3.53.50 \(\int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log (e^{10} x)}{1+2 x+x^2+(-2 x^2-2 x^3) \log ^2(4 x)+x^4 \log ^4(4 x)+(8+8 x-8 x^2 \log ^2(4 x)) \log (e^{10} x)+16 \log ^2(e^{10} x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3 x}{1+x-x^2 \log ^2(4 x)+4 \log \left (e^{10} x\right )} \]

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Rubi [F]  time = 0.85, 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 {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-9 + 6*x^2*Log[4*x] + 3*x^2*Log[4*x]^2 + 12*Log[E^10*x])/(1 + 2*x + x^2 + (-2*x^2 - 2*x^3)*Log[4*x]^2 + x
^4*Log[4*x]^4 + (8 + 8*x - 8*x^2*Log[4*x]^2)*Log[E^10*x] + 16*Log[E^10*x]^2),x]

[Out]

24*Defer[Int][Log[x]/(41 + x + 4*Log[x] - x^2*Log[4*x]^2)^2, x] + 234*Defer[Int][(-41 - x - 4*Log[x] + x^2*Log
[4*x]^2)^(-2), x] + 3*Defer[Int][x/(-41 - x - 4*Log[x] + x^2*Log[4*x]^2)^2, x] + 6*Defer[Int][(x^2*Log[4*x])/(
-41 - x - 4*Log[x] + x^2*Log[4*x]^2)^2, x] + 3*Defer[Int][(-41 - x - 4*Log[x] + x^2*Log[4*x]^2)^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (37+4 \log (x)+2 x^2 \log (4 x)+x^2 \log ^2(4 x)\right )}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx\\ &=3 \int \frac {37+4 \log (x)+2 x^2 \log (4 x)+x^2 \log ^2(4 x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx\\ &=3 \int \left (\frac {78+x+8 \log (x)+2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)}\right ) \, dx\\ &=3 \int \frac {78+x+8 \log (x)+2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx\\ &=3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx+3 \int \left (\frac {8 \log (x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2}+\frac {78}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {x}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}+\frac {2 x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2}\right ) \, dx\\ &=3 \int \frac {x}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+3 \int \frac {1}{-41-x-4 \log (x)+x^2 \log ^2(4 x)} \, dx+6 \int \frac {x^2 \log (4 x)}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx+24 \int \frac {\log (x)}{\left (41+x+4 \log (x)-x^2 \log ^2(4 x)\right )^2} \, dx+234 \int \frac {1}{\left (-41-x-4 \log (x)+x^2 \log ^2(4 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 23, normalized size = 0.85 \begin {gather*} \frac {3 x}{41+x+4 \log (x)-x^2 \log ^2(4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + 6*x^2*Log[4*x] + 3*x^2*Log[4*x]^2 + 12*Log[E^10*x])/(1 + 2*x + x^2 + (-2*x^2 - 2*x^3)*Log[4*x]
^2 + x^4*Log[4*x]^4 + (8 + 8*x - 8*x^2*Log[4*x]^2)*Log[E^10*x] + 16*Log[E^10*x]^2),x]

[Out]

(3*x)/(41 + x + 4*Log[x] - x^2*Log[4*x]^2)

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fricas [A]  time = 0.45, size = 30, normalized size = 1.11 \begin {gather*} -\frac {3 \, x}{x^{2} \log \left (4 \, x\right )^{2} - x + 8 \, \log \relax (2) - 4 \, \log \left (4 \, x\right ) - 41} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x*exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+
8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2*x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="fricas")

[Out]

-3*x/(x^2*log(4*x)^2 - x + 8*log(2) - 4*log(4*x) - 41)

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giac [A]  time = 0.19, size = 40, normalized size = 1.48 \begin {gather*} -\frac {3 \, x}{4 \, x^{2} \log \relax (2)^{2} + 4 \, x^{2} \log \relax (2) \log \relax (x) + x^{2} \log \relax (x)^{2} - x - 4 \, \log \relax (x) - 41} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x*exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+
8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2*x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="giac")

