\frac {1}{2} \log (1+2 x)-4 \int \frac {x \log (x) \log (2 x)}{(1+2 x)^2} \, dx-4 \int \frac {x^2 \log (x) \log (2 x)}{(1+2 x)^2} \, dx-\int \frac {\log (2 x)}{(1+2 x)^2} \, dx-\int \frac {\log (x) \log (2 x)}{(1+2 x)^2} \, dx\\ &=2 x-\log (x)-x \log (x)-\frac {x \log (2 x)}{1+2 x}+\frac {\log (x) \log (2 x)}{2 (1+2 x)}-\frac {1}{2} \log (1+2 x)-4 \int \left (\frac {1}{4} \log (x) \log (2 x)+\frac {\log (x) \log (2 x)}{4 (1+2 x)^2}-\frac {\log (x) \log (2 x)}{2 (1+2 x)}\right ) \, dx-4 \int \left (-\frac {\log (x) \log (2 x)}{2 (1+2 x)^2}+\frac {\log (x) \log (2 x)}{2 (1+2 x)}\right ) \, dx+\int \frac {1}{1+2 x} \, dx+\int \frac {\log (x)}{(-2-4 x) x} \, dx+\int \frac {\log (2 x)}{(-2-4 x) x} \, dx\\ &=2 x-\log (x)-x \log (x)-\frac {x \log (2 x)}{1+2 x}+\frac {\log (x) \log (2 x)}{2 (1+2 x)}-\frac {1}{2} \int \frac {\log (x)}{x} \, dx-\frac {1}{2} \int \frac {\log (2 x)}{x} \, dx-2 \int \frac {\log (x)}{-2-4 x} \, dx-2 \int \frac {\log (2 x)}{-2-4 x} \, dx+2 \int \frac {\log (x) \log (2 x)}{(1+2 x)^2} \, dx-\int \log (x) \log (2 x) \, dx-\int \frac {\log (x) \log (2 x)}{(1+2 x)^2} \, dx\\ &=2 x-\log (x)-x \log (x)-\frac {\log ^2(x)}{4}+x \log (2 x)-\frac {x \log (2 x)}{1+2 x}-x \log (x) \log (2 x)-\frac {1}{4} \log ^2(2 x)+\frac {1}{2} \log (x) \log (1+2 x)+\frac {1}{2} \log (2 x) \log (1+2 x)-2 \left (\frac {1}{2} \int \frac {\log (1+2 x)}{x} \, dx\right )-2 \int \frac {\log (x)}{(-2-4 x) x} \, dx-2 \int \frac {\log (2 x)}{(-2-4 x) x} \, dx+\int (-1+\log (x)) \, dx+\int \frac {\log (x)}{(-2-4 x) x} \, dx+\int \frac {\log (2 x)}{(-2-4 x) x} \, dx\\ &=x-\log (x)-x \log (x)-\frac {\log ^2(x)}{4}+x \log (2 x)-\frac {x \log (2 x)}{1+2 x}-x \log (x) \log (2 x)-\frac {1}{4} \log ^2(2 x)+\frac {1}{2} \log (x) \log (1+2 x)+\frac {1}{2} \log (2 x) \log (1+2 x)+\text {Li}_2(-2 x)-\frac {1}{2} \int \frac {\log (x)}{x} \, dx-\frac {1}{2} \int \frac {\log (2 x)}{x} \, dx-2 \int \frac {\log (x)}{-2-4 x} \, dx-2 \int \frac {\log (2 x)}{-2-4 x} \, dx+4 \int \frac {\log (x)}{-2-4 x} \, dx+4 \int \frac {\log (2 x)}{-2-4 x} \, dx+\int \log (x) \, dx+\int \frac {\log (x)}{x} \, dx+\int \frac {\log (2 x)}{x} \, dx\\ &=-\log (x)+x \log (2 x)-\frac {x \log (2 x)}{1+2 x}-x \log (x) \log (2 x)+\text {Li}_2(-2 x)-2 \left (\frac {1}{2} \int \frac {\log (1+2 x)}{x} \, dx\right )+2 \int \frac {\log (1+2 x)}{x} \, dx\\ &=-\log (x)+x \log (2 x)-\frac {x \log (2 x)}{1+2 x}-x \log (x) \log (2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 43, normalized size = 1.65 \begin {gather*} -\frac {-\left (\left (1+2 x+4 x^2\right ) \log (2 x)\right )+(1+2 x) \log (x) (3+2 x \log (2 x))}{2+4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 4*x - 2*x^2 + 4*x^3 + (-x - 4*x^2 - 4*x^3)*Log[x] + (-x + (-x - 4*x^2 - 4*x^3)*Log[x])*Log[2*x
])/(x + 4*x^2 + 4*x^3),x]

[Out]

