3.85.5 \(\int \frac {(1-6 x^2) \log (2)+\log (2) \log (3)}{5+(1+x-2 x^3) \log (2)+x \log (2) \log (3)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (-1-x+2 x^3-\frac {5}{\log (2)}-x \log (3)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1587} \begin {gather*} \log \left (\left (-2 x^3+x+1\right ) \log (2)+x \log (2) \log (3)+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 6*x^2)*Log[2] + Log[2]*Log[3])/(5 + (1 + x - 2*x^3)*Log[2] + x*Log[2]*Log[3]),x]

[Out]

Log[5 + (1 + x - 2*x^3)*Log[2] + x*Log[2]*Log[3]]

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]

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 80, normalized size = 3.64 \begin {gather*} \log (2) \text {RootSum}\left [5+\log (2)+\log (2) \text {$\#$1}+\log (2) \log (3) \text {$\#$1}-\log (4) \text {$\#$1}^3\&,\frac {\log (x-\text {$\#$1})+\log (3) \log (x-\text {$\#$1})-6 \log (x-\text {$\#$1}) \text {$\#$1}^2}{\log (2)+\log (2) \log (3)-3 \log (4) \text {$\#$1}^2}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 6*x^2)*Log[2] + Log[2]*Log[3])/(5 + (1 + x - 2*x^3)*Log[2] + x*Log[2]*Log[3]),x]

[Out]

Log[2]*RootSum[5 + Log[2] + Log[2]*#1 + Log[2]*Log[3]*#1 - Log[4]*#1^3 & , (Log[x - #1] + Log[3]*Log[x - #1] -
6*Log[x - #1]*#1^2)/(Log[2] + Log[2]*Log[3] - 3*Log[4]*#1^2) & ]

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fricas [A]  time = 1.05, size = 23, normalized size = 1.05 \begin {gather*} \log \left (-x \log \relax (3) \log \relax (2) + {\left (2 \, x^{3} - x - 1\right )} \log \relax (2) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)*log(3)+(-6*x^2+1)*log(2))/(x*log(2)*log(3)+(-2*x^3+x+1)*log(2)+5),x, algorithm="fricas")

[Out]

log(-x*log(3)*log(2) + (2*x^3 - x - 1)*log(2) - 5)

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giac [A]  time = 0.23, size = 25, normalized size = 1.14 \begin {gather*} \log \left ({\left | x \log \relax (3) \log \relax (2) - {\left (2 \, x^{3} - x\right )} \log \relax (2) + \log \relax (2) + 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)*log(3)+(-6*x^2+1)*log(2))/(x*log(2)*log(3)+(-2*x^3+x+1)*log(2)+5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 21, normalized size = 0.95

method	result	size
derivativedivides	\(\ln \left (x \ln \relax (2) \ln \relax (3)+\left (-2 x^{3}+x +1\right ) \ln \relax (2)+5\right )\)	\(21\)
default	\(\ln \left (-2 x^{3} \ln \relax (2)+x \ln \relax (2) \ln \relax (3)+x \ln \relax (2)+\ln \relax (2)+5\right )\)	\(23\)
norman	\(\ln \left (-2 x^{3} \ln \relax (2)+x \ln \relax (2) \ln \relax (3)+x \ln \relax (2)+\ln \relax (2)+5\right )\)	\(23\)
risch	\(\ln \left (-2 x^{3} \ln \relax (2)+\left (\ln \relax (2) \ln \relax (3)+\ln \relax (2)\right ) x +\ln \relax (2)+5\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2)*ln(3)+(-6*x^2+1)*ln(2))/(x*ln(2)*ln(3)+(-2*x^3+x+1)*ln(2)+5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 23, normalized size = 1.05 \begin {gather*} \log \left (x \log \relax (3) \log \relax (2) - {\left (2 \, x^{3} - x - 1\right )} \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)*log(3)+(-6*x^2+1)*log(2))/(x*log(2)*log(3)+(-2*x^3+x+1)*log(2)+5),x, algorithm="maxima")

[Out]

log(x*log(3)*log(2) - (2*x^3 - x - 1)*log(2) + 5)

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mupad [B]  time = 5.31, size = 23, normalized size = 1.05 \begin {gather*} \ln \left (2\,\ln \relax (2)\,x^3-\ln \relax (2)\,\left (\ln \relax (3)+1\right )\,x-\ln \relax (2)-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)*log(3) - log(2)*(6*x^2 - 1))/(log(2)*(x - 2*x^3 + 1) + x*log(2)*log(3) + 5),x)

[Out]

log(2*x^3*log(2) - log(2) - x*log(2)*(log(3) + 1) - 5)

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sympy [A]  time = 0.56, size = 27, normalized size = 1.23 \begin {gather*} \log {\left (2 x^{3} \log {\relax (2 )} + x \left (- \log {\relax (2 )} \log {\relax (3 )} - \log {\relax (2 )}\right ) - 5 - \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2)*ln(3)+(-6*x**2+1)*ln(2))/(x*ln(2)*ln(3)+(-2*x**3+x+1)*ln(2)+5),x)

[Out]

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

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