[Out]

-3*x/(4*x^2*log(2)^2 + 4*x^2*log(2)*log(x) + x^2*log(x)^2 - x - 4*log(x) - 41)

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maple [A]  time = 1.00, size = 42, normalized size = 1.56

method	result	size
risch	\(\frac {12 x}{164+4 x -16 x^{2} \ln \relax (2) \ln \relax (x )-4 x^{2} \ln \relax (x )^{2}+16 \ln \relax (x )-16 x^{2} \ln \relax (2)^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*ln(x*exp(5)^2)+3*x^2*ln(4*x)^2+6*x^2*ln(4*x)-9)/(16*ln(x*exp(5)^2)^2+(-8*x^2*ln(4*x)^2+8*x+8)*ln(x*exp
(5)^2)+x^4*ln(4*x)^4+(-2*x^3-2*x^2)*ln(4*x)^2+x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

12*x/(164+4*x-16*x^2*ln(2)*ln(x)-4*x^2*ln(x)^2+16*ln(x)-16*x^2*ln(2)^2)

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maxima [A]  time = 0.50, size = 39, normalized size = 1.44 \begin {gather*} -\frac {3 \, x}{4 \, x^{2} \log \relax (2)^{2} + x^{2} \log \relax (x)^{2} + 4 \, {\left (x^{2} \log \relax (2) - 1\right )} \log \relax (x) - x - 41} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x*exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+
8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2*x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="maxima")

[Out]

-3*x/(4*x^2*log(2)^2 + x^2*log(x)^2 + 4*(x^2*log(2) - 1)*log(x) - x - 41)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {12\,\ln \left (x\,{\mathrm {e}}^{10}\right )+6\,x^2\,\ln \left (4\,x\right )+3\,x^2\,{\ln \left (4\,x\right )}^2-9}{2\,x+16\,{\ln \left (x\,{\mathrm {e}}^{10}\right )}^2+\ln \left (x\,{\mathrm {e}}^{10}\right )\,\left (-8\,x^2\,{\ln \left (4\,x\right )}^2+8\,x+8\right )-{\ln \left (4\,x\right )}^2\,\left (2\,x^3+2\,x^2\right )+x^2+x^4\,{\ln \left (4\,x\right )}^4+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*log(x*exp(10)) + 6*x^2*log(4*x) + 3*x^2*log(4*x)^2 - 9)/(2*x + 16*log(x*exp(10))^2 + log(x*exp(10))*(8
*x - 8*x^2*log(4*x)^2 + 8) - log(4*x)^2*(2*x^2 + 2*x^3) + x^2 + x^4*log(4*x)^4 + 1),x)

[Out]

int((12*log(x*exp(10)) + 6*x^2*log(4*x) + 3*x^2*log(4*x)^2 - 9)/(2*x + 16*log(x*exp(10))^2 + log(x*exp(10))*(8
*x - 8*x^2*log(4*x)^2 + 8) - log(4*x)^2*(2*x^2 + 2*x^3) + x^2 + x^4*log(4*x)^4 + 1), x)

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sympy [A]  time = 0.38, size = 29, normalized size = 1.07 \begin {gather*} - \frac {3 x}{x^{2} \log {\left (4 x \right )}^{2} - x - 4 \log {\left (4 x \right )} - 41 + 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*ln(x*exp(5)**2)+3*x**2*ln(4*x)**2+6*x**2*ln(4*x)-9)/(16*ln(x*exp(5)**2)**2+(-8*x**2*ln(4*x)**2+8
*x+8)*ln(x*exp(5)**2)+x**4*ln(4*x)**4+(-2*x**3-2*x**2)*ln(4*x)**2+x**2+2*x+1),x)

[Out]

-3*x/(x**2*log(4*x)**2 - x - 4*log(4*x) - 41 + 8*log(2))

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