3.2.89 \(\int \frac {12 x-18 x^2+e^2 (3-6 x+3 x^2)}{e^2 (25 x^4-50 x^5+25 x^6)} \, dx\)

Optimal. Leaf size=26 \[ \frac {\frac {6}{-1+x}-\frac {e^2}{x}}{25 e^2 x^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.62, number of steps used = 6, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 1594, 27, 1820} \begin {gather*} -\frac {1}{25 x^3}-\frac {6}{25 e^2 x^2}-\frac {6}{25 e^2 x}-\frac {6}{25 e^2 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*x - 18*x^2 + E^2*(3 - 6*x + 3*x^2))/(E^2*(25*x^4 - 50*x^5 + 25*x^6)),x]

[Out]

-6/(25*E^2*(1 - x)) - 1/(25*x^3) - 6/(25*E^2*x^2) - 6/(25*E^2*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 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 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {12 x-18 x^2+e^2 \left (3-6 x+3 x^2\right )}{25 x^4-50 x^5+25 x^6} \, dx}{e^2}\\ &=\frac {\int \frac {12 x-18 x^2+e^2 \left (3-6 x+3 x^2\right )}{x^4 \left (25-50 x+25 x^2\right )} \, dx}{e^2}\\ &=\frac {\int \frac {12 x-18 x^2+e^2 \left (3-6 x+3 x^2\right )}{25 (-1+x)^2 x^4} \, dx}{e^2}\\ &=\frac {\int \frac {12 x-18 x^2+e^2 \left (3-6 x+3 x^2\right )}{(-1+x)^2 x^4} \, dx}{25 e^2}\\ &=\frac {\int \left (-\frac {6}{(-1+x)^2}+\frac {3 e^2}{x^4}+\frac {12}{x^3}+\frac {6}{x^2}\right ) \, dx}{25 e^2}\\ &=-\frac {6}{25 e^2 (1-x)}-\frac {1}{25 x^3}-\frac {6}{25 e^2 x^2}-\frac {6}{25 e^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.77 \begin {gather*} \frac {-1+\frac {6 x}{e^2 (-1+x)}}{25 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*x - 18*x^2 + E^2*(3 - 6*x + 3*x^2))/(E^2*(25*x^4 - 50*x^5 + 25*x^6)),x]

[Out]

(-1 + (6*x)/(E^2*(-1 + x)))/(25*x^3)

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fricas [A]  time = 0.95, size = 25, normalized size = 0.96 \begin {gather*} -\frac {{\left ({\left (x - 1\right )} e^{2} - 6 \, x\right )} e^{\left (-2\right )}}{25 \, {\left (x^{4} - x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x+3)*exp(2)-18*x^2+12*x)/(25*x^6-50*x^5+25*x^4)/exp(1)^2,x, algorithm="fricas")

[Out]

-1/25*((x - 1)*e^2 - 6*x)*e^(-2)/(x^4 - x^3)

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giac [A]  time = 0.36, size = 28, normalized size = 1.08 \begin {gather*} \frac {1}{25} \, {\left (\frac {6}{x - 1} - \frac {6 \, x^{2} + 6 \, x + e^{2}}{x^{3}}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x+3)*exp(2)-18*x^2+12*x)/(25*x^6-50*x^5+25*x^4)/exp(1)^2,x, algorithm="giac")

[Out]

1/25*(6/(x - 1) - (6*x^2 + 6*x + e^2)/x^3)*e^(-2)

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maple [A]  time = 0.06, size = 25, normalized size = 0.96

method	result	size
risch	\(\frac {{\mathrm e}^{-2} \left (\left (-\frac {{\mathrm e}^{2}}{25}+\frac {6}{25}\right ) x +\frac {{\mathrm e}^{2}}{25}\right )}{x^{3} \left (x -1\right )}\)	\(25\)
gosper	\(-\frac {\left ({\mathrm e}^{2} x -{\mathrm e}^{2}-6 x \right ) {\mathrm e}^{-2}}{25 x^{3} \left (x -1\right )}\)	\(27\)
default	\(\frac {{\mathrm e}^{-2} \left (-\frac {{\mathrm e}^{2}}{x^{3}}-\frac {6}{x^{2}}-\frac {6}{x}+\frac {6}{x -1}\right )}{25}\)	\(32\)
norman	\(\frac {\left (\frac {{\mathrm e}^{-1} {\mathrm e}^{2}}{25}-\frac {\left ({\mathrm e}^{2}-6\right ) {\mathrm e}^{-1} x}{25}\right ) {\mathrm e}^{-1}}{x^{3} \left (x -1\right )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2-6*x+3)*exp(2)-18*x^2+12*x)/(25*x^6-50*x^5+25*x^4)/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-2)*((-1/25*exp(2)+6/25)*x+1/25*exp(2))/x^3/(x-1)

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maxima [A]  time = 0.42, size = 26, normalized size = 1.00 \begin {gather*} -\frac {{\left (x {\left (e^{2} - 6\right )} - e^{2}\right )} e^{\left (-2\right )}}{25 \, {\left (x^{4} - x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x+3)*exp(2)-18*x^2+12*x)/(25*x^6-50*x^5+25*x^4)/exp(1)^2,x, algorithm="maxima")

[Out]

-1/25*(x*(e^2 - 6) - e^2)*e^(-2)/(x^4 - x^3)

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mupad [B]  time = 0.13, size = 29, normalized size = 1.12 \begin {gather*} -\frac {{\mathrm {e}}^2-x\,\left ({\mathrm {e}}^2-6\right )}{25\,x^3\,{\mathrm {e}}^2-25\,x^4\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(12*x + exp(2)*(3*x^2 - 6*x + 3) - 18*x^2))/(25*x^4 - 50*x^5 + 25*x^6),x)

[Out]

-(exp(2) - x*(exp(2) - 6))/(25*x^3*exp(2) - 25*x^4*exp(2))

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sympy [A]  time = 0.47, size = 27, normalized size = 1.04 \begin {gather*} - \frac {x \left (-6 + e^{2}\right ) - e^{2}}{25 x^{4} e^{2} - 25 x^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2-6*x+3)*exp(2)-18*x**2+12*x)/(25*x**6-50*x**5+25*x**4)/exp(1)**2,x)

[Out]

-(x*(-6 + exp(2)) - exp(2))/(25*x**4*exp(2) - 25*x**3*exp(2))

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