3.28.44 \(\int \frac {e^{-\frac {5}{-3+e^2}} (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)))}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx\)

Optimal. Leaf size=33 \[ -x+\frac {e^{\frac {5}{3-e^2}}}{x \left (x^2+\frac {\log (3)}{4}\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 56, normalized size of antiderivative = 1.70, number of steps used = 7, number of rules used = 6, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {12, 1594, 28, 1805, 1253, 14} \begin {gather*} -\frac {64 e^{\frac {5}{3-e^2}} x}{\log (3) \left (16 x^2+\log (81)\right )}-x+\frac {4 e^{\frac {5}{3-e^2}}}{x \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48*x^2 - 4*Log[3] + E^(5/(-3 + E^2))*(-16*x^6 - 8*x^4*Log[3] - x^2*Log[3]^2))/(E^(5/(-3 + E^2))*(16*x^6
+ 8*x^4*Log[3] + x^2*Log[3]^2)),x]

[Out]

-x + (4*E^(5/(3 - E^2)))/(x*Log[3]) - (64*E^(5/(3 - E^2))*x)/(Log[3]*(16*x^2 + Log[81]))

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1253

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[(f*x)^m*(d + e*x^2)^(q + p)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{\frac {5}{3-e^2}} \int \frac {-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx\\ &=e^{\frac {5}{3-e^2}} \int \frac {-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )}{x^2 \left (16 x^4+8 x^2 \log (3)+\log ^2(3)\right )} \, dx\\ &=\left (16 e^{\frac {5}{3-e^2}}\right ) \int \frac {-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )}{x^2 \left (16 x^2+4 \log (3)\right )^2} \, dx\\ &=-\frac {64 e^{\frac {5}{3-e^2}} x}{\log (3) \left (16 x^2+\log (81)\right )}-\frac {\left (2 e^{\frac {5}{3-e^2}}\right ) \int \frac {8 \log (3)+8 e^{-\frac {5}{3-e^2}} x^4 \log (3)+2 x^2 \left (16+e^{-\frac {5}{3-e^2}} \log ^2(3)\right )}{x^2 \left (16 x^2+4 \log (3)\right )} \, dx}{\log (3)}\\ &=-\frac {64 e^{\frac {5}{3-e^2}} x}{\log (3) \left (16 x^2+\log (81)\right )}-\frac {\left (2 e^{\frac {5}{3-e^2}}\right ) \int \frac {2+\frac {1}{2} e^{-\frac {5}{3-e^2}} x^2 \log (3)}{x^2} \, dx}{\log (3)}\\ &=-\frac {64 e^{\frac {5}{3-e^2}} x}{\log (3) \left (16 x^2+\log (81)\right )}-\frac {\left (2 e^{\frac {5}{3-e^2}}\right ) \int \left (\frac {2}{x^2}+\frac {1}{2} e^{\frac {5}{-3+e^2}} \log (3)\right ) \, dx}{\log (3)}\\ &=-x+\frac {4 e^{\frac {5}{3-e^2}}}{x \log (3)}-\frac {64 e^{\frac {5}{3-e^2}} x}{\log (3) \left (16 x^2+\log (81)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 41, normalized size = 1.24 \begin {gather*} e^{-\frac {5}{-3+e^2}} \left (-e^{\frac {5}{-3+e^2}} x+\frac {4}{4 x^3+x \log (3)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48*x^2 - 4*Log[3] + E^(5/(-3 + E^2))*(-16*x^6 - 8*x^4*Log[3] - x^2*Log[3]^2))/(E^(5/(-3 + E^2))*(1
6*x^6 + 8*x^4*Log[3] + x^2*Log[3]^2)),x]

[Out]

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

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fricas [A]  time = 0.52, size = 47, normalized size = 1.42 \begin {gather*} -\frac {{\left ({\left (4 \, x^{4} + x^{2} \log \relax (3)\right )} e^{\left (\frac {5}{e^{2} - 3}\right )} - 4\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )}}{4 \, x^{3} + x \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)-48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+1
6*x^6)/exp(5/(exp(2)-3)),x, algorithm="fricas")

[Out]

-((4*x^4 + x^2*log(3))*e^(5/(e^2 - 3)) - 4)*e^(-5/(e^2 - 3))/(4*x^3 + x*log(3))

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giac [A]  time = 0.29, size = 37, normalized size = 1.12 \begin {gather*} -{\left (x e^{\left (\frac {5}{e^{2} - 3}\right )} - \frac {4}{4 \, x^{3} + x \log \relax (3)}\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)-48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+1
6*x^6)/exp(5/(exp(2)-3)),x, algorithm="giac")

[Out]

-(x*e^(5/(e^2 - 3)) - 4/(4*x^3 + x*log(3)))*e^(-5/(e^2 - 3))

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maple [A]  time = 0.07, size = 29, normalized size = 0.88

method	result	size
risch	\(-x +\frac {4 \,{\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}}}{x \left (4 x^{2}+\ln \relax (3)\right )}\)	\(29\)
default	\({\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}} \left (-{\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x +\frac {4}{x \ln \relax (3)}-\frac {16 x}{\ln \relax (3) \left (4 x^{2}+\ln \relax (3)\right )}\right )\)	\(52\)
gosper	\(-\frac {\left (4 \,{\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{4}-4+\ln \relax (3) {\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{2}\right ) {\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}}}{x \left (4 x^{2}+\ln \relax (3)\right )}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2*ln(3)^2-8*x^4*ln(3)-16*x^6)*exp(5/(exp(2)-3))-4*ln(3)-48*x^2)/(x^2*ln(3)^2+8*x^4*ln(3)+16*x^6)/exp(
5/(exp(2)-3)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 37, normalized size = 1.12 \begin {gather*} -{\left (x e^{\left (\frac {5}{e^{2} - 3}\right )} - \frac {4}{4 \, x^{3} + x \log \relax (3)}\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)-48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+1
6*x^6)/exp(5/(exp(2)-3)),x, algorithm="maxima")

[Out]

-(x*e^(5/(e^2 - 3)) - 4/(4*x^3 + x*log(3)))*e^(-5/(e^2 - 3))

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mupad [B]  time = 1.86, size = 28, normalized size = 0.85 \begin {gather*} \frac {4\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^2-3}}}{x\,\left (4\,x^2+\ln \relax (3)\right )}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5/(exp(2) - 3))*(4*log(3) + exp(5/(exp(2) - 3))*(x^2*log(3)^2 + 8*x^4*log(3) + 16*x^6) + 48*x^2))/(
x^2*log(3)^2 + 8*x^4*log(3) + 16*x^6),x)

[Out]

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

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sympy [A]  time = 0.39, size = 29, normalized size = 0.88 \begin {gather*} - x + \frac {4}{4 x^{3} e^{\frac {5}{-3 + e^{2}}} + x e^{\frac {5}{-3 + e^{2}}} \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2*ln(3)**2-8*x**4*ln(3)-16*x**6)*exp(5/(exp(2)-3))-4*ln(3)-48*x**2)/(x**2*ln(3)**2+8*x**4*ln(3
)+16*x**6)/exp(5/(exp(2)-3)),x)

[Out]

-x + 4/(4*x**3*exp(5/(-3 + exp(2))) + x*exp(5/(-3 + exp(2)))*log(3))

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