3.53.78 \(\int \frac {6 x+4 x^2+(-6-4 x) \log (\frac {1}{-4-8 x+(1+2 x) \log (8)})}{1+2 x} \, dx\)

Optimal. Leaf size=26 \[ 5+\left (-x+\log \left (\frac {x}{\left (x+2 x^2\right ) (-4+\log (8))}\right )\right )^2 \]

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Rubi [A]  time = 0.10, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6741, 12, 6686} \begin {gather*} \left (x-\log \left (-\frac {1}{(2 x+1) (4-\log (8))}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*x + 4*x^2 + (-6 - 4*x)*Log[(-4 - 8*x + (1 + 2*x)*Log[8])^(-1)])/(1 + 2*x),x]

[Out]

(x - Log[-(1/((1 + 2*x)*(4 - Log[8])))])^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 (3+2 x) \left (x-\log \left (\frac {1}{(1+2 x) (-4+\log (8))}\right )\right )}{1+2 x} \, dx\\ &=2 \int \frac {(3+2 x) \left (x-\log \left (\frac {1}{(1+2 x) (-4+\log (8))}\right )\right )}{1+2 x} \, dx\\ &=\left (x-\log \left (-\frac {1}{(1+2 x) (4-\log (8))}\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.81 \begin {gather*} \left (x-\log \left (\frac {1}{(1+2 x) (-4+\log (8))}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*x + 4*x^2 + (-6 - 4*x)*Log[(-4 - 8*x + (1 + 2*x)*Log[8])^(-1)])/(1 + 2*x),x]

[Out]

(x - Log[1/((1 + 2*x)*(-4 + Log[8]))])^2

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fricas [A]  time = 0.57, size = 43, normalized size = 1.65 \begin {gather*} x^{2} - 2 \, x \log \left (\frac {1}{3 \, {\left (2 \, x + 1\right )} \log \relax (2) - 8 \, x - 4}\right ) + \log \left (\frac {1}{3 \, {\left (2 \, x + 1\right )} \log \relax (2) - 8 \, x - 4}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(1/(3*(2*x+1)*log(2)-8*x-4))+4*x^2+6*x)/(2*x+1),x, algorithm="fricas")

[Out]

x^2 - 2*x*log(1/(3*(2*x + 1)*log(2) - 8*x - 4)) + log(1/(3*(2*x + 1)*log(2) - 8*x - 4))^2

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giac [B]  time = 2.31, size = 405, normalized size = 15.58 \begin {gather*} -\frac {1}{144} \, {\left (\frac {24 \, {\left (\frac {9 \, \log \relax (2)}{6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4} - \frac {12}{6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4} + 1\right )} \log \left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}{\frac {81 \, \log \relax (2)^{4}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{3}} - \frac {432 \, \log \relax (2)^{3}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{3}} + \frac {864 \, \log \relax (2)^{2}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{3}} - \frac {768 \, \log \relax (2)}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{3}} + \frac {256}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{3}}} + \frac {\frac {12 \, \log \relax (2)}{6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4} - \frac {648 \, \log \relax (2)^{2}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{2}} - \frac {16}{6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4} + \frac {1728 \, \log \relax (2)}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{2}} - \frac {1152}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{2}} + 9}{\frac {243 \, \log \relax (2)^{5}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}} - \frac {1620 \, \log \relax (2)^{4}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}} + \frac {4320 \, \log \relax (2)^{3}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}} - \frac {5760 \, \log \relax (2)^{2}}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}} + \frac {3840 \, \log \relax (2)}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}} - \frac {1024}{{\left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right )}^{4}}}\right )} {\left (3 \, \log \relax (2) - 4\right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(1/(3*(2*x+1)*log(2)-8*x-4))+4*x^2+6*x)/(2*x+1),x, algorithm="giac")

[Out]

-1/144*(24*(9*log(2)/(6*x*log(2) - 8*x + 3*log(2) - 4) - 12/(6*x*log(2) - 8*x + 3*log(2) - 4) + 1)*log(6*x*log
(2) - 8*x + 3*log(2) - 4)/(81*log(2)^4/(6*x*log(2) - 8*x + 3*log(2) - 4)^3 - 432*log(2)^3/(6*x*log(2) - 8*x +
3*log(2) - 4)^3 + 864*log(2)^2/(6*x*log(2) - 8*x + 3*log(2) - 4)^3 - 768*log(2)/(6*x*log(2) - 8*x + 3*log(2) -
4)^3 + 256/(6*x*log(2) - 8*x + 3*log(2) - 4)^3) + (12*log(2)/(6*x*log(2) - 8*x + 3*log(2) - 4) - 648*log(2)^2
/(6*x*log(2) - 8*x + 3*log(2) - 4)^2 - 16/(6*x*log(2) - 8*x + 3*log(2) - 4) + 1728*log(2)/(6*x*log(2) - 8*x +
3*log(2) - 4)^2 - 1152/(6*x*log(2) - 8*x + 3*log(2) - 4)^2 + 9)/(243*log(2)^5/(6*x*log(2) - 8*x + 3*log(2) - 4
)^4 - 1620*log(2)^4/(6*x*log(2) - 8*x + 3*log(2) - 4)^4 + 4320*log(2)^3/(6*x*log(2) - 8*x + 3*log(2) - 4)^4 -
5760*log(2)^2/(6*x*log(2) - 8*x + 3*log(2) - 4)^4 + 3840*log(2)/(6*x*log(2) - 8*x + 3*log(2) - 4)^4 - 1024/(6*
x*log(2) - 8*x + 3*log(2) - 4)^4))*(3*log(2) - 4)^2

