3.87.44 \(\int \frac {e^{1+e} (-24 x+12 \log ^2(4))}{25+10 x^2+x^4+(-10 x-2 x^3) \log ^2(4)+x^2 \log ^4(4)} \, dx\)

Optimal. Leaf size=23 \[ 3+\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \]

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Rubi [A]  time = 0.04, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6, 12, 1680, 261} \begin {gather*} \frac {12 e^{1+e}}{x^2-x \log ^2(4)+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + E)*(-24*x + 12*Log[4]^2))/(25 + 10*x^2 + x^4 + (-10*x - 2*x^3)*Log[4]^2 + x^2*Log[4]^4),x]

[Out]

(12*E^(1 + E))/(5 + x^2 - x*Log[4]^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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1+e} \left (-24 x+12 \log ^2(4)\right )}{25+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \left (10+\log ^4(4)\right )} \, dx\\ &=e^{1+e} \int \frac {-24 x+12 \log ^2(4)}{25+x^4+\left (-10 x-2 x^3\right ) \log ^2(4)+x^2 \left (10+\log ^4(4)\right )} \, dx\\ &=e^{1+e} \operatorname {Subst}\left (\int -\frac {384 x}{\left (20+4 x^2-\log ^4(4)\right )^2} \, dx,x,x-\frac {\log ^2(4)}{2}\right )\\ &=-\left (\left (384 e^{1+e}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (20+4 x^2-\log ^4(4)\right )^2} \, dx,x,x-\frac {\log ^2(4)}{2}\right )\right )\\ &=\frac {12 e^{1+e}}{5+x^2-x \log ^2(4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.91 \begin {gather*} \frac {12 e^{1+e}}{5+x^2-x \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + E)*(-24*x + 12*Log[4]^2))/(25 + 10*x^2 + x^4 + (-10*x - 2*x^3)*Log[4]^2 + x^2*Log[4]^4),x]

[Out]

(12*E^(1 + E))/(5 + x^2 - x*Log[4]^2)

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fricas [A]  time = 1.11, size = 23, normalized size = 1.00 \begin {gather*} -\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \relax (2)^{2} - x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x)*log(2)^2+x^4+10*x^2+25),x, algorit
hm="fricas")

[Out]

-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)

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giac [A]  time = 0.18, size = 23, normalized size = 1.00 \begin {gather*} -\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \relax (2)^{2} - x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x)*log(2)^2+x^4+10*x^2+25),x, algorit
hm="giac")

[Out]

-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)

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maple [A]  time = 0.06, size = 23, normalized size = 1.00

method	result	size
risch	\(-\frac {3 \,{\mathrm e}^{1+{\mathrm e}}}{x \ln \relax (2)^{2}-\frac {x^{2}}{4}-\frac {5}{4}}\)	\(23\)
gosper	\(-\frac {12 \,{\mathrm e}^{1+{\mathrm e}}}{4 x \ln \relax (2)^{2}-x^{2}-5}\)	\(24\)
norman	\(-\frac {12 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}}}{4 x \ln \relax (2)^{2}-x^{2}-5}\)	\(24\)
default	\(\frac {24 \,{\mathrm e}^{1+{\mathrm e}} \left (10-8 \ln \relax (2)^{4}\right )}{\left (20-16 \ln \relax (2)^{4}\right ) \left (x^{2}-4 x \ln \relax (2)^{2}+5\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*ln(2)^2-24*x)*exp(1+exp(1))/(16*x^2*ln(2)^4+4*(-2*x^3-10*x)*ln(2)^2+x^4+10*x^2+25),x,method=_RETURNVER
BOSE)

[Out]

-3/(x*ln(2)^2-1/4*x^2-5/4)*exp(1+exp(1))

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maxima [A]  time = 0.37, size = 23, normalized size = 1.00 \begin {gather*} -\frac {12 \, e^{\left (e + 1\right )}}{4 \, x \log \relax (2)^{2} - x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*log(2)^2-24*x)*exp(1+exp(1))/(16*x^2*log(2)^4+4*(-2*x^3-10*x)*log(2)^2+x^4+10*x^2+25),x, algorit
hm="maxima")

[Out]

-12*e^(e + 1)/(4*x*log(2)^2 - x^2 - 5)

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mupad [B]  time = 5.32, size = 21, normalized size = 0.91 \begin {gather*} \frac {12\,{\mathrm {e}}^{\mathrm {e}+1}}{x^2-4\,{\ln \relax (2)}^2\,x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(1) + 1)*(24*x - 48*log(2)^2))/(16*x^2*log(2)^4 - 4*log(2)^2*(10*x + 2*x^3) + 10*x^2 + x^4 + 25),
x)

[Out]

(12*exp(exp(1) + 1))/(x^2 - 4*x*log(2)^2 + 5)

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sympy [A]  time = 0.32, size = 22, normalized size = 0.96 \begin {gather*} \frac {12 e e^{e}}{x^{2} - 4 x \log {\relax (2 )}^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*ln(2)**2-24*x)*exp(1+exp(1))/(16*x**2*ln(2)**4+4*(-2*x**3-10*x)*ln(2)**2+x**4+10*x**2+25),x)

[Out]

12*E*exp(E)/(x**2 - 4*x*log(2)**2 + 5)

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