3.36.43 \(\int \frac {2 x^2-4 x^3+(-16 x+4 e^4 x-4 x^2+4 x^3) \log (-4+e^4-x+x^2)}{-4 x^4+e^4 x^4-x^5+x^6+(16 x^2-4 e^4 x^2+4 x^3-4 x^4) \log (-4+e^4-x+x^2)+(-16+4 e^4-4 x+4 x^2) \log ^2(-4+e^4-x+x^2)} \, dx\)

Optimal. Leaf size=29 \[ \frac {x}{x+\frac {2 \left (-x^2+\log \left (-4+e^4-x+x^2\right )\right )}{x}} \]

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Rubi [A]  time = 0.42, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 6688, 12, 6711, 32} \begin {gather*} \frac {2}{2-\frac {x^2}{\log \left (x^2-x+e^4-4\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 - 4*x^3 + (-16*x + 4*E^4*x - 4*x^2 + 4*x^3)*Log[-4 + E^4 - x + x^2])/(-4*x^4 + E^4*x^4 - x^5 + x^6
+ (16*x^2 - 4*E^4*x^2 + 4*x^3 - 4*x^4)*Log[-4 + E^4 - x + x^2] + (-16 + 4*E^4 - 4*x + 4*x^2)*Log[-4 + E^4 - x
+ x^2]^2),x]

[Out]

2/(2 - x^2/Log[-4 + E^4 - x + x^2])

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{\left (-4+e^4\right ) x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx\\ &=\int \frac {2 x \left (-x+2 x^2-2 \left (-4+e^4-x+x^2\right ) \log \left (-4+e^4-x+x^2\right )\right )}{\left (4-e^4+x-x^2\right ) \left (x^2-2 \log \left (-4+e^4-x+x^2\right )\right )^2} \, dx\\ &=2 \int \frac {x \left (-x+2 x^2-2 \left (-4+e^4-x+x^2\right ) \log \left (-4+e^4-x+x^2\right )\right )}{\left (4-e^4+x-x^2\right ) \left (x^2-2 \log \left (-4+e^4-x+x^2\right )\right )^2} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{(-2+x)^2} \, dx,x,\frac {x^2}{\log \left (-4+e^4-x+x^2\right )}\right )\\ &=\frac {2}{2-\frac {x^2}{\log \left (-4+e^4-x+x^2\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 3.49, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^2}{-x^2+2 \log \left (-4+e^4-x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 - 4*x^3 + (-16*x + 4*E^4*x - 4*x^2 + 4*x^3)*Log[-4 + E^4 - x + x^2])/(-4*x^4 + E^4*x^4 - x^5
+ x^6 + (16*x^2 - 4*E^4*x^2 + 4*x^3 - 4*x^4)*Log[-4 + E^4 - x + x^2] + (-16 + 4*E^4 - 4*x + 4*x^2)*Log[-4 + E^
4 - x + x^2]^2),x]

[Out]

x^2/(-x^2 + 2*Log[-4 + E^4 - x + x^2])

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fricas [A]  time = 0.61, size = 24, normalized size = 0.83 \begin {gather*} -\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(4)+4*x^3-4*x^2-16*x)*log(exp(4)+x^2-x-4)-4*x^3+2*x^2)/((4*exp(4)+4*x^2-4*x-16)*log(exp(4)+
x^2-x-4)^2+(-4*x^2*exp(4)-4*x^4+4*x^3+16*x^2)*log(exp(4)+x^2-x-4)+x^4*exp(4)+x^6-x^5-4*x^4),x, algorithm="fric
as")

[Out]

-x^2/(x^2 - 2*log(x^2 - x + e^4 - 4))

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giac [A]  time = 0.27, size = 24, normalized size = 0.83 \begin {gather*} -\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(4)+4*x^3-4*x^2-16*x)*log(exp(4)+x^2-x-4)-4*x^3+2*x^2)/((4*exp(4)+4*x^2-4*x-16)*log(exp(4)+
x^2-x-4)^2+(-4*x^2*exp(4)-4*x^4+4*x^3+16*x^2)*log(exp(4)+x^2-x-4)+x^4*exp(4)+x^6-x^5-4*x^4),x, algorithm="giac
")

[Out]

-x^2/(x^2 - 2*log(x^2 - x + e^4 - 4))

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maple [A]  time = 0.21, size = 25, normalized size = 0.86

method	result	size
risch	\(-\frac {x^{2}}{x^{2}-2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}\)	\(25\)
norman	\(-\frac {2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}{x^{2}-2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x*exp(4)+4*x^3-4*x^2-16*x)*ln(exp(4)+x^2-x-4)-4*x^3+2*x^2)/((4*exp(4)+4*x^2-4*x-16)*ln(exp(4)+x^2-x-4)
^2+(-4*x^2*exp(4)-4*x^4+4*x^3+16*x^2)*ln(exp(4)+x^2-x-4)+x^4*exp(4)+x^6-x^5-4*x^4),x,method=_RETURNVERBOSE)

[Out]

-x^2/(x^2-2*ln(exp(4)+x^2-x-4))

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maxima [A]  time = 0.61, size = 24, normalized size = 0.83 \begin {gather*} -\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(4)+4*x^3-4*x^2-16*x)*log(exp(4)+x^2-x-4)-4*x^3+2*x^2)/((4*exp(4)+4*x^2-4*x-16)*log(exp(4)+
x^2-x-4)^2+(-4*x^2*exp(4)-4*x^4+4*x^3+16*x^2)*log(exp(4)+x^2-x-4)+x^4*exp(4)+x^6-x^5-4*x^4),x, algorithm="maxi
ma")

[Out]

-x^2/(x^2 - 2*log(x^2 - x + e^4 - 4))

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mupad [B]  time = 15.52, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^2}{2\,\ln \left (x^2-x+{\mathrm {e}}^4-4\right )-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(4) - x + x^2 - 4)*(16*x - 4*x*exp(4) + 4*x^2 - 4*x^3) - 2*x^2 + 4*x^3)/(log(exp(4) - x + x^2 - 4)
^2*(4*x - 4*exp(4) - 4*x^2 + 16) - x^4*exp(4) + 4*x^4 + x^5 - x^6 + log(exp(4) - x + x^2 - 4)*(4*x^2*exp(4) -
16*x^2 - 4*x^3 + 4*x^4)),x)

[Out]

x^2/(2*log(exp(4) - x + x^2 - 4) - x^2)

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sympy [A]  time = 0.25, size = 19, normalized size = 0.66 \begin {gather*} \frac {x^{2}}{- x^{2} + 2 \log {\left (x^{2} - x - 4 + e^{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(4)+4*x**3-4*x**2-16*x)*ln(exp(4)+x**2-x-4)-4*x**3+2*x**2)/((4*exp(4)+4*x**2-4*x-16)*ln(exp
(4)+x**2-x-4)**2+(-4*x**2*exp(4)-4*x**4+4*x**3+16*x**2)*ln(exp(4)+x**2-x-4)+x**4*exp(4)+x**6-x**5-4*x**4),x)

[Out]

x**2/(-x**2 + 2*log(x**2 - x - 4 + exp(4)))

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