3.77.23 \(\int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+(240+120 x+15 x^2) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx\)

Optimal. Leaf size=12 \[ -\frac {x}{(4+x+5 \log (x))^2} \]

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Rubi [F]  time = 0.19, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(6 + x - 5*Log[x])/(64 + 48*x + 12*x^2 + x^3 + (240 + 120*x + 15*x^2)*Log[x] + (300 + 75*x)*Log[x]^2 + 125
*Log[x]^3),x]

[Out]

10*Defer[Int][(4 + x + 5*Log[x])^(-3), x] + 2*Defer[Int][x/(4 + x + 5*Log[x])^3, x] - Defer[Int][(4 + x + 5*Lo
g[x])^(-2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+x-5 \log (x)}{(4+x+5 \log (x))^3} \, dx\\ &=\int \left (\frac {2 (5+x)}{(4+x+5 \log (x))^3}-\frac {1}{(4+x+5 \log (x))^2}\right ) \, dx\\ &=2 \int \frac {5+x}{(4+x+5 \log (x))^3} \, dx-\int \frac {1}{(4+x+5 \log (x))^2} \, dx\\ &=2 \int \left (\frac {5}{(4+x+5 \log (x))^3}+\frac {x}{(4+x+5 \log (x))^3}\right ) \, dx-\int \frac {1}{(4+x+5 \log (x))^2} \, dx\\ &=2 \int \frac {x}{(4+x+5 \log (x))^3} \, dx+10 \int \frac {1}{(4+x+5 \log (x))^3} \, dx-\int \frac {1}{(4+x+5 \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 12, normalized size = 1.00 \begin {gather*} -\frac {x}{(4+x+5 \log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + x - 5*Log[x])/(64 + 48*x + 12*x^2 + x^3 + (240 + 120*x + 15*x^2)*Log[x] + (300 + 75*x)*Log[x]^2
+ 125*Log[x]^3),x]

[Out]

-(x/(4 + x + 5*Log[x])^2)

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fricas [B]  time = 1.99, size = 26, normalized size = 2.17 \begin {gather*} -\frac {x}{x^{2} + 10 \, {\left (x + 4\right )} \log \relax (x) + 25 \, \log \relax (x)^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="fricas")

[Out]

-x/(x^2 + 10*(x + 4)*log(x) + 25*log(x)^2 + 8*x + 16)

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giac [B]  time = 0.19, size = 53, normalized size = 4.42 \begin {gather*} -\frac {x^{2} + 5 \, x}{x^{3} + 10 \, x^{2} \log \relax (x) + 25 \, x \log \relax (x)^{2} + 13 \, x^{2} + 90 \, x \log \relax (x) + 125 \, \log \relax (x)^{2} + 56 \, x + 200 \, \log \relax (x) + 80} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="giac")

[Out]

-(x^2 + 5*x)/(x^3 + 10*x^2*log(x) + 25*x*log(x)^2 + 13*x^2 + 90*x*log(x) + 125*log(x)^2 + 56*x + 200*log(x) +
80)

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maple [A]  time = 0.04, size = 13, normalized size = 1.08

method	result	size
norman	\(-\frac {x}{\left (5 \ln \relax (x )+4+x \right )^{2}}\)	\(13\)
risch	\(-\frac {x}{\left (5 \ln \relax (x )+4+x \right )^{2}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(x)+x+6)/(125*ln(x)^3+(75*x+300)*ln(x)^2+(15*x^2+120*x+240)*ln(x)+x^3+12*x^2+48*x+64),x,method=_RETU
RNVERBOSE)

[Out]

-x/(5*ln(x)+4+x)^2

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maxima [B]  time = 0.39, size = 26, normalized size = 2.17 \begin {gather*} -\frac {x}{x^{2} + 10 \, {\left (x + 4\right )} \log \relax (x) + 25 \, \log \relax (x)^{2} + 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="maxima")

[Out]

-x/(x^2 + 10*(x + 4)*log(x) + 25*log(x)^2 + 8*x + 16)

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mupad [B]  time = 5.01, size = 12, normalized size = 1.00 \begin {gather*} -\frac {x}{{\left (x+5\,\ln \relax (x)+4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 5*log(x) + 6)/(48*x + 125*log(x)^3 + log(x)*(120*x + 15*x^2 + 240) + 12*x^2 + x^3 + log(x)^2*(75*x +
300) + 64),x)

[Out]

-x/(x + 5*log(x) + 4)^2

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sympy [B]  time = 0.18, size = 26, normalized size = 2.17 \begin {gather*} - \frac {x}{x^{2} + 8 x + \left (10 x + 40\right ) \log {\relax (x )} + 25 \log {\relax (x )}^{2} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(x)+x+6)/(125*ln(x)**3+(75*x+300)*ln(x)**2+(15*x**2+120*x+240)*ln(x)+x**3+12*x**2+48*x+64),x)

[Out]

-x/(x**2 + 8*x + (10*x + 40)*log(x) + 25*log(x)**2 + 16)

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