3.99.50 \(\int \frac {400+150 x}{-100 x-25 x^2+(64 x^5+48 x^6+12 x^7+x^8) \log ^2(\log (4))} \, dx\)

Optimal. Leaf size=20 \[ \log \left (8 \left (-1+\frac {25}{x^4 (4+x)^2 \log ^2(\log (4))}\right )\right ) \]

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Rubi [B]  time = 0.11, antiderivative size = 48, normalized size of antiderivative = 2.40, number of steps used = 4, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2074, 1587} \begin {gather*} \log \left (x^3 (-\log (\log (4)))-4 x^2 \log (\log (4))+5\right )+\log \left (x^3 \log (\log (4))+4 x^2 \log (\log (4))+5\right )-4 \log (x)-2 \log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(400 + 150*x)/(-100*x - 25*x^2 + (64*x^5 + 48*x^6 + 12*x^7 + x^8)*Log[Log[4]]^2),x]

[Out]

-4*Log[x] - 2*Log[4 + x] + Log[5 - 4*x^2*Log[Log[4]] - x^3*Log[Log[4]]] + Log[5 + 4*x^2*Log[Log[4]] + x^3*Log[
Log[4]]]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, 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 {4}{x}-\frac {2}{4+x}+\frac {x (8+3 x) \log (\log (4))}{-5+4 x^2 \log (\log (4))+x^3 \log (\log (4))}+\frac {x (8+3 x) \log (\log (4))}{5+4 x^2 \log (\log (4))+x^3 \log (\log (4))}\right ) \, dx\\ &=-4 \log (x)-2 \log (4+x)+\log (\log (4)) \int \frac {x (8+3 x)}{-5+4 x^2 \log (\log (4))+x^3 \log (\log (4))} \, dx+\log (\log (4)) \int \frac {x (8+3 x)}{5+4 x^2 \log (\log (4))+x^3 \log (\log (4))} \, dx\\ &=-4 \log (x)-2 \log (4+x)+\log \left (5-4 x^2 \log (\log (4))-x^3 \log (\log (4))\right )+\log \left (5+4 x^2 \log (\log (4))+x^3 \log (\log (4))\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 54, normalized size = 2.70 \begin {gather*} 50 \left (-\frac {2 \log (x)}{25}-\frac {1}{25} \log (4+x)+\frac {1}{50} \log \left (25-16 x^4 \log ^2(\log (4))-8 x^5 \log ^2(\log (4))-x^6 \log ^2(\log (4))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(400 + 150*x)/(-100*x - 25*x^2 + (64*x^5 + 48*x^6 + 12*x^7 + x^8)*Log[Log[4]]^2),x]

[Out]

50*((-2*Log[x])/25 - Log[4 + x]/25 + Log[25 - 16*x^4*Log[Log[4]]^2 - 8*x^5*Log[Log[4]]^2 - x^6*Log[Log[4]]^2]/
50)

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fricas [A]  time = 0.58, size = 36, normalized size = 1.80 \begin {gather*} \log \left ({\left (x^{6} + 8 \, x^{5} + 16 \, x^{4}\right )} \log \left (2 \, \log \relax (2)\right )^{2} - 25\right ) - 2 \, \log \left (x + 4\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((150*x+400)/((x^8+12*x^7+48*x^6+64*x^5)*log(2*log(2))^2-25*x^2-100*x),x, algorithm="fricas")

[Out]

log((x^6 + 8*x^5 + 16*x^4)*log(2*log(2))^2 - 25) - 2*log(x + 4) - 4*log(x)

