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

Optimal. Leaf size=23 \[ x-x \left (3+\log \left (x+\frac {x \left (-x+x^3\right )}{\log (4)}\right )\right ) \]

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Rubi [A]  time = 8.92, antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 21, number of rules used = 8, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6742, 2079, 800, 634, 618, 204, 628, 2523} \begin {gather*} x \left (-\log \left (-x \left (-x^3+x-\log (4)\right )\right )\right )-2 x+x \log (\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2079

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9*a*d^2 + S
qrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Dist[1/d^(2*p), Int[(e + f*x)^m*Simp[(18^(1/3)*b*d)/(3*r) - r/18^(1/
3) + d*x, x]^p*Simp[(b*d)/3 + (12^(1/3)*b^2*d^2)/(3*r^2) + r^2/(3*12^(1/3)) - d*((2^(1/3)*b*d)/(3^(1/3)*r) - r
/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0
]

Rule 2523

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

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 \left (\frac {4 x-6 x^3-3 \log (4)}{-x+x^3+\log (4)}+\log (\log (4))-\log \left (x \left (-x+x^3+\log (4)\right )\right )\right ) \, dx\\ &=x \log (\log (4))+\int \frac {4 x-6 x^3-3 \log (4)}{-x+x^3+\log (4)} \, dx-\int \log \left (x \left (-x+x^3+\log (4)\right )\right ) \, dx\\ &=-x \log \left (-x \left (x-x^3-\log (4)\right )\right )+x \log (\log (4))+\int \frac {2 x-4 x^3-\log (4)}{x-x^3-\log (4)} \, dx+\int \left (-6+\frac {-2 x+\log (64)}{-x+x^3+\log (4)}\right ) \, dx\\ &=-6 x-x \log \left (-x \left (x-x^3-\log (4)\right )\right )+x \log (\log (4))+\int \frac {-2 x+\log (64)}{-x+x^3+\log (4)} \, dx+\int \left (4+\frac {-2 x+\log (64)}{x-x^3-\log (4)}\right ) \, dx\\ &=-2 x-x \log \left (-x \left (x-x^3-\log (4)\right )\right )+x \log (\log (4))+\int \frac {-2 x+\log (64)}{x-x^3-\log (4)} \, dx+\int \frac {-2 x+\log (64)}{\left (x+\frac {2 \sqrt [3]{\frac {3}{9 \log (4)-\sqrt {-12+81 \log ^2(4)}}}+\sqrt [3]{2 \left (9 \log (4)-\sqrt {-12+81 \log ^2(4)}\right )}}{6^{2/3}}\right ) \left (x^2-\frac {1}{3} x \left (3^{2/3} \sqrt [3]{\frac {2}{9 \log (4)-\sqrt {-12+81 \log ^2(4)}}}+\sqrt [3]{\frac {3}{2} \left (9 \log (4)-\sqrt {-12+81 \log ^2(4)}\right )}\right )+\frac {1}{18} \left (-6+6 \sqrt [3]{3} \left (\frac {2}{9 \log (4)-\sqrt {-12+81 \log ^2(4)}}\right )^{2/3}+\sqrt [3]{2} \left (3 \left (9 \log (4)-\sqrt {-12+81 \log ^2(4)}\right )\right )^{2/3}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 23, normalized size = 1.00 \begin {gather*} -2 x-x \log \left (\frac {x \left (-x+x^3+\log (4)\right )}{\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 28, normalized size = 1.22 \begin {gather*} -x \log \left (\frac {x^{4} - x^{2} + 2 \, x \log \relax (2)}{2 \, \log \relax (2)}\right ) - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2)-x^3+x)*log(1/2*(2*x*log(2)+x^4-x^2)/log(2))-6*log(2)-6*x^3+4*x)/(2*log(2)+x^3-x),x, algo
rithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 28, normalized size = 1.22 \begin {gather*} x {\left (\log \relax (2) + \log \left (\log \relax (2)\right ) - 2\right )} - x \log \left (x^{4} - x^{2} + 2 \, x \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2)-x^3+x)*log(1/2*(2*x*log(2)+x^4-x^2)/log(2))-6*log(2)-6*x^3+4*x)/(2*log(2)+x^3-x),x, algo
rithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 29, normalized size = 1.26

method	result	size
norman	\(-2 x -x \ln \left (\frac {2 x \ln \relax (2)+x^{4}-x^{2}}{2 \ln \relax (2)}\right )\)	\(29\)
risch	\(-2 x -x \ln \left (\frac {2 x \ln \relax (2)+x^{4}-x^{2}}{2 \ln \relax (2)}\right )\)	\(29\)
default	\(x \ln \relax (2)-2 x +2 \left (\munderset {\textit {\_R} =\RootOf \left (2 \ln \relax (2)+\textit {\_Z}^{3}-\textit {\_Z} \right )}{\sum }\frac {\left (-\textit {\_R} +3 \ln \relax (2)\right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}-1}\right )-x \ln \left (x \left (2 \ln \relax (2)+x^{3}-x \right )\right )+2 \left (\munderset {\textit {\_R} =\RootOf \left (2 \ln \relax (2)+\textit {\_Z}^{3}-\textit {\_Z} \right )}{\sum }\frac {\left (\textit {\_R} -3 \ln \relax (2)\right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}-1}\right )+x \ln \left (\ln \relax (2)\right )\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(2)-x^3+x)*ln(1/2*(2*x*ln(2)+x^4-x^2)/ln(2))-6*ln(2)-6*x^3+4*x)/(2*ln(2)+x^3-x),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 30, normalized size = 1.30 \begin {gather*} x {\left (\log \relax (2) + \log \left (\log \relax (2)\right ) - 2\right )} - x \log \left (x^{3} - x + 2 \, \log \relax (2)\right ) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2)-x^3+x)*log(1/2*(2*x*log(2)+x^4-x^2)/log(2))-6*log(2)-6*x^3+4*x)/(2*log(2)+x^3-x),x, algo
rithm="maxima")

[Out]

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

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mupad [B]  time = 2.43, size = 31, normalized size = 1.35 \begin {gather*} x\,\ln \relax (2)-2\,x+x\,\ln \left (\ln \relax (2)\right )-x\,\ln \left (x^4-x^2+2\,\ln \relax (2)\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.18, size = 26, normalized size = 1.13 \begin {gather*} - x \log {\left (\frac {\frac {x^{4}}{2} - \frac {x^{2}}{2} + x \log {\relax (2 )}}{\log {\relax (2 )}} \right )} - 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(2)-x**3+x)*ln(1/2*(2*x*ln(2)+x**4-x**2)/ln(2))-6*ln(2)-6*x**3+4*x)/(2*ln(2)+x**3-x),x)

[Out]

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

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