3.31.2 \(\int \frac {e^e x^2 \log (2)+(1-3 x^2-2 x^3) \log (2)}{-x+3 x^2-3 x^3+e^e x^3-x^4} \, dx\)

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

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Rubi [A]  time = 0.21, antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6, 2074, 1587} \begin {gather*} \log (2) \log \left (x^3+\left (3-e^e\right ) x^2-3 x+1\right )-\log (2) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E*x^2*Log[2] + (1 - 3*x^2 - 2*x^3)*Log[2])/(-x + 3*x^2 - 3*x^3 + E^E*x^3 - x^4),x]

[Out]

-(Log[2]*Log[x]) + Log[2]*Log[1 - 3*x + (3 - E^E)*x^2 + x^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^e x^2 \log (2)+\left (1-3 x^2-2 x^3\right ) \log (2)}{-x+3 x^2+\left (-3+e^e\right ) x^3-x^4} \, dx\\ &=\int \left (-\frac {\log (2)}{x}+\frac {\left (-3+2 \left (3-e^e\right ) x+3 x^2\right ) \log (2)}{1-3 x+\left (3-e^e\right ) x^2+x^3}\right ) \, dx\\ &=-\log (2) \log (x)+\log (2) \int \frac {-3+2 \left (3-e^e\right ) x+3 x^2}{1-3 x+\left (3-e^e\right ) x^2+x^3} \, dx\\ &=-\log (2) \log (x)+\log (2) \log \left (1-3 x+\left (3-e^e\right ) x^2+x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.36 \begin {gather*} \log (2) \left (-\log (x)+\log \left (1-3 x+3 x^2-e^e x^2+x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E*x^2*Log[2] + (1 - 3*x^2 - 2*x^3)*Log[2])/(-x + 3*x^2 - 3*x^3 + E^E*x^3 - x^4),x]

[Out]

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

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fricas [A]  time = 0.53, size = 32, normalized size = 1.45 \begin {gather*} \log \relax (2) \log \left (x^{3} - x^{2} e^{e} + 3 \, x^{2} - 3 \, x + 1\right ) - \log \relax (2) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)*exp(exp(1))+(-2*x^3-3*x^2+1)*log(2))/(x^3*exp(exp(1))-x^4-3*x^3+3*x^2-x),x, algorithm="f
ricas")

[Out]

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

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giac [A]  time = 0.39, size = 34, normalized size = 1.55 \begin {gather*} \log \relax (2) \log \left ({\left | x^{3} - x^{2} e^{e} + 3 \, x^{2} - 3 \, x + 1 \right |}\right ) - \log \relax (2) \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)*exp(exp(1))+(-2*x^3-3*x^2+1)*log(2))/(x^3*exp(exp(1))-x^4-3*x^3+3*x^2-x),x, algorithm="g
iac")

[Out]

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

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maple [A]  time = 0.06, size = 31, normalized size = 1.41

method	result	size
default	\(\ln \relax (2) \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}}+x^{3}+3 x^{2}-3 x +1\right )-\ln \relax (x )\right )\)	\(31\)
risch	\(-\ln \relax (2) \ln \relax (x )+\ln \relax (2) \ln \left (1+x^{3}+\left (3-{\mathrm e}^{{\mathrm e}}\right ) x^{2}-3 x \right )\)	\(31\)
norman	\(\ln \relax (2) \ln \left (x^{2} {\mathrm e}^{{\mathrm e}}-x^{3}-3 x^{2}+3 x -1\right )-\ln \relax (2) \ln \relax (x )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(2)*exp(exp(1))+(-2*x^3-3*x^2+1)*ln(2))/(x^3*exp(exp(1))-x^4-3*x^3+3*x^2-x),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [A]  time = 0.56, size = 29, normalized size = 1.32 \begin {gather*} \log \relax (2) \log \left (x^{3} - x^{2} {\left (e^{e} - 3\right )} - 3 \, x + 1\right ) - \log \relax (2) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)*exp(exp(1))+(-2*x^3-3*x^2+1)*log(2))/(x^3*exp(exp(1))-x^4-3*x^3+3*x^2-x),x, algorithm="m
axima")

[Out]

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

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mupad [B]  time = 1.87, size = 30, normalized size = 1.36 \begin {gather*} \ln \relax (2)\,\left (\ln \left (3\,x^2-x^2\,{\mathrm {e}}^{\mathrm {e}}-3\,x+x^3+1\right )-\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)*(3*x^2 + 2*x^3 - 1) - x^2*exp(exp(1))*log(2))/(x - x^3*exp(exp(1)) - 3*x^2 + 3*x^3 + x^4),x)

[Out]

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

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sympy [A]  time = 1.28, size = 29, normalized size = 1.32 \begin {gather*} - \log {\relax (2 )} \log {\relax (x )} + \log {\relax (2 )} \log {\left (x^{3} + x^{2} \left (3 - e^{e}\right ) - 3 x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(2)*exp(exp(1))+(-2*x**3-3*x**2+1)*ln(2))/(x**3*exp(exp(1))-x**4-3*x**3+3*x**2-x),x)

[Out]

-log(2)*log(x) + log(2)*log(x**3 + x**2*(3 - exp(E)) - 3*x + 1)

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