3.21.72 \(\int \frac {51 x+17 e^{2 e^2} x}{-8+12 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=26 \[ 4-\frac {\left (3+e^{2 e^2}\right ) \left (-1+x+4 x^2\right )}{(-2+x)^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.42, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6, 12, 2074} \begin {gather*} \frac {17 \left (3+e^{2 e^2}\right )}{2-x}-\frac {17 \left (3+e^{2 e^2}\right )}{(2-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(51*x + 17*E^(2*E^2)*x)/(-8 + 12*x - 6*x^2 + x^3),x]

[Out]

(-17*(3 + E^(2*E^2)))/(2 - x)^2 + (17*(3 + E^(2*E^2)))/(2 - x)

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 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (51+17 e^{2 e^2}\right ) x}{-8+12 x-6 x^2+x^3} \, dx\\ &=\left (17 \left (3+e^{2 e^2}\right )\right ) \int \frac {x}{-8+12 x-6 x^2+x^3} \, dx\\ &=\left (17 \left (3+e^{2 e^2}\right )\right ) \int \left (\frac {2}{(-2+x)^3}+\frac {1}{(-2+x)^2}\right ) \, dx\\ &=-\frac {17 \left (3+e^{2 e^2}\right )}{(2-x)^2}+\frac {17 \left (3+e^{2 e^2}\right )}{2-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.81 \begin {gather*} \frac {17 \left (3+e^{2 e^2}\right ) (1-x)}{(-2+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(51*x + 17*E^(2*E^2)*x)/(-8 + 12*x - 6*x^2 + x^3),x]

[Out]

(17*(3 + E^(2*E^2))*(1 - x))/(-2 + x)^2

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fricas [A]  time = 0.95, size = 26, normalized size = 1.00 \begin {gather*} -\frac {17 \, {\left ({\left (x - 1\right )} e^{\left (2 \, e^{2}\right )} + 3 \, x - 3\right )}}{x^{2} - 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((17*x*exp(exp(1)^2)^2+51*x)/(x^3-6*x^2+12*x-8),x, algorithm="fricas")

[Out]

-17*((x - 1)*e^(2*e^2) + 3*x - 3)/(x^2 - 4*x + 4)

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giac [A]  time = 0.29, size = 26, normalized size = 1.00 \begin {gather*} -\frac {17 \, {\left (x e^{\left (2 \, e^{2}\right )} + 3 \, x - e^{\left (2 \, e^{2}\right )} - 3\right )}}{{\left (x - 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((17*x*exp(exp(1)^2)^2+51*x)/(x^3-6*x^2+12*x-8),x, algorithm="giac")

[Out]

-17*(x*e^(2*e^2) + 3*x - e^(2*e^2) - 3)/(x - 2)^2

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maple [A]  time = 0.03, size = 25, normalized size = 0.96

method	result	size
gosper	\(-\frac {17 \left ({\mathrm e}^{2 \,{\mathrm e}^{2}}+3\right ) \left (x -1\right )}{x^{2}-4 x +4}\)	\(25\)
default	\(17 \left ({\mathrm e}^{2 \,{\mathrm e}^{2}}+3\right ) \left (-\frac {1}{\left (x -2\right )^{2}}-\frac {1}{x -2}\right )\)	\(27\)
norman	\(\frac {\left (-17 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}-51\right ) x +17 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}+51}{\left (x -2\right )^{2}}\)	\(31\)
risch	\(\frac {\left (-17 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}-51\right ) x +17 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}+51}{x^{2}-4 x +4}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((17*x*exp(exp(1)^2)^2+51*x)/(x^3-6*x^2+12*x-8),x,method=_RETURNVERBOSE)

[Out]

-17*(exp(exp(1)^2)^2+3)*(x-1)/(x^2-4*x+4)

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maxima [A]  time = 0.69, size = 30, normalized size = 1.15 \begin {gather*} -\frac {17 \, {\left (x {\left (e^{\left (2 \, e^{2}\right )} + 3\right )} - e^{\left (2 \, e^{2}\right )} - 3\right )}}{x^{2} - 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((17*x*exp(exp(1)^2)^2+51*x)/(x^3-6*x^2+12*x-8),x, algorithm="maxima")

[Out]

-17*(x*(e^(2*e^2) + 3) - e^(2*e^2) - 3)/(x^2 - 4*x + 4)

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mupad [B]  time = 0.06, size = 17, normalized size = 0.65 \begin {gather*} -\frac {17\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^2}+3\right )\,\left (x-1\right )}{{\left (x-2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((51*x + 17*x*exp(2*exp(2)))/(12*x - 6*x^2 + x^3 - 8),x)

[Out]

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

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sympy [A]  time = 0.09, size = 20, normalized size = 0.77 \begin {gather*} \frac {\left (1 - x\right ) \left (51 + 17 e^{2 e^{2}}\right )}{x^{2} - 4 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((17*x*exp(exp(1)**2)**2+51*x)/(x**3-6*x**2+12*x-8),x)

[Out]

(1 - x)*(51 + 17*exp(2*exp(2)))/(x**2 - 4*x + 4)

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