3.88.23 \(\int \frac {-16+30 x+(-25 x^3-10 x^4) \log (4)}{-8 x+30 x^2-25 x^3+(25 x^4+5 x^5) \log (4)} \, dx\) [8723]

Optimal. Leaf size=28 \[ 1-\log \left (-\left (5-\frac {3}{x}\right )^2+\frac {1}{x^2}+5 x (5+x) \log (4)\right ) \]

[Out]

1-ln(10*ln(2)*(5+x)*x+1/x^2-(5-3/x)^2)

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Rubi [A]
time = 0.12, antiderivative size = 32, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 2, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2099, 1601} \begin {gather*} 2 \log (x)-\log \left (-5 x^4 \log (4)-25 x^3 \log (4)+25 x^2-30 x+8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 30*x + (-25*x^3 - 10*x^4)*Log[4])/(-8*x + 30*x^2 - 25*x^3 + (25*x^4 + 5*x^5)*Log[4]),x]

[Out]

2*Log[x] - Log[8 - 30*x + 25*x^2 - 25*x^3*Log[4] - 5*x^4*Log[4]]

Rule 1601

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]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2099

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 {2}{x}-\frac {5 \left (6-10 x+15 x^2 \log (4)+4 x^3 \log (4)\right )}{-8+30 x-25 x^2+25 x^3 \log (4)+5 x^4 \log (4)}\right ) \, dx\\ &=2 \log (x)-5 \int \frac {6-10 x+15 x^2 \log (4)+4 x^3 \log (4)}{-8+30 x-25 x^2+25 x^3 \log (4)+5 x^4 \log (4)} \, dx\\ &=2 \log (x)-\log \left (8-30 x+25 x^2-25 x^3 \log (4)-5 x^4 \log (4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 32, normalized size = 1.14 \begin {gather*} 2 \log (x)-\log \left (8-30 x+25 x^2-25 x^3 \log (4)-5 x^4 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 30*x + (-25*x^3 - 10*x^4)*Log[4])/(-8*x + 30*x^2 - 25*x^3 + (25*x^4 + 5*x^5)*Log[4]),x]

[Out]

2*Log[x] - Log[8 - 30*x + 25*x^2 - 25*x^3*Log[4] - 5*x^4*Log[4]]

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Maple [A]
time = 0.37, size = 33, normalized size = 1.18

method	result	size
default	\(2 \ln \left (x \right )-\ln \left (10 x^{4} \ln \left (2\right )+50 x^{3} \ln \left (2\right )-25 x^{2}+30 x -8\right )\)	\(33\)
norman	\(2 \ln \left (x \right )-\ln \left (10 x^{4} \ln \left (2\right )+50 x^{3} \ln \left (2\right )-25 x^{2}+30 x -8\right )\)	\(33\)
risch	\(2 \ln \left (-x \right )-\ln \left (10 x^{4} \ln \left (2\right )+50 x^{3} \ln \left (2\right )-25 x^{2}+30 x -8\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-10*x^4-25*x^3)*ln(2)+30*x-16)/(2*(5*x^5+25*x^4)*ln(2)-25*x^3+30*x^2-8*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)-ln(10*x^4*ln(2)+50*x^3*ln(2)-25*x^2+30*x-8)

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Maxima [A]
time = 0.26, size = 32, normalized size = 1.14 \begin {gather*} -\log \left (10 \, x^{4} \log \left (2\right ) + 50 \, x^{3} \log \left (2\right ) - 25 \, x^{2} + 30 \, x - 8\right ) + 2 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-10*x^4-25*x^3)*log(2)+30*x-16)/(2*(5*x^5+25*x^4)*log(2)-25*x^3+30*x^2-8*x),x, algorithm="maxima
")

[Out]

-log(10*x^4*log(2) + 50*x^3*log(2) - 25*x^2 + 30*x - 8) + 2*log(x)

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Fricas [A]
time = 0.38, size = 31, normalized size = 1.11 \begin {gather*} -\log \left (-25 \, x^{2} + 10 \, {\left (x^{4} + 5 \, x^{3}\right )} \log \left (2\right ) + 30 \, x - 8\right ) + 2 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-10*x^4-25*x^3)*log(2)+30*x-16)/(2*(5*x^5+25*x^4)*log(2)-25*x^3+30*x^2-8*x),x, algorithm="fricas
")

[Out]

-log(-25*x^2 + 10*(x^4 + 5*x^3)*log(2) + 30*x - 8) + 2*log(x)

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Sympy [A]
time = 1.71, size = 37, normalized size = 1.32 \begin {gather*} 2 \log {\left (x \right )} - \log {\left (x^{4} + 5 x^{3} - \frac {5 x^{2}}{2 \log {\left (2 \right )}} + \frac {3 x}{\log {\left (2 \right )}} - \frac {4}{5 \log {\left (2 \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-10*x**4-25*x**3)*ln(2)+30*x-16)/(2*(5*x**5+25*x**4)*ln(2)-25*x**3+30*x**2-8*x),x)

[Out]

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

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Giac [A]
time = 0.42, size = 34, normalized size = 1.21 \begin {gather*} -\log \left ({\left | 10 \, x^{4} \log \left (2\right ) + 50 \, x^{3} \log \left (2\right ) - 25 \, x^{2} + 30 \, x - 8 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-10*x^4-25*x^3)*log(2)+30*x-16)/(2*(5*x^5+25*x^4)*log(2)-25*x^3+30*x^2-8*x),x, algorithm="giac")

[Out]

-log(abs(10*x^4*log(2) + 50*x^3*log(2) - 25*x^2 + 30*x - 8)) + 2*log(abs(x))

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Mupad [B]
time = 5.40, size = 32, normalized size = 1.14 \begin {gather*} 2\,\ln \left (x\right )-\ln \left (30\,\ln \left (2\right )\,x^4+150\,\ln \left (2\right )\,x^3-75\,x^2+90\,x-24\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2)*(25*x^3 + 10*x^4) - 30*x + 16)/(8*x - 2*log(2)*(25*x^4 + 5*x^5) - 30*x^2 + 25*x^3),x)

[Out]

2*log(x) - log(90*x + 150*x^3*log(2) + 30*x^4*log(2) - 75*x^2 - 24)

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