3.1.63 \(\int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{(25 x^2+20 x^3+4 x^4) \log ^2(x)+(10 x^2+4 x^3) \log ^3(x)+x^2 \log ^4(x)} \, dx\) [63]

Optimal. Leaf size=21 \[ 28+\frac {\log (4)}{x \log (x) (5+2 x+\log (x))} \]

[Out]

28+2*ln(2)/x/ln(x)/(2*x+5+ln(x))

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Rubi [F]
time = 0.89, 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 {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-5 - 2*x)*Log[4] + (-7 - 4*x)*Log[4]*Log[x] - Log[4]*Log[x]^2)/((25*x^2 + 20*x^3 + 4*x^4)*Log[x]^2 + (10
*x^2 + 4*x^3)*Log[x]^3 + x^2*Log[x]^4),x]

[Out]

-(Log[4]*Defer[Int][1/(x^2*(5 + 2*x)*Log[x]^2), x]) + Log[4]*Defer[Int][(-5 - 4*x)/(x^2*(5 + 2*x)^2*Log[x]), x
] + (Log[4]*Defer[Int][1/(x^2*(5 + 2*x + Log[x])^2), x])/5 + (8*Log[4]*Defer[Int][1/(x*(5 + 2*x + Log[x])^2),
x])/25 - (16*Log[4]*Defer[Int][1/((5 + 2*x)*(5 + 2*x + Log[x])^2), x])/25 + (Log[4]*Defer[Int][1/(x^2*(5 + 2*x
+ Log[x])), x])/5 - (4*Log[4]*Defer[Int][1/((5 + 2*x)^2*(5 + 2*x + Log[x])), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log (4) \left (-5-2 x-(7+4 x) \log (x)-\log ^2(x)\right )}{x^2 \log ^2(x) (5+2 x+\log (x))^2} \, dx\\ &=\log (4) \int \frac {-5-2 x-(7+4 x) \log (x)-\log ^2(x)}{x^2 \log ^2(x) (5+2 x+\log (x))^2} \, dx\\ &=\log (4) \int \left (-\frac {1}{x^2 (5+2 x) \log ^2(x)}+\frac {-5-4 x}{x^2 (5+2 x)^2 \log (x)}+\frac {1+2 x}{x^2 (5+2 x) (5+2 x+\log (x))^2}+\frac {5+4 x}{x^2 (5+2 x)^2 (5+2 x+\log (x))}\right ) \, dx\\ &=-\left (\log (4) \int \frac {1}{x^2 (5+2 x) \log ^2(x)} \, dx\right )+\log (4) \int \frac {-5-4 x}{x^2 (5+2 x)^2 \log (x)} \, dx+\log (4) \int \frac {1+2 x}{x^2 (5+2 x) (5+2 x+\log (x))^2} \, dx+\log (4) \int \frac {5+4 x}{x^2 (5+2 x)^2 (5+2 x+\log (x))} \, dx\\ &=-\left (\log (4) \int \frac {1}{x^2 (5+2 x) \log ^2(x)} \, dx\right )+\log (4) \int \frac {-5-4 x}{x^2 (5+2 x)^2 \log (x)} \, dx+\log (4) \int \left (\frac {1}{5 x^2 (5+2 x+\log (x))^2}+\frac {8}{25 x (5+2 x+\log (x))^2}-\frac {16}{25 (5+2 x) (5+2 x+\log (x))^2}\right ) \, dx+\log (4) \int \left (\frac {1}{5 x^2 (5+2 x+\log (x))}-\frac {4}{5 (5+2 x)^2 (5+2 x+\log (x))}\right ) \, dx\\ &=\frac {1}{5} \log (4) \int \frac {1}{x^2 (5+2 x+\log (x))^2} \, dx+\frac {1}{5} \log (4) \int \frac {1}{x^2 (5+2 x+\log (x))} \, dx+\frac {1}{25} (8 \log (4)) \int \frac {1}{x (5+2 x+\log (x))^2} \, dx-\frac {1}{25} (16 \log (4)) \int \frac {1}{(5+2 x) (5+2 x+\log (x))^2} \, dx-\frac {1}{5} (4 \log (4)) \int \frac {1}{(5+2 x)^2 (5+2 x+\log (x))} \, dx-\log (4) \int \frac {1}{x^2 (5+2 x) \log ^2(x)} \, dx+\log (4) \int \frac {-5-4 x}{x^2 (5+2 x)^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 19, normalized size = 0.90 \begin {gather*} \frac {\log (4)}{x \log (x) (5+2 x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-5 - 2*x)*Log[4] + (-7 - 4*x)*Log[4]*Log[x] - Log[4]*Log[x]^2)/((25*x^2 + 20*x^3 + 4*x^4)*Log[x]^2
+ (10*x^2 + 4*x^3)*Log[x]^3 + x^2*Log[x]^4),x]

