3.63.6 \(\int \frac {-16 x^2+(36-212 x-8 x^2) \log ^2(5)+(-616-53 x-x^2) \log ^4(5)}{-48 x^2+16 x^3+(-600 x+176 x^2+8 x^3) \log ^2(5)+(-1875+475 x+47 x^2+x^3) \log ^4(5)} \, dx\)

Optimal. Leaf size=23 \[ \frac {3}{25+x+\frac {4 x}{\log ^2(5)}}-\log (3-x) \]

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Rubi [A]  time = 0.09, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 1, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2074} \begin {gather*} \frac {3 \log ^2(5)}{x \left (4+\log ^2(5)\right )+25 \log ^2(5)}-\log (3-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x^2 + (36 - 212*x - 8*x^2)*Log[5]^2 + (-616 - 53*x - x^2)*Log[5]^4)/(-48*x^2 + 16*x^3 + (-600*x + 176
*x^2 + 8*x^3)*Log[5]^2 + (-1875 + 475*x + 47*x^2 + x^3)*Log[5]^4),x]

[Out]

(3*Log[5]^2)/(25*Log[5]^2 + x*(4 + Log[5]^2)) - Log[3 - x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{3-x}+\frac {3 \log ^2(5) \left (-4-\log ^2(5)\right )}{\left (25 \log ^2(5)+x \left (4+\log ^2(5)\right )\right )^2}\right ) \, dx\\ &=\frac {3 \log ^2(5)}{25 \log ^2(5)+x \left (4+\log ^2(5)\right )}-\log (3-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 30, normalized size = 1.30 \begin {gather*} \frac {3 \log ^2(5)}{25 \log ^2(5)+x \left (4+\log ^2(5)\right )}-\log (-3+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + (36 - 212*x - 8*x^2)*Log[5]^2 + (-616 - 53*x - x^2)*Log[5]^4)/(-48*x^2 + 16*x^3 + (-600*x
+ 176*x^2 + 8*x^3)*Log[5]^2 + (-1875 + 475*x + 47*x^2 + x^3)*Log[5]^4),x]

[Out]

(3*Log[5]^2)/(25*Log[5]^2 + x*(4 + Log[5]^2)) - Log[-3 + x]

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fricas [A]  time = 0.79, size = 40, normalized size = 1.74 \begin {gather*} \frac {3 \, \log \relax (5)^{2} - {\left ({\left (x + 25\right )} \log \relax (5)^{2} + 4 \, x\right )} \log \left (x - 3\right )}{{\left (x + 25\right )} \log \relax (5)^{2} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-53*x-616)*log(5)^4+(-8*x^2-212*x+36)*log(5)^2-16*x^2)/((x^3+47*x^2+475*x-1875)*log(5)^4+(8*x^
3+176*x^2-600*x)*log(5)^2+16*x^3-48*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 32, normalized size = 1.39 \begin {gather*} \frac {3 \, \log \relax (5)^{2}}{x \log \relax (5)^{2} + 25 \, \log \relax (5)^{2} + 4 \, x} - \log \left ({\left | x - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-53*x-616)*log(5)^4+(-8*x^2-212*x+36)*log(5)^2-16*x^2)/((x^3+47*x^2+475*x-1875)*log(5)^4+(8*x^
3+176*x^2-600*x)*log(5)^2+16*x^3-48*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 32, normalized size = 1.39

method	result	size
default	\(-\ln \left (x -3\right )+\frac {3 \ln \relax (5)^{2}}{x \ln \relax (5)^{2}+25 \ln \relax (5)^{2}+4 x}\)	\(32\)
risch	\(-\ln \left (x -3\right )+\frac {3 \ln \relax (5)^{2}}{x \ln \relax (5)^{2}+25 \ln \relax (5)^{2}+4 x}\)	\(32\)
norman	\(-\frac {\left (3 \ln \relax (5)^{4}+12 \ln \relax (5)^{2}\right ) x}{25 \ln \relax (5)^{2} \left (x \ln \relax (5)^{2}+25 \ln \relax (5)^{2}+4 x \right )}-\ln \left (x -3\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-53*x-616)*ln(5)^4+(-8*x^2-212*x+36)*ln(5)^2-16*x^2)/((x^3+47*x^2+475*x-1875)*ln(5)^4+(8*x^3+176*x^2
-600*x)*ln(5)^2+16*x^3-48*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 30, normalized size = 1.30 \begin {gather*} \frac {3 \, \log \relax (5)^{2}}{{\left (\log \relax (5)^{2} + 4\right )} x + 25 \, \log \relax (5)^{2}} - \log \left (x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-53*x-616)*log(5)^4+(-8*x^2-212*x+36)*log(5)^2-16*x^2)/((x^3+47*x^2+475*x-1875)*log(5)^4+(8*x^
3+176*x^2-600*x)*log(5)^2+16*x^3-48*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.16, size = 31, normalized size = 1.35 \begin {gather*} \frac {3\,{\ln \relax (5)}^2}{4\,x+x\,{\ln \relax (5)}^2+25\,{\ln \relax (5)}^2}-\ln \left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)^4*(53*x + x^2 + 616) + log(5)^2*(212*x + 8*x^2 - 36) + 16*x^2)/(log(5)^4*(475*x + 47*x^2 + x^3 -
1875) + log(5)^2*(176*x^2 - 600*x + 8*x^3) - 48*x^2 + 16*x^3),x)

[Out]

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

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sympy [A]  time = 0.93, size = 26, normalized size = 1.13 \begin {gather*} - \log {\left (x - 3 \right )} + \frac {3 \log {\relax (5 )}^{2}}{x \left (\log {\relax (5 )}^{2} + 4\right ) + 25 \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-53*x-616)*ln(5)**4+(-8*x**2-212*x+36)*ln(5)**2-16*x**2)/((x**3+47*x**2+475*x-1875)*ln(5)**4+
(8*x**3+176*x**2-600*x)*ln(5)**2+16*x**3-48*x**2),x)

[Out]

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

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