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maple [A]  time = 0.24, size = 44, normalized size = 1.69

method	result	size
norman	\(x^{2}+\ln \left (\frac {1}{3 \left (2 x +1\right ) \ln \relax (2)-8 x -4}\right )^{2}-2 x \ln \left (\frac {1}{3 \left (2 x +1\right ) \ln \relax (2)-8 x -4}\right )\)	\(44\)
risch	\(x^{2}+2 x -\ln \left (2 x +1\right )+\frac {3 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{3 \ln \relax (2)-4}-\frac {4 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{3 \ln \relax (2)-4}-\frac {6 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right ) x}{3 \ln \relax (2)-4}-\frac {3 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{3 \ln \relax (2)-4}-\frac {6 \ln \relax (2) x}{3 \ln \relax (2)-4}+\frac {8 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right ) x}{3 \ln \relax (2)-4}-\frac {3 \ln \relax (2)}{3 \ln \relax (2)-4}+\frac {4 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{3 \ln \relax (2)-4}+\frac {8 x}{3 \ln \relax (2)-4}+\frac {4}{3 \ln \relax (2)-4}\)	\(236\)
derivativedivides	\(\frac {9 \ln \relax (2)^{2} \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {9 \ln \relax (2)^{2} \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {24 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {3 \ln \relax (2) \left (-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right ) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {24 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {16 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {3 \left (2 x +1\right ) \ln \relax (2)}{2 \left (3 \ln \relax (2)-4\right )}-\frac {4 \left (-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right ) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )\right )}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {16 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {2 \left (2 x +1\right )}{3 \ln \relax (2)-4}+\frac {\left (2 x +1\right )^{2}}{4}\)	\(333\)
default	\(\frac {9 \ln \relax (2)^{2} \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {9 \ln \relax (2)^{2} \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {24 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {3 \ln \relax (2) \left (-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right ) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {24 \ln \relax (2) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {16 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )^{2}}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {3 \left (2 x +1\right ) \ln \relax (2)}{2 \left (3 \ln \relax (2)-4\right )}-\frac {4 \left (-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right ) \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )-\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )\right )}{\left (3 \ln \relax (2)-4\right )^{2}}+\frac {16 \ln \left (\frac {1}{\left (2 x +1\right ) \left (3 \ln \relax (2)-4\right )}\right )}{\left (3 \ln \relax (2)-4\right )^{2}}-\frac {2 \left (2 x +1\right )}{3 \ln \relax (2)-4}+\frac {\left (2 x +1\right )^{2}}{4}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-6)*ln(1/(3*(2*x+1)*ln(2)-8*x-4))+4*x^2+6*x)/(2*x+1),x,method=_RETURNVERBOSE)

[Out]

x^2+ln(1/(3*(2*x+1)*ln(2)-8*x-4))^2-2*x*ln(1/(3*(2*x+1)*ln(2)-8*x-4))

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maxima [B]  time = 0.46, size = 57, normalized size = 2.19 \begin {gather*} x^{2} + {\left (2 \, x - \log \left (2 \, x + 1\right )\right )} \log \left (6 \, x \log \relax (2) - 8 \, x + 3 \, \log \relax (2) - 4\right ) + 2 \, \log \left (2 \, x + 1\right )^{2} + 3 \, \log \left (2 \, x + 1\right ) \log \left (3 \, \log \relax (2) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(1/(3*(2*x+1)*log(2)-8*x-4))+4*x^2+6*x)/(2*x+1),x, algorithm="maxima")

[Out]

x^2 + (2*x - log(2*x + 1))*log(6*x*log(2) - 8*x + 3*log(2) - 4) + 2*log(2*x + 1)^2 + 3*log(2*x + 1)*log(3*log(
2) - 4)

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mupad [B]  time = 6.11, size = 25, normalized size = 0.96 \begin {gather*} {\left (x-\ln \left (-\frac {1}{8\,x-3\,\ln \relax (2)\,\left (2\,x+1\right )+4}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + 4*x^2 - log(-1/(8*x - 3*log(2)*(2*x + 1) + 4))*(4*x + 6))/(2*x + 1),x)

[Out]

(x - log(-1/(8*x - 3*log(2)*(2*x + 1) + 4)))^2

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sympy [B]  time = 0.14, size = 41, normalized size = 1.58 \begin {gather*} x^{2} - 2 x \log {\left (\frac {1}{- 8 x + \left (6 x + 3\right ) \log {\relax (2 )} - 4} \right )} + \log {\left (\frac {1}{- 8 x + \left (6 x + 3\right ) \log {\relax (2 )} - 4} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*ln(1/(3*(2*x+1)*ln(2)-8*x-4))+4*x**2+6*x)/(2*x+1),x)

[Out]

x**2 - 2*x*log(1/(-8*x + (6*x + 3)*log(2) - 4)) + log(1/(-8*x + (6*x + 3)*log(2) - 4))**2

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