3.63.69 \(\int \frac {4 x+32 x^2+8 x \log (2 x)}{1+24 x+192 x^2+512 x^3} \, dx\)

Optimal. Leaf size=16 \[ \frac {4 x^2 \log (2 x)}{(1+8 x)^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6688, 12, 6742, 43, 2335} \begin {gather*} \frac {4 x^2 \log (2 x)}{(8 x+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + 32*x^2 + 8*x*Log[2*x])/(1 + 24*x + 192*x^2 + 512*x^3),x]

[Out]

(4*x^2*Log[2*x])/(1 + 8*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rule 6688

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

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

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Mathematica [A]  time = 0.06, size = 32, normalized size = 2.00 \begin {gather*} \frac {-x \log (65536)+\log (x)+64 x^2 \log (x)-\log (2 x)}{16 (1+8 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + 32*x^2 + 8*x*Log[2*x])/(1 + 24*x + 192*x^2 + 512*x^3),x]

[Out]

(-(x*Log[65536]) + Log[x] + 64*x^2*Log[x] - Log[2*x])/(16*(1 + 8*x)^2)

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fricas [A]  time = 0.75, size = 21, normalized size = 1.31 \begin {gather*} \frac {4 \, x^{2} \log \left (2 \, x\right )}{64 \, x^{2} + 16 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)+32*x^2+4*x)/(512*x^3+192*x^2+24*x+1),x, algorithm="fricas")

[Out]

4*x^2*log(2*x)/(64*x^2 + 16*x + 1)

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giac [A]  time = 0.20, size = 28, normalized size = 1.75 \begin {gather*} -\frac {{\left (16 \, x + 1\right )} \log \left (2 \, x\right )}{16 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} + \frac {1}{16} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)+32*x^2+4*x)/(512*x^3+192*x^2+24*x+1),x, algorithm="giac")

[Out]

-1/16*(16*x + 1)*log(2*x)/(64*x^2 + 16*x + 1) + 1/16*log(x)

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maple [A]  time = 0.08, size = 17, normalized size = 1.06

method	result	size
norman	\(\frac {4 x^{2} \ln \left (2 x \right )}{\left (8 x +1\right )^{2}}\)	\(17\)
risch	\(-\frac {\left (16 x +1\right ) \ln \left (2 x \right )}{16 \left (64 x^{2}+16 x +1\right )}+\frac {\ln \relax (x )}{16}\)	\(29\)
derivativedivides	\(\frac {\ln \left (2 x \right ) x}{8 x +1}-\frac {\ln \left (2 x \right ) x \left (4 x +1\right )}{\left (8 x +1\right )^{2}}\)	\(34\)
default	\(\frac {\ln \left (2 x \right ) x}{8 x +1}-\frac {\ln \left (2 x \right ) x \left (4 x +1\right )}{\left (8 x +1\right )^{2}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*ln(2*x)+32*x^2+4*x)/(512*x^3+192*x^2+24*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.38, size = 75, normalized size = 4.69 \begin {gather*} -\frac {{\left (16 \, x + 1\right )} \log \left (2 \, x\right )}{16 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} + \frac {32 \, x + 3}{32 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} - \frac {16 \, x + 1}{32 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} - \frac {1}{16 \, {\left (8 \, x + 1\right )}} + \frac {1}{16} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)+32*x^2+4*x)/(512*x^3+192*x^2+24*x+1),x, algorithm="maxima")

[Out]

-1/16*(16*x + 1)*log(2*x)/(64*x^2 + 16*x + 1) + 1/32*(32*x + 3)/(64*x^2 + 16*x + 1) - 1/32*(16*x + 1)/(64*x^2
+ 16*x + 1) - 1/16/(8*x + 1) + 1/16*log(x)

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mupad [B]  time = 4.36, size = 16, normalized size = 1.00 \begin {gather*} \frac {4\,x^2\,\ln \left (2\,x\right )}{{\left (8\,x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 8*x*log(2*x) + 32*x^2)/(24*x + 192*x^2 + 512*x^3 + 1),x)

[Out]

(4*x^2*log(2*x))/(8*x + 1)^2

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sympy [A]  time = 0.16, size = 26, normalized size = 1.62 \begin {gather*} \frac {\left (- 16 x - 1\right ) \log {\left (2 x \right )}}{1024 x^{2} + 256 x + 16} + \frac {\log {\relax (x )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*ln(2*x)+32*x**2+4*x)/(512*x**3+192*x**2+24*x+1),x)

[Out]

(-16*x - 1)*log(2*x)/(1024*x**2 + 256*x + 16) + log(x)/16

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