3.27.40 \(\int \frac {12 e^4}{13-26 x+13 x^2} \, dx\)

Optimal. Leaf size=22 \[ e^4 \left (2+\frac {12 x^2}{13 \left (x-x^2\right )}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 14, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 27, 32} \begin {gather*} \frac {12 e^4}{13 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*E^4)/(13 - 26*x + 13*x^2),x]

[Out]

(12*E^4)/(13*(1 - 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (12 e^4\right ) \int \frac {1}{13-26 x+13 x^2} \, dx\\ &=\left (12 e^4\right ) \int \frac {1}{13 (-1+x)^2} \, dx\\ &=\frac {1}{13} \left (12 e^4\right ) \int \frac {1}{(-1+x)^2} \, dx\\ &=\frac {12 e^4}{13 (1-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.55 \begin {gather*} -\frac {12 e^4}{13 (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^4)/(13 - 26*x + 13*x^2),x]

[Out]

(-12*E^4)/(13*(-1 + x))

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fricas [A]  time = 0.59, size = 9, normalized size = 0.41 \begin {gather*} -\frac {12 \, e^{4}}{13 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(12*exp(4)/(13*x^2-26*x+13),x, algorithm="fricas")

[Out]

-12/13*e^4/(x - 1)

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giac [A]  time = 0.27, size = 9, normalized size = 0.41 \begin {gather*} -\frac {12 \, e^{4}}{13 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(12*exp(4)/(13*x^2-26*x+13),x, algorithm="giac")

[Out]

-12/13*e^4/(x - 1)

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maple [A]  time = 0.42, size = 10, normalized size = 0.45

method	result	size
gosper	\(-\frac {12 \,{\mathrm e}^{4}}{13 \left (x -1\right )}\)	\(10\)
default	\(-\frac {12 \,{\mathrm e}^{4}}{13 \left (x -1\right )}\)	\(10\)
norman	\(-\frac {12 \,{\mathrm e}^{4}}{13 \left (x -1\right )}\)	\(10\)
risch	\(-\frac {12 \,{\mathrm e}^{4}}{13 \left (x -1\right )}\)	\(10\)
meijerg	\(\frac {12 \,{\mathrm e}^{4} x}{13 \left (1-x \right )}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(12*exp(4)/(13*x^2-26*x+13),x,method=_RETURNVERBOSE)

[Out]

-12/13*exp(4)/(x-1)

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maxima [A]  time = 0.35, size = 9, normalized size = 0.41 \begin {gather*} -\frac {12 \, e^{4}}{13 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(12*exp(4)/(13*x^2-26*x+13),x, algorithm="maxima")

[Out]

-12/13*e^4/(x - 1)

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mupad [B]  time = 0.04, size = 11, normalized size = 0.50 \begin {gather*} -\frac {12\,{\mathrm {e}}^4}{13\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(4))/(13*x^2 - 26*x + 13),x)

[Out]

-(12*exp(4))/(13*(x - 1))

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sympy [A]  time = 0.08, size = 10, normalized size = 0.45 \begin {gather*} - \frac {12 e^{4}}{13 x - 13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(12*exp(4)/(13*x**2-26*x+13),x)

[Out]

-12*exp(4)/(13*x - 13)

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