3.90.28 \(\int \frac {2016+1260 x+252 x^2-320 x^3+(504+189 x+80 x^3) \log (4)}{256 x^3+128 x^4+16 x^5} \, dx\)

Optimal. Leaf size=22 \[ \frac {\left (-5-\frac {63}{16 x^2}\right ) (4+2 x+\log (4))}{4+x} \]

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Rubi [A]  time = 0.10, antiderivative size = 40, normalized size of antiderivative = 1.82, number of steps used = 5, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {1594, 27, 12, 1620} \begin {gather*} -\frac {63 (4+\log (4))}{64 x^2}+\frac {1343 (4-\log (4))}{256 (x+4)}-\frac {63 (4-\log (4))}{256 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2016 + 1260*x + 252*x^2 - 320*x^3 + (504 + 189*x + 80*x^3)*Log[4])/(256*x^3 + 128*x^4 + 16*x^5),x]

[Out]

(-63*(4 - Log[4]))/(256*x) + (1343*(4 - Log[4]))/(256*(4 + x)) - (63*(4 + Log[4]))/(64*x^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 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 1620

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} &=\int \frac {2016+1260 x+252 x^2-320 x^3+\left (504+189 x+80 x^3\right ) \log (4)}{x^3 \left (256+128 x+16 x^2\right )} \, dx\\ &=\int \frac {2016+1260 x+252 x^2-320 x^3+\left (504+189 x+80 x^3\right ) \log (4)}{16 x^3 (4+x)^2} \, dx\\ &=\frac {1}{16} \int \frac {2016+1260 x+252 x^2-320 x^3+\left (504+189 x+80 x^3\right ) \log (4)}{x^3 (4+x)^2} \, dx\\ &=\frac {1}{16} \int \left (-\frac {63 (-4+\log (4))}{16 x^2}+\frac {1343 (-4+\log (4))}{16 (4+x)^2}+\frac {63 (4+\log (4))}{2 x^3}\right ) \, dx\\ &=-\frac {63 (4-\log (4))}{256 x}+\frac {1343 (4-\log (4))}{256 (4+x)}-\frac {63 (4+\log (4))}{64 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 1.64 \begin {gather*} \frac {-1008 x-504 (4+\log (4))+x^2 (2560-451 \log (4)-63 \log (64))}{128 x^2 (4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2016 + 1260*x + 252*x^2 - 320*x^3 + (504 + 189*x + 80*x^3)*Log[4])/(256*x^3 + 128*x^4 + 16*x^5),x]

[Out]

(-1008*x - 504*(4 + Log[4]) + x^2*(2560 - 451*Log[4] - 63*Log[64]))/(128*x^2*(4 + x))

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fricas [A]  time = 0.46, size = 34, normalized size = 1.55 \begin {gather*} \frac {160 \, x^{2} - {\left (80 \, x^{2} + 63\right )} \log \relax (2) - 63 \, x - 126}{8 \, {\left (x^{3} + 4 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(80*x^3+189*x+504)*log(2)-320*x^3+252*x^2+1260*x+2016)/(16*x^5+128*x^4+256*x^3),x, algorithm="fri
cas")

[Out]

1/8*(160*x^2 - (80*x^2 + 63)*log(2) - 63*x - 126)/(x^3 + 4*x^2)

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giac [A]  time = 0.14, size = 30, normalized size = 1.36 \begin {gather*} -\frac {1343 \, {\left (\log \relax (2) - 2\right )}}{128 \, {\left (x + 4\right )}} + \frac {63 \, {\left (x \log \relax (2) - 2 \, x - 4 \, \log \relax (2) - 8\right )}}{128 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(80*x^3+189*x+504)*log(2)-320*x^3+252*x^2+1260*x+2016)/(16*x^5+128*x^4+256*x^3),x, algorithm="gia
c")

[Out]

-1343/128*(log(2) - 2)/(x + 4) + 63/128*(x*log(2) - 2*x - 4*log(2) - 8)/x^2

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maple [A]  time = 0.06, size = 29, normalized size = 1.32

method	result	size
norman	\(\frac {\left (20-10 \ln \relax (2)\right ) x^{2}-\frac {63 x}{8}-\frac {63 \ln \relax (2)}{8}-\frac {63}{4}}{x^{2} \left (4+x \right )}\)	\(29\)
risch	\(\frac {\left (20-10 \ln \relax (2)\right ) x^{2}-\frac {63 x}{8}-\frac {63 \ln \relax (2)}{8}-\frac {63}{4}}{x^{2} \left (4+x \right )}\)	\(29\)
gosper	\(-\frac {80 x^{2} \ln \relax (2)-160 x^{2}+63 \ln \relax (2)+63 x +126}{8 x^{2} \left (4+x \right )}\)	\(32\)
default	\(-\frac {-\frac {63 \ln \relax (2)}{16}+\frac {63}{8}}{8 x}-\frac {\frac {63 \ln \relax (2)}{2}+63}{16 x^{2}}-\frac {\frac {1343 \ln \relax (2)}{16}-\frac {1343}{8}}{8 \left (4+x \right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(80*x^3+189*x+504)*ln(2)-320*x^3+252*x^2+1260*x+2016)/(16*x^5+128*x^4+256*x^3),x,method=_RETURNVERBOSE)

[Out]

((20-10*ln(2))*x^2-63/8*x-63/8*ln(2)-63/4)/x^2/(4+x)

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maxima [A]  time = 0.36, size = 31, normalized size = 1.41 \begin {gather*} -\frac {80 \, x^{2} {\left (\log \relax (2) - 2\right )} + 63 \, x + 63 \, \log \relax (2) + 126}{8 \, {\left (x^{3} + 4 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(80*x^3+189*x+504)*log(2)-320*x^3+252*x^2+1260*x+2016)/(16*x^5+128*x^4+256*x^3),x, algorithm="max
ima")

[Out]

-1/8*(80*x^2*(log(2) - 2) + 63*x + 63*log(2) + 126)/(x^3 + 4*x^2)

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mupad [B]  time = 0.14, size = 34, normalized size = 1.55 \begin {gather*} -\frac {\left (80\,\ln \relax (2)-160\right )\,x^2+63\,x+63\,\ln \relax (2)+126}{8\,x^3+32\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1260*x + 2*log(2)*(189*x + 80*x^3 + 504) + 252*x^2 - 320*x^3 + 2016)/(256*x^3 + 128*x^4 + 16*x^5),x)

[Out]

-(63*x + 63*log(2) + x^2*(80*log(2) - 160) + 126)/(32*x^2 + 8*x^3)

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sympy [A]  time = 0.62, size = 31, normalized size = 1.41 \begin {gather*} - \frac {x^{2} \left (-160 + 80 \log {\relax (2 )}\right ) + 63 x + 63 \log {\relax (2 )} + 126}{8 x^{3} + 32 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(80*x**3+189*x+504)*ln(2)-320*x**3+252*x**2+1260*x+2016)/(16*x**5+128*x**4+256*x**3),x)

[Out]

-(x**2*(-160 + 80*log(2)) + 63*x + 63*log(2) + 126)/(8*x**3 + 32*x**2)

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