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giac [B]  time = 0.16, size = 59, normalized size = 2.95 \begin {gather*} \log \left ({\left | x^{3} \log \left (2 \, \log \relax (2)\right ) + 4 \, x^{2} \log \left (2 \, \log \relax (2)\right ) + 5 \right |}\right ) + \log \left ({\left | x^{3} \log \left (2 \, \log \relax (2)\right ) + 4 \, x^{2} \log \left (2 \, \log \relax (2)\right ) - 5 \right |}\right ) - 2 \, \log \left ({\left | x + 4 \right |}\right ) - 4 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((150*x+400)/((x^8+12*x^7+48*x^6+64*x^5)*log(2*log(2))^2-25*x^2-100*x),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.30, size = 56, normalized size = 2.80

method	result	size
default	\(-4 \ln \relax (x )+\ln \left (x^{3} \ln \left (2 \ln \relax (2)\right )+4 x^{2} \ln \left (2 \ln \relax (2)\right )-5\right )-2 \ln \left (4+x \right )+\ln \left (x^{3} \ln \left (2 \ln \relax (2)\right )+4 x^{2} \ln \left (2 \ln \relax (2)\right )+5\right )\)	\(56\)
norman	\(-4 \ln \relax (x )+\ln \left (x^{3} \ln \left (2 \ln \relax (2)\right )+4 x^{2} \ln \left (2 \ln \relax (2)\right )-5\right )-2 \ln \left (4+x \right )+\ln \left (x^{3} \ln \left (2 \ln \relax (2)\right )+4 x^{2} \ln \left (2 \ln \relax (2)\right )+5\right )\)	\(56\)
risch	\(-2 \ln \left (4+x \right )-4 \ln \left (-x \right )+\ln \left (\left (-\ln \relax (2)^{2}-2 \ln \relax (2) \ln \left (\ln \relax (2)\right )-\ln \left (\ln \relax (2)\right )^{2}\right ) x^{6}+\left (-8 \ln \relax (2)^{2}-16 \ln \relax (2) \ln \left (\ln \relax (2)\right )-8 \ln \left (\ln \relax (2)\right )^{2}\right ) x^{5}+\left (-16 \ln \relax (2)^{2}-32 \ln \relax (2) \ln \left (\ln \relax (2)\right )-16 \ln \left (\ln \relax (2)\right )^{2}\right ) x^{4}+25\right )\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((150*x+400)/((x^8+12*x^7+48*x^6+64*x^5)*ln(2*ln(2))^2-25*x^2-100*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.35, size = 55, normalized size = 2.75 \begin {gather*} \log \left (x^{3} \log \left (2 \, \log \relax (2)\right ) + 4 \, x^{2} \log \left (2 \, \log \relax (2)\right ) + 5\right ) + \log \left (x^{3} \log \left (2 \, \log \relax (2)\right ) + 4 \, x^{2} \log \left (2 \, \log \relax (2)\right ) - 5\right ) - 2 \, \log \left (x + 4\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((150*x+400)/((x^8+12*x^7+48*x^6+64*x^5)*log(2*log(2))^2-25*x^2-100*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.83, size = 43, normalized size = 2.15 \begin {gather*} \ln \left ({\ln \left (\ln \relax (4)\right )}^2\,x^6+8\,{\ln \left (\ln \relax (4)\right )}^2\,x^5+16\,{\ln \left (\ln \relax (4)\right )}^2\,x^4-25\right )-2\,\ln \left (x+4\right )-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(150*x + 400)/(100*x - log(2*log(2))^2*(64*x^5 + 48*x^6 + 12*x^7 + x^8) + 25*x^2),x)

[Out]

log(16*x^4*log(log(4))^2 + 8*x^5*log(log(4))^2 + x^6*log(log(4))^2 - 25) - 2*log(x + 4) - 4*log(x)

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sympy [B]  time = 15.88, size = 49, normalized size = 2.45 \begin {gather*} - 4 \log {\relax (x )} - 2 \log {\left (x + 4 \right )} + \log {\left (x^{6} + 8 x^{5} + 16 x^{4} - \frac {25}{2 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + \log {\left (\log {\relax (2 )} \right )}^{2} + \log {\relax (2 )}^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((150*x+400)/((x**8+12*x**7+48*x**6+64*x**5)*ln(2*ln(2))**2-25*x**2-100*x),x)

[Out]

-4*log(x) - 2*log(x + 4) + log(x**6 + 8*x**5 + 16*x**4 - 25/(2*log(2)*log(log(2)) + log(log(2))**2 + log(2)**2
))

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