[Out]

Log[4]/(x*Log[x]*(5 + 2*x + Log[x]))

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Maple [A]
time = 0.19, size = 21, normalized size = 1.00

method	result	size
default	\(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\)	\(21\)
norman	\(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\)	\(21\)
risch	\(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(2)*ln(x)^2+2*(-4*x-7)*ln(2)*ln(x)+2*(-2*x-5)*ln(2))/(x^2*ln(x)^4+(4*x^3+10*x^2)*ln(x)^3+(4*x^4+20*x
^3+25*x^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

2*ln(2)/x/ln(x)/(2*x+5+ln(x))

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Maxima [A]
time = 0.51, size = 25, normalized size = 1.19 \begin {gather*} \frac {2 \, \log \left (2\right )}{x \log \left (x\right )^{2} + {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/(x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^
3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, algorithm="maxima")

[Out]

2*log(2)/(x*log(x)^2 + (2*x^2 + 5*x)*log(x))

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Fricas [A]
time = 0.32, size = 25, normalized size = 1.19 \begin {gather*} \frac {2 \, \log \left (2\right )}{x \log \left (x\right )^{2} + {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/(x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^
3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, algorithm="fricas")

[Out]

2*log(2)/(x*log(x)^2 + (2*x^2 + 5*x)*log(x))

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Sympy [A]
time = 0.06, size = 22, normalized size = 1.05 \begin {gather*} \frac {2 \log {\left (2 \right )}}{x \log {\left (x \right )}^{2} + \left (2 x^{2} + 5 x\right ) \log {\left (x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(2)*ln(x)**2+2*(-4*x-7)*ln(2)*ln(x)+2*(-2*x-5)*ln(2))/(x**2*ln(x)**4+(4*x**3+10*x**2)*ln(x)**3
+(4*x**4+20*x**3+25*x**2)*ln(x)**2),x)

[Out]

2*log(2)/(x*log(x)**2 + (2*x**2 + 5*x)*log(x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
time = 0.39, size = 52, normalized size = 2.48 \begin {gather*} -\frac {2 \, \log \left (2\right )}{4 \, x^{3} + 2 \, x^{2} \log \left (x\right ) + 20 \, x^{2} + 5 \, x \log \left (x\right ) + 25 \, x} + \frac {2 \, \log \left (2\right )}{2 \, x^{2} \log \left (x\right ) + 5 \, x \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/(x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^
3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, algorithm="giac")

[Out]

-2*log(2)/(4*x^3 + 2*x^2*log(x) + 20*x^2 + 5*x*log(x) + 25*x) + 2*log(2)/(2*x^2*log(x) + 5*x*log(x))

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Mupad [B]
time = 0.34, size = 37, normalized size = 1.76 \begin {gather*} \frac {2\,\left (5\,\ln \left (2\right )+2\,x\,\ln \left (2\right )\right )}{x\,\left ({\ln \left (x\right )}^2+\left (2\,x+5\right )\,\ln \left (x\right )\right )\,\left (2\,x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(2*x + 5) + 2*log(2)*log(x)^2 + 2*log(2)*log(x)*(4*x + 7))/(log(x)^3*(10*x^2 + 4*x^3) + x^2*log
(x)^4 + log(x)^2*(25*x^2 + 20*x^3 + 4*x^4)),x)

[Out]

(2*(5*log(2) + 2*x*log(2)))/(x*(log(x)^2 + log(x)*(2*x + 5))*(2*x + 5))

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