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

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fricas [B]  time = 0.74, size = 62, normalized size = 2.38 \begin {gather*} -\frac {2 \, {\left (2 \, x^{2} + x\right )} \log \relax (x)^{2} - {\left (4 \, x^{2} + 2 \, x + 1\right )} \log \relax (2) - 2 \, {\left (2 \, x^{2} - {\left (2 \, x^{2} + x\right )} \log \relax (2) - 2 \, x - 1\right )} \log \relax (x)}{2 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-4*x^2-x)*log(x)-x)*log(2*x)+(-4*x^3-4*x^2-x)*log(x)+4*x^3-2*x^2-4*x-1)/(4*x^3+4*x^2+x),x,
algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 48, normalized size = 1.85 \begin {gather*} -x \log \relax (x)^{2} + x \log \relax (2) - \frac {1}{2} \, {\left (2 \, x {\left (\log \relax (2) - 1\right )} - \frac {1}{2 \, x + 1}\right )} \log \relax (x) + \frac {\log \relax (2)}{2 \, {\left (2 \, x + 1\right )}} - \frac {3}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-4*x^2-x)*log(x)-x)*log(2*x)+(-4*x^3-4*x^2-x)*log(x)+4*x^3-2*x^2-4*x-1)/(4*x^3+4*x^2+x),x,
algorithm="giac")

[Out]

-x*log(x)^2 + x*log(2) - 1/2*(2*x*(log(2) - 1) - 1/(2*x + 1))*log(x) + 1/2*log(2)/(2*x + 1) - 3/2*log(x)

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maple [A]  time = 0.17, size = 51, normalized size = 1.96

method	result	size
default	\(\frac {\ln \relax (2)}{4 x +2}-x \ln \relax (2) \ln \relax (x )+x \ln \relax (2)-\ln \relax (x )-x \ln \relax (x )^{2}+x \ln \relax (x )-\frac {\ln \relax (x ) x}{2 x +1}\)	\(51\)
risch	\(-x \ln \relax (x )^{2}-\frac {\left (-1+4 x^{2} \ln \relax (2)+2 x \ln \relax (2)-4 x^{2}-2 x \right ) \ln \relax (x )}{2 \left (2 x +1\right )}-\frac {-8 x^{2} \ln \relax (2)+12 x \ln \relax (x )-4 x \ln \relax (2)+6 \ln \relax (x )-2 \ln \relax (2)}{4 \left (2 x +1\right )}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3-4*x^2-x)*ln(x)-x)*ln(2*x)+(-4*x^3-4*x^2-x)*ln(x)+4*x^3-2*x^2-4*x-1)/(4*x^3+4*x^2+x),x,method=_RE
TURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.49, size = 78, normalized size = 3.00 \begin {gather*} x + \frac {4 \, x^{2} {\left (\log \relax (2) - 1\right )} - 2 \, {\left (2 \, x^{2} + x\right )} \log \relax (x)^{2} + 2 \, x {\left (\log \relax (2) - 1\right )} - 2 \, {\left (2 \, x^{2} {\left (\log \relax (2) - 1\right )} + x {\left (\log \relax (2) - 1\right )}\right )} \log \relax (x) + \log \relax (2)}{2 \, {\left (2 \, x + 1\right )}} + \frac {\log \relax (x)}{2 \, {\left (2 \, x + 1\right )}} - \frac {3}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-4*x^2-x)*log(x)-x)*log(2*x)+(-4*x^3-4*x^2-x)*log(x)+4*x^3-2*x^2-4*x-1)/(4*x^3+4*x^2+x),x,
algorithm="maxima")

[Out]

x + 1/2*(4*x^2*(log(2) - 1) - 2*(2*x^2 + x)*log(x)^2 + 2*x*(log(2) - 1) - 2*(2*x^2*(log(2) - 1) + x*(log(2) -
1))*log(x) + log(2))/(2*x + 1) + 1/2*log(x)/(2*x + 1) - 3/2*log(x)

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mupad [B]  time = 4.24, size = 32, normalized size = 1.23 \begin {gather*} x\,\ln \left (2\,x\right )-\frac {3\,\ln \relax (x)}{2}+\frac {\ln \left (2\,x\right )}{2\,\left (2\,x+1\right )}-x\,\ln \left (2\,x\right )\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + log(2*x)*(x + log(x)*(x + 4*x^2 + 4*x^3)) + log(x)*(x + 4*x^2 + 4*x^3) + 2*x^2 - 4*x^3 + 1)/(x + 4
*x^2 + 4*x^3),x)

[Out]

x*log(2*x) - (3*log(x))/2 + log(2*x)/(2*(2*x + 1)) - x*log(2*x)*log(x)

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sympy [B]  time = 0.51, size = 60, normalized size = 2.31 \begin {gather*} - x \log {\relax (x )}^{2} + x \log {\relax (2 )} - \frac {3 \log {\relax (x )}}{2} + \frac {\left (- 4 x^{2} \log {\relax (2 )} + 4 x^{2} - 2 x \log {\relax (2 )} + 2 x + 1\right ) \log {\relax (x )}}{4 x + 2} + \frac {\log {\relax (2 )}}{4 x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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