3.93.41 \(\int \frac {-2-4 x^2+e^3 (-2+4 x-4 x^2+12 x^3)}{e^3 (3 x^3+6 x^5+3 x^7)} \, dx\)

Optimal. Leaf size=22 \[ \frac {-4+\frac {1+\frac {1}{e^3}}{x}}{3 \left (x+x^3\right )} \]

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Rubi [B]  time = 0.10, antiderivative size = 56, normalized size of antiderivative = 2.55, number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 1594, 28, 1805, 1586, 37} \begin {gather*} \frac {\left (-2 e^3 x+e^3+1\right )^2}{3 e^3 \left (1+e^3\right ) x^2}-\frac {-4 e^3 x+e^3+1}{3 e^3 \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 4*x^2 + E^3*(-2 + 4*x - 4*x^2 + 12*x^3))/(E^3*(3*x^3 + 6*x^5 + 3*x^7)),x]

[Out]

(1 + E^3 - 2*E^3*x)^2/(3*E^3*(1 + E^3)*x^2) - (1 + E^3 - 4*E^3*x)/(3*E^3*(1 + x^2))

Rule 12

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

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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} &=\frac {\int \frac {-2-4 x^2+e^3 \left (-2+4 x-4 x^2+12 x^3\right )}{3 x^3+6 x^5+3 x^7} \, dx}{e^3}\\ &=\frac {\int \frac {-2-4 x^2+e^3 \left (-2+4 x-4 x^2+12 x^3\right )}{x^3 \left (3+6 x^2+3 x^4\right )} \, dx}{e^3}\\ &=\frac {3 \int \frac {-2-4 x^2+e^3 \left (-2+4 x-4 x^2+12 x^3\right )}{x^3 \left (3+3 x^2\right )^2} \, dx}{e^3}\\ &=-\frac {1+e^3-4 e^3 x}{3 e^3 \left (1+x^2\right )}-\frac {\int \frac {4 \left (1+e^3\right )-8 e^3 x+4 \left (1+e^3\right ) x^2-8 e^3 x^3}{x^3 \left (3+3 x^2\right )} \, dx}{2 e^3}\\ &=-\frac {1+e^3-4 e^3 x}{3 e^3 \left (1+x^2\right )}-\frac {\int \frac {\frac {4}{3}+\frac {4 e^3}{3}-\frac {8 e^3 x}{3}}{x^3} \, dx}{2 e^3}\\ &=\frac {\left (1+e^3-2 e^3 x\right )^2}{3 e^3 \left (1+e^3\right ) x^2}-\frac {1+e^3-4 e^3 x}{3 e^3 \left (1+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.23 \begin {gather*} \frac {1+e^3-4 e^3 x}{3 e^3 \left (x^2+x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 4*x^2 + E^3*(-2 + 4*x - 4*x^2 + 12*x^3))/(E^3*(3*x^3 + 6*x^5 + 3*x^7)),x]

[Out]

(1 + E^3 - 4*E^3*x)/(3*E^3*(x^2 + x^4))

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fricas [A]  time = 0.64, size = 23, normalized size = 1.05 \begin {gather*} -\frac {{\left ({\left (4 \, x - 1\right )} e^{3} - 1\right )} e^{\left (-3\right )}}{3 \, {\left (x^{4} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-4*x^2+4*x-2)*exp(3)-4*x^2-2)/(3*x^7+6*x^5+3*x^3)/exp(3),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 40, normalized size = 1.82 \begin {gather*} \frac {1}{3} \, {\left (\frac {4 \, x e^{3} - e^{3} - 1}{x^{2} + 1} - \frac {4 \, x e^{3} - e^{3} - 1}{x^{2}}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-4*x^2+4*x-2)*exp(3)-4*x^2-2)/(3*x^7+6*x^5+3*x^3)/exp(3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 25, normalized size = 1.14

method	result	size
risch	\(\frac {{\mathrm e}^{-3} \left (\frac {1}{3}-\frac {4 x \,{\mathrm e}^{3}}{3}+\frac {{\mathrm e}^{3}}{3}\right )}{x^{2} \left (x^{2}+1\right )}\)	\(25\)
norman	\(\frac {-\frac {4 x}{3}+\frac {\left ({\mathrm e}^{3}+1\right ) {\mathrm e}^{-3}}{3}}{x^{2} \left (x^{2}+1\right )}\)	\(26\)
gosper	\(-\frac {\left (4 x \,{\mathrm e}^{3}-{\mathrm e}^{3}-1\right ) {\mathrm e}^{-3}}{3 x^{2} \left (x^{2}+1\right )}\)	\(28\)
default	\(\frac {{\mathrm e}^{-3} \left (-\frac {-{\mathrm e}^{3}-1}{x^{2}}-\frac {4 \,{\mathrm e}^{3}}{x}-\frac {2 \left (\frac {1}{2}-2 x \,{\mathrm e}^{3}+\frac {{\mathrm e}^{3}}{2}\right )}{x^{2}+1}\right )}{3}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^3-4*x^2+4*x-2)*exp(3)-4*x^2-2)/(3*x^7+6*x^5+3*x^3)/exp(3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 24, normalized size = 1.09 \begin {gather*} -\frac {{\left (4 \, x e^{3} - e^{3} - 1\right )} e^{\left (-3\right )}}{3 \, {\left (x^{4} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-4*x^2+4*x-2)*exp(3)-4*x^2-2)/(3*x^7+6*x^5+3*x^3)/exp(3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.17, size = 27, normalized size = 1.23 \begin {gather*} \frac {{\mathrm {e}}^3-4\,x\,{\mathrm {e}}^3+1}{3\,{\mathrm {e}}^3\,x^4+3\,{\mathrm {e}}^3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3)*(4*x^2 - exp(3)*(4*x - 4*x^2 + 12*x^3 - 2) + 2))/(3*x^3 + 6*x^5 + 3*x^7),x)

[Out]

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

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sympy [A]  time = 1.13, size = 27, normalized size = 1.23 \begin {gather*} \frac {- 4 x e^{3} + 1 + e^{3}}{3 x^{4} e^{3} + 3 x^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**3-4*x**2+4*x-2)*exp(3)-4*x**2-2)/(3*x**7+6*x**5+3*x**3)/exp(3),x)

[Out]

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

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