\(\int \frac {3 x^4+(2-6 x^3) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+(2 x-2 x^4) \log (4)+x^3 \log ^2(4)} \, dx\) [6448]

Optimal result

Integrand size = 57, antiderivative size = 18 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log \left (30 x \left (x^2+\frac {2}{-x+\log (4)}\right )\right ) \]

[Out]

ln(30*(2/(2*ln(2)-x)+x^2)*x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {2099, 1601} \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log \left (-x^3+x^2 \log (4)+2\right )+\log (x)-\log (x-\log (4)) \]

[In]

Int[(3*x^4 + (2 - 6*x^3)*Log[4] + 3*x^2*Log[4]^2)/(-2*x^2 + x^5 + (2*x - 2*x^4)*Log[4] + x^3*Log[4]^2),x]

[Out]

Log[x] - Log[x - Log[4]] + Log[2 - x^3 + x^2*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{align*} \text {integral}& = \int \left (\frac {1}{x}+\frac {1}{-x+\log (4)}+\frac {x (3 x-2 \log (4))}{-2+x^3-x^2 \log (4)}\right ) \, dx \\ & = \log (x)-\log (x-\log (4))+\int \frac {x (3 x-2 \log (4))}{-2+x^3-x^2 \log (4)} \, dx \\ & = \log (x)-\log (x-\log (4))+\log \left (2-x^3+x^2 \log (4)\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.05 (sec) , antiderivative size = 109, normalized size of antiderivative = 6.06 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log (x)+\text {RootSum}\left [\log (16)-2 \text {$\#$1}+\log ^2(4) \text {$\#$1}^2-2 \log (4) \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})+\log ^2(4) \log (x-\text {$\#$1}) \text {$\#$1}-2 \log (4) \log (x-\text {$\#$1}) \text {$\#$1}^2+\log (x-\text {$\#$1}) \text {$\#$1}^3}{-1+\log ^2(4) \text {$\#$1}-3 \log (4) \text {$\#$1}^2+2 \text {$\#$1}^3}\&\right ] \]

[In]

Integrate[(3*x^4 + (2 - 6*x^3)*Log[4] + 3*x^2*Log[4]^2)/(-2*x^2 + x^5 + (2*x - 2*x^4)*Log[4] + x^3*Log[4]^2),x
]

[Out]

Log[x] + RootSum[Log[16] - 2*#1 + Log[4]^2*#1^2 - 2*Log[4]*#1^3 + #1^4 & , (Log[x - #1] + Log[4]^2*Log[x - #1]
*#1 - 2*Log[4]*Log[x - #1]*#1^2 + Log[x - #1]*#1^3)/(-1 + Log[4]^2*#1 - 3*Log[4]*#1^2 + 2*#1^3) & ]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44

method	result	size
default	\(\ln \left (-2 x^{2} \ln \left (2\right )+x^{3}-2\right )-\ln \left (x -2 \ln \left (2\right )\right )+\ln \left (x \right )\)	\(26\)
risch	\(-\ln \left (x -2 \ln \left (2\right )\right )+\ln \left (-2 x^{3} \ln \left (2\right )+x^{4}-2 x \right )\)	\(26\)
parallelrisch	\(\ln \left (-2 x^{2} \ln \left (2\right )+x^{3}-2\right )-\ln \left (x -2 \ln \left (2\right )\right )+\ln \left (x \right )\)	\(26\)
norman	\(-\ln \left (2 \ln \left (2\right )-x \right )+\ln \left (x \right )+\ln \left (2 x^{2} \ln \left (2\right )-x^{3}+2\right )\)	\(30\)

[In]

int((12*x^2*ln(2)^2+2*(-6*x^3+2)*ln(2)+3*x^4)/(4*x^3*ln(2)^2+2*(-2*x^4+2*x)*ln(2)+x^5-2*x^2),x,method=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log \left (x^{4} - 2 \, x^{3} \log \left (2\right ) - 2 \, x\right ) - \log \left (x - 2 \, \log \left (2\right )\right ) \]

[In]

integrate((12*x^2*log(2)^2+2*(-6*x^3+2)*log(2)+3*x^4)/(4*x^3*log(2)^2+2*(-2*x^4+2*x)*log(2)+x^5-2*x^2),x, algo
rithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=- \log {\left (x - 2 \log {\left (2 \right )} \right )} + \log {\left (x^{4} - 2 x^{3} \log {\left (2 \right )} - 2 x \right )} \]

[In]

integrate((12*x**2*ln(2)**2+2*(-6*x**3+2)*ln(2)+3*x**4)/(4*x**3*ln(2)**2+2*(-2*x**4+2*x)*ln(2)+x**5-2*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log \left (x^{3} - 2 \, x^{2} \log \left (2\right ) - 2\right ) - \log \left (x - 2 \, \log \left (2\right )\right ) + \log \left (x\right ) \]

[In]

integrate((12*x^2*log(2)^2+2*(-6*x^3+2)*log(2)+3*x^4)/(4*x^3*log(2)^2+2*(-2*x^4+2*x)*log(2)+x^5-2*x^2),x, algo
rithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\log \left ({\left | x^{3} - 2 \, x^{2} \log \left (2\right ) - 2 \right |}\right ) - \log \left ({\left | x - 2 \, \log \left (2\right ) \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((12*x^2*log(2)^2+2*(-6*x^3+2)*log(2)+3*x^4)/(4*x^3*log(2)^2+2*(-2*x^4+2*x)*log(2)+x^5-2*x^2),x, algo
rithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {3 x^4+\left (2-6 x^3\right ) \log (4)+3 x^2 \log ^2(4)}{-2 x^2+x^5+\left (2 x-2 x^4\right ) \log (4)+x^3 \log ^2(4)} \, dx=\ln \left (x^4-2\,\ln \left (2\right )\,x^3-2\,x\right )-\ln \left (x-\ln \left (4\right )\right ) \]

[In]

int((12*x^2*log(2)^2 - 2*log(2)*(6*x^3 - 2) + 3*x^4)/(4*x^3*log(2)^2 + 2*log(2)*(2*x - 2*x^4) - 2*x^2 + x^5),x
)

[Out]

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