3.45.91 \(\int \frac {-e^3-x^2-16 x^3+3 x^4+x^2 \log (\log (3))}{e^3 x-x^2-x^3-8 x^4+x^5+x^3 \log (\log (3))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (1+x+x \left (\left (8-\frac {e^3}{x^3}-x\right ) x-\log (\log (3))\right )\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 33, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 3, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6, 2074, 1587} \begin {gather*} \log \left (x^4-8 x^3-x^2 (1-\log (\log (3)))-x+e^3\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] + Log[E^3 - x - 8*x^3 + x^4 - x^2*(1 - Log[Log[3]])]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 \frac {-e^3-16 x^3+3 x^4+x^2 (-1+\log (\log (3)))}{e^3 x-x^2-x^3-8 x^4+x^5+x^3 \log (\log (3))} \, dx\\ &=\int \frac {-e^3-16 x^3+3 x^4+x^2 (-1+\log (\log (3)))}{e^3 x-x^2-8 x^4+x^5+x^3 (-1+\log (\log (3)))} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {-1-24 x^2+4 x^3-2 x (1-\log (\log (3)))}{e^3-x-8 x^3+x^4-x^2 (1-\log (\log (3)))}\right ) \, dx\\ &=-\log (x)+\int \frac {-1-24 x^2+4 x^3-2 x (1-\log (\log (3)))}{e^3-x-8 x^3+x^4-x^2 (1-\log (\log (3)))} \, dx\\ &=-\log (x)+\log \left (e^3-x-8 x^3+x^4-x^2 (1-\log (\log (3)))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 33, normalized size = 1.22 \begin {gather*} -\log (x)+\log \left (e^3-x-x^2-8 x^3+x^4+x^2 \log (\log (3))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] + Log[E^3 - x - x^2 - 8*x^3 + x^4 + x^2*Log[Log[3]]]

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fricas [A]  time = 0.67, size = 32, normalized size = 1.19 \begin {gather*} \log \left (x^{4} - 8 \, x^{3} + x^{2} \log \left (\log \relax (3)\right ) - x^{2} - x + e^{3}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^4 - 8*x^3 + x^2*log(log(3)) - x^2 - x + e^3) - log(x)

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giac [A]  time = 0.35, size = 34, normalized size = 1.26 \begin {gather*} \log \left ({\left | x^{4} - 8 \, x^{3} + x^{2} \log \left (\log \relax (3)\right ) - x^{2} - x + e^{3} \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(x^4 - 8*x^3 + x^2*log(log(3)) - x^2 - x + e^3)) - log(abs(x))

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maple [A]  time = 0.07, size = 32, normalized size = 1.19

method	result	size
risch	\(-\ln \left (-x \right )+\ln \left (x^{4}-8 x^{3}+\left (\ln \left (\ln \relax (3)\right )-1\right ) x^{2}-x +{\mathrm e}^{3}\right )\)	\(32\)
default	\(-\ln \relax (x )+\ln \left (x^{4}+x^{2} \ln \left (\ln \relax (3)\right )-8 x^{3}-x^{2}+{\mathrm e}^{3}-x \right )\)	\(33\)
norman	\(-\ln \relax (x )+\ln \left (x^{4}+x^{2} \ln \left (\ln \relax (3)\right )-8 x^{3}-x^{2}+{\mathrm e}^{3}-x \right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(-x)+ln(x^4-8*x^3+(ln(ln(3))-1)*x^2-x+exp(3))

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maxima [A]  time = 0.37, size = 29, normalized size = 1.07 \begin {gather*} \log \left (x^{4} - 8 \, x^{3} + x^{2} {\left (\log \left (\log \relax (3)\right ) - 1\right )} - x + e^{3}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^4 - 8*x^3 + x^2*(log(log(3)) - 1) - x + e^3) - log(x)

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mupad [B]  time = 3.60, size = 32, normalized size = 1.19 \begin {gather*} \ln \left ({\mathrm {e}}^3-x+x^2\,\ln \left (\ln \relax (3)\right )-x^2-8\,x^3+x^4\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(3) - x + x^2*log(log(3)) - x^2 - 8*x^3 + x^4) - log(x)

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sympy [A]  time = 3.95, size = 27, normalized size = 1.00 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{4} - 8 x^{3} + x^{2} \left (-1 + \log {\left (\log {\relax (3 )} \right )}\right ) - x + e^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**4 - 8*x**3 + x**2*(-1 + log(log(3))) - x + exp(3))

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