3.47.59 \(\int \frac {192 x^2+36 x^3-3 x^4+(-576 x^2-48 x^3+9 x^4) \log (x)+(16-1528 x-47 x^2+24 x^3) \log ^2(x)}{(16+8 x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ x+\frac {3 (-16+x) x^2 \left (4+\frac {x}{\log (x)}\right )}{4+x} \]

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Rubi [F]  time = 0.37, 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 {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(192*x^2 + 36*x^3 - 3*x^4 + (-576*x^2 - 48*x^3 + 9*x^4)*Log[x] + (16 - 1528*x - 47*x^2 + 24*x^3)*Log[x]^2)
/((16 + 8*x + x^2)*Log[x]^2),x]

[Out]

-239*x + 12*x^2 - 3840/(4 + x) - 3*Defer[Int][((-16 + x)*x^2)/((4 + x)*Log[x]^2), x] + 3*Defer[Int][(x^2*(-192
- 16*x + 3*x^2))/((4 + x)^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{(4+x)^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {16-1528 x-47 x^2+24 x^3}{(4+x)^2}-\frac {3 (-16+x) x^2}{(4+x) \log ^2(x)}+\frac {3 x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)}\right ) \, dx\\ &=-\left (3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx\right )+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx+\int \frac {16-1528 x-47 x^2+24 x^3}{(4+x)^2} \, dx\\ &=-\left (3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx\right )+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx+\int \left (-239+24 x+\frac {3840}{(4+x)^2}\right ) \, dx\\ &=-239 x+12 x^2-\frac {3840}{4+x}-3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 33, normalized size = 1.43 \begin {gather*} -239 x+12 x^2-\frac {3840}{4+x}+\frac {3 (-16+x) x^3}{(4+x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(192*x^2 + 36*x^3 - 3*x^4 + (-576*x^2 - 48*x^3 + 9*x^4)*Log[x] + (16 - 1528*x - 47*x^2 + 24*x^3)*Log
[x]^2)/((16 + 8*x + x^2)*Log[x]^2),x]

[Out]

-239*x + 12*x^2 - 3840/(4 + x) + (3*(-16 + x)*x^3)/((4 + x)*Log[x])

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fricas [A]  time = 0.62, size = 39, normalized size = 1.70 \begin {gather*} \frac {3 \, x^{4} - 48 \, x^{3} + {\left (12 \, x^{3} - 191 \, x^{2} - 956 \, x - 3840\right )} \log \relax (x)}{{\left (x + 4\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="fricas")

[Out]

(3*x^4 - 48*x^3 + (12*x^3 - 191*x^2 - 956*x - 3840)*log(x))/((x + 4)*log(x))

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giac [A]  time = 0.12, size = 38, normalized size = 1.65 \begin {gather*} 12 \, x^{2} - 239 \, x + \frac {3 \, {\left (x^{4} - 16 \, x^{3}\right )}}{x \log \relax (x) + 4 \, \log \relax (x)} - \frac {3840}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="giac")

[Out]

12*x^2 - 239*x + 3*(x^4 - 16*x^3)/(x*log(x) + 4*log(x)) - 3840/(x + 4)

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maple [A]  time = 0.08, size = 40, normalized size = 1.74

method	result	size
norman	\(\frac {-16 \ln \relax (x )-48 x^{3}+3 x^{4}-191 x^{2} \ln \relax (x )+12 x^{3} \ln \relax (x )}{\left (4+x \right ) \ln \relax (x )}\)	\(40\)
risch	\(\frac {12 x^{3}-191 x^{2}-956 x -3840}{4+x}+\frac {3 x^{3} \left (x -16\right )}{\left (4+x \right ) \ln \relax (x )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*x^3-47*x^2-1528*x+16)*ln(x)^2+(9*x^4-48*x^3-576*x^2)*ln(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

(-16*ln(x)-48*x^3+3*x^4-191*x^2*ln(x)+12*x^3*ln(x))/(4+x)/ln(x)

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maxima [A]  time = 0.41, size = 39, normalized size = 1.70 \begin {gather*} \frac {3 \, x^{4} - 48 \, x^{3} + {\left (12 \, x^{3} - 191 \, x^{2} - 956 \, x - 3840\right )} \log \relax (x)}{{\left (x + 4\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="maxima")

[Out]

(3*x^4 - 48*x^3 + (12*x^3 - 191*x^2 - 956*x - 3840)*log(x))/((x + 4)*log(x))

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mupad [B]  time = 3.62, size = 41, normalized size = 1.78 \begin {gather*} \frac {x\,\left (12\,x^2-191\,x+4\right )}{x+4}-\frac {x\,\left (48\,x^2-3\,x^3\right )}{\ln \relax (x)\,\left (x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(576*x^2 + 48*x^3 - 9*x^4) + log(x)^2*(1528*x + 47*x^2 - 24*x^3 - 16) - 192*x^2 - 36*x^3 + 3*x^4)
/(log(x)^2*(8*x + x^2 + 16)),x)

[Out]

(x*(12*x^2 - 191*x + 4))/(x + 4) - (x*(48*x^2 - 3*x^3))/(log(x)*(x + 4))

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sympy [A]  time = 0.15, size = 29, normalized size = 1.26 \begin {gather*} 12 x^{2} - 239 x + \frac {3 x^{4} - 48 x^{3}}{\left (x + 4\right ) \log {\relax (x )}} - \frac {3840}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x**3-47*x**2-1528*x+16)*ln(x)**2+(9*x**4-48*x**3-576*x**2)*ln(x)-3*x**4+36*x**3+192*x**2)/(x**2
+8*x+16)/ln(x)**2,x)

[Out]

12*x**2 - 239*x + (3*x**4 - 48*x**3)/((x + 4)*log(x)) - 3840/(x + 4)

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