3.36.44 \(\int \frac {-8 x-32 x^2-16 x^3+(2-12 x-62 x^2-32 x^3) \log (2)}{(x+2 x^2+x^3) \log (2)} \, dx\) [3544]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 5, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 1608, 27, 1634} \begin {gather*} -\frac {16 x (1+\log (4))}{\log (2)}+\frac {\log (4) \log (x)}{\log (2)}-\frac {8 (1+\log (4))}{(x+1) \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x - 32*x^2 - 16*x^3 + (2 - 12*x - 62*x^2 - 32*x^3)*Log[2])/((x + 2*x^2 + x^3)*Log[2]),x]

[Out]

(-16*x*(1 + Log[4]))/Log[2] - (8*(1 + Log[4]))/((1 + x)*Log[2]) + (Log[4]*Log[x])/Log[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 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 1608

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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x+2 x^2+x^3} \, dx}{\log (2)}\\ &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x \left (1+2 x+x^2\right )} \, dx}{\log (2)}\\ &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x (1+x)^2} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {\log (4)}{x}-16 (1+\log (4))+\frac {8 (1+\log (4))}{(1+x)^2}\right ) \, dx}{\log (2)}\\ &=-\frac {16 x (1+\log (4))}{\log (2)}-\frac {8 (1+\log (4))}{(1+x) \log (2)}+\frac {\log (4) \log (x)}{\log (2)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 30, normalized size = 1.20 \begin {gather*} -\frac {2 \left (8 x (1+\log (4))+\frac {4+\log (256)}{1+x}-\log (2) \log (x)\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x - 32*x^2 - 16*x^3 + (2 - 12*x - 62*x^2 - 32*x^3)*Log[2])/((x + 2*x^2 + x^3)*Log[2]),x]

[Out]

(-2*(8*x*(1 + Log[4]) + (4 + Log[256])/(1 + x) - Log[2]*Log[x]))/Log[2]

________________________________________________________________________________________

Maple [A]
time = 0.40, size = 34, normalized size = 1.36

method	result	size
default	\(\frac {-32 x \ln \left (2\right )-16 x -\frac {2 \left (8 \ln \left (2\right )+4\right )}{x +1}+2 \ln \left (2\right ) \ln \left (x \right )}{\ln \left (2\right )}\)	\(34\)
risch	\(-32 x -\frac {16 x}{\ln \left (2\right )}-\frac {16}{x +1}-\frac {8}{\ln \left (2\right ) \left (x +1\right )}+2 \ln \left (-x \right )\)	\(36\)
norman	\(\frac {-\frac {8 \left (1+2 \ln \left (2\right )\right ) x}{\ln \left (2\right )}-\frac {16 \left (1+2 \ln \left (2\right )\right ) x^{2}}{\ln \left (2\right )}}{x +1}+2 \ln \left (x \right )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^3-62*x^2-12*x+2)*ln(2)-16*x^3-32*x^2-8*x)/(x^3+2*x^2+x)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/ln(2)*(-32*x*ln(2)-16*x-2*(8*ln(2)+4)/(x+1)+2*ln(2)*ln(x))

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 35, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (8 \, x {\left (2 \, \log \left (2\right ) + 1\right )} - \log \left (2\right ) \log \left (x\right ) + \frac {4 \, {\left (2 \, \log \left (2\right ) + 1\right )}}{x + 1}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-62*x^2-12*x+2)*log(2)-16*x^3-32*x^2-8*x)/(x^3+2*x^2+x)/log(2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 43, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left ({\left (x + 1\right )} \log \left (2\right ) \log \left (x\right ) - 8 \, x^{2} - 8 \, {\left (2 \, x^{2} + 2 \, x + 1\right )} \log \left (2\right ) - 8 \, x - 4\right )}}{{\left (x + 1\right )} \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-62*x^2-12*x+2)*log(2)-16*x^3-32*x^2-8*x)/(x^3+2*x^2+x)/log(2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.18, size = 27, normalized size = 1.08 \begin {gather*} - x \left (\frac {16}{\log {\left (2 \right )}} + 32\right ) + 2 \log {\left (x \right )} - \frac {8 + 16 \log {\left (2 \right )}}{x \log {\left (2 \right )} + \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**3-62*x**2-12*x+2)*ln(2)-16*x**3-32*x**2-8*x)/(x**3+2*x**2+x)/ln(2),x)

[Out]

-x*(16/log(2) + 32) + 2*log(x) - (8 + 16*log(2))/(x*log(2) + log(2))

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 35, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (16 \, x \log \left (2\right ) - \log \left (2\right ) \log \left ({\left | x \right |}\right ) + 8 \, x + \frac {4 \, {\left (2 \, \log \left (2\right ) + 1\right )}}{x + 1}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3-62*x^2-12*x+2)*log(2)-16*x^3-32*x^2-8*x)/(x^3+2*x^2+x)/log(2),x, algorithm="giac")

[Out]

-2*(16*x*log(2) - log(2)*log(abs(x)) + 8*x + 4*(2*log(2) + 1)/(x + 1))/log(2)

________________________________________________________________________________________

Mupad [B]
time = 0.11, size = 35, normalized size = 1.40 \begin {gather*} 2\,\ln \left (x\right )-\frac {16\,\ln \left (2\right )+8}{\ln \left (2\right )+x\,\ln \left (2\right )}-\frac {x\,\left (32\,\ln \left (2\right )+16\right )}{\ln \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + log(2)*(12*x + 62*x^2 + 32*x^3 - 2) + 32*x^2 + 16*x^3)/(log(2)*(x + 2*x^2 + x^3)),x)

[Out]

2*log(x) - (16*log(2) + 8)/(log(2) + x*log(2)) - (x*(32*log(2) + 16))/log(2)

________________________________________________________________________________________