3.54.23 \(\int \frac {-40-392 x+320 x \log (2 x)}{x+16 x^2+64 x^3} \, dx\)

Optimal. Leaf size=19 \[ \frac {1-x-5 \log (2 x)}{\frac {1}{8}+x} \]

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Rubi [A]  time = 0.26, antiderivative size = 28, normalized size of antiderivative = 1.47, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1594, 27, 6741, 12, 6742, 77, 2314, 31} \begin {gather*} \frac {9}{8 x+1}-40 \log (x)+\frac {320 x \log (2 x)}{8 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 - 392*x + 320*x*Log[2*x])/(x + 16*x^2 + 64*x^3),x]

[Out]

9/(1 + 8*x) - 40*Log[x] + (320*x*Log[2*x])/(1 + 8*x)

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 31

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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 6741

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

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 {-40-392 x+320 x \log (2 x)}{x \left (1+16 x+64 x^2\right )} \, dx\\ &=\int \frac {-40-392 x+320 x \log (2 x)}{x (1+8 x)^2} \, dx\\ &=\int \frac {8 (-5-49 x+40 x \log (2 x))}{x (1+8 x)^2} \, dx\\ &=8 \int \frac {-5-49 x+40 x \log (2 x)}{x (1+8 x)^2} \, dx\\ &=8 \int \left (\frac {-5-49 x}{x (1+8 x)^2}+\frac {40 \log (2 x)}{(1+8 x)^2}\right ) \, dx\\ &=8 \int \frac {-5-49 x}{x (1+8 x)^2} \, dx+320 \int \frac {\log (2 x)}{(1+8 x)^2} \, dx\\ &=\frac {320 x \log (2 x)}{1+8 x}+8 \int \left (-\frac {5}{x}-\frac {9}{(1+8 x)^2}+\frac {40}{1+8 x}\right ) \, dx-320 \int \frac {1}{1+8 x} \, dx\\ &=\frac {9}{1+8 x}-40 \log (x)+\frac {320 x \log (2 x)}{1+8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.89 \begin {gather*} \frac {8 (9-40 \log (2 x))}{8+64 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 - 392*x + 320*x*Log[2*x])/(x + 16*x^2 + 64*x^3),x]

[Out]

(8*(9 - 40*Log[2*x]))/(8 + 64*x)

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fricas [A]  time = 0.61, size = 17, normalized size = 0.89 \begin {gather*} -\frac {40 \, \log \left (2 \, x\right ) - 9}{8 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((320*x*log(2*x)-392*x-40)/(64*x^3+16*x^2+x),x, algorithm="fricas")

[Out]

-(40*log(2*x) - 9)/(8*x + 1)

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giac [A]  time = 0.16, size = 23, normalized size = 1.21 \begin {gather*} -\frac {40 \, \log \left (2 \, x\right )}{8 \, x + 1} + \frac {9}{8 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((320*x*log(2*x)-392*x-40)/(64*x^3+16*x^2+x),x, algorithm="giac")

[Out]

-40*log(2*x)/(8*x + 1) + 9/(8*x + 1)

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maple [A]  time = 0.06, size = 19, normalized size = 1.00

method	result	size
norman	\(\frac {-72 x -40 \ln \left (2 x \right )}{8 x +1}\)	\(19\)
risch	\(-\frac {40 \ln \left (2 x \right )}{8 x +1}+\frac {9}{8 x +1}\)	\(24\)
derivativedivides	\(\frac {320 \ln \left (2 x \right ) x}{8 x +1}-40 \ln \left (2 x \right )+\frac {9}{8 x +1}\)	\(31\)
default	\(\frac {320 \ln \left (2 x \right ) x}{8 x +1}-40 \ln \left (2 x \right )+\frac {9}{8 x +1}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((320*x*ln(2*x)-392*x-40)/(64*x^3+16*x^2+x),x,method=_RETURNVERBOSE)

[Out]

(-72*x-40*ln(2*x))/(8*x+1)

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maxima [A]  time = 0.38, size = 23, normalized size = 1.21 \begin {gather*} -\frac {40 \, \log \left (2 \, x\right )}{8 \, x + 1} + \frac {9}{8 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((320*x*log(2*x)-392*x-40)/(64*x^3+16*x^2+x),x, algorithm="maxima")

[Out]

-40*log(2*x)/(8*x + 1) + 9/(8*x + 1)

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mupad [B]  time = 3.56, size = 17, normalized size = 0.89 \begin {gather*} -\frac {40\,\ln \left (2\,x\right )-9}{8\,x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(392*x - 320*x*log(2*x) + 40)/(x + 16*x^2 + 64*x^3),x)

[Out]

-(40*log(2*x) - 9)/(8*x + 1)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.89 \begin {gather*} \frac {72}{64 x + 8} - \frac {40 \log {\left (2 x \right )}}{8 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((320*x*ln(2*x)-392*x-40)/(64*x**3+16*x**2+x),x)

[Out]

72/(64*x + 8) - 40*log(2*x)/(8*x + 1)

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