3.45.56 \(\int \frac {-24-6 x-48 x^3-3 x^4+e (-45-90 x^3)}{e (1-2 x^3+x^6)} \, dx\)

Optimal. Leaf size=22 \[ \frac {9 x \left (5+\frac {8+x}{3 e}\right )}{-1+x^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 28, 1858, 8} \begin {gather*} -\frac {3 x (x+15 e+8)}{e \left (1-x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 - 6*x - 48*x^3 - 3*x^4 + E*(-45 - 90*x^3))/(E*(1 - 2*x^3 + x^6)),x]

[Out]

(-3*x*(8 + 15*E + x))/(E*(1 - x^3))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-24-6 x-48 x^3-3 x^4+e \left (-45-90 x^3\right )}{1-2 x^3+x^6} \, dx}{e}\\ &=\frac {\int \frac {-24-6 x-48 x^3-3 x^4+e \left (-45-90 x^3\right )}{\left (-1+x^3\right )^2} \, dx}{e}\\ &=-\frac {3 x (8+15 e+x)}{e \left (1-x^3\right )}+\frac {\int 0 \, dx}{3 e}\\ &=-\frac {3 x (8+15 e+x)}{e \left (1-x^3\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.86 \begin {gather*} \frac {3 x (8+15 e+x)}{e \left (-1+x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 - 6*x - 48*x^3 - 3*x^4 + E*(-45 - 90*x^3))/(E*(1 - 2*x^3 + x^6)),x]

[Out]

(3*x*(8 + 15*E + x))/(E*(-1 + x^3))

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fricas [A]  time = 0.51, size = 23, normalized size = 1.05 \begin {gather*} \frac {3 \, {\left (x^{2} + 15 \, x e + 8 \, x\right )} e^{\left (-1\right )}}{x^{3} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x^3-45)*exp(1)-3*x^4-48*x^3-6*x-24)/(x^6-2*x^3+1)/exp(1),x, algorithm="fricas")

[Out]

3*(x^2 + 15*x*e + 8*x)*e^(-1)/(x^3 - 1)

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giac [A]  time = 0.13, size = 23, normalized size = 1.05 \begin {gather*} \frac {3 \, {\left (x^{2} + 15 \, x e + 8 \, x\right )} e^{\left (-1\right )}}{x^{3} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x^3-45)*exp(1)-3*x^4-48*x^3-6*x-24)/(x^6-2*x^3+1)/exp(1),x, algorithm="giac")

[Out]

3*(x^2 + 15*x*e + 8*x)*e^(-1)/(x^3 - 1)

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maple [A]  time = 0.07, size = 22, normalized size = 1.00

method	result	size
gosper	\(\frac {3 x \left (x +15 \,{\mathrm e}+8\right ) {\mathrm e}^{-1}}{x^{3}-1}\)	\(22\)
risch	\(\frac {{\mathrm e}^{-1} \left (3 x^{2}+\left (24+45 \,{\mathrm e}\right ) x \right )}{x^{3}-1}\)	\(25\)
norman	\(\frac {3 x^{2} {\mathrm e}^{-1}+3 \left (15 \,{\mathrm e}+8\right ) {\mathrm e}^{-1} x}{x^{3}-1}\)	\(32\)
default	\({\mathrm e}^{-1} \left (-\frac {3 \left (-5 \,{\mathrm e}-3\right )}{x -1}+\frac {3 \left (-2-5 \,{\mathrm e}\right ) x +15 \,{\mathrm e}+9}{x^{2}+x +1}\right )\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-90*x^3-45)*exp(1)-3*x^4-48*x^3-6*x-24)/(x^6-2*x^3+1)/exp(1),x,method=_RETURNVERBOSE)

[Out]

3*x*(x+15*exp(1)+8)/exp(1)/(x^3-1)

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maxima [A]  time = 0.37, size = 23, normalized size = 1.05 \begin {gather*} \frac {3 \, {\left (x^{2} + x {\left (15 \, e + 8\right )}\right )} e^{\left (-1\right )}}{x^{3} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x^3-45)*exp(1)-3*x^4-48*x^3-6*x-24)/(x^6-2*x^3+1)/exp(1),x, algorithm="maxima")

[Out]

3*(x^2 + x*(15*e + 8))*e^(-1)/(x^3 - 1)

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mupad [B]  time = 0.10, size = 28, normalized size = 1.27 \begin {gather*} -\frac {3\,x^2+\left (45\,\mathrm {e}+24\right )\,x}{\mathrm {e}-x^3\,\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*(6*x + exp(1)*(90*x^3 + 45) + 48*x^3 + 3*x^4 + 24))/(x^6 - 2*x^3 + 1),x)

[Out]

-(3*x^2 + x*(45*exp(1) + 24))/(exp(1) - x^3*exp(1))

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sympy [A]  time = 0.84, size = 26, normalized size = 1.18 \begin {gather*} - \frac {- 3 x^{2} + x \left (- 45 e - 24\right )}{e x^{3} - e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-90*x**3-45)*exp(1)-3*x**4-48*x**3-6*x-24)/(x**6-2*x**3+1)/exp(1),x)

[Out]

-(-3*x**2 + x*(-45*E - 24))/(E*x**3 - E)

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