3.55.63 \(\int \frac {65 e^{-\frac {1}{-1+65 x}}}{1-130 x+4225 x^2} \, dx\)

Optimal. Leaf size=13 \[ e^{\frac {x}{x-65 x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 9, normalized size of antiderivative = 0.69, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 27, 2209} \begin {gather*} e^{\frac {1}{1-65 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[65/(E^(-1 + 65*x)^(-1)*(1 - 130*x + 4225*x^2)),x]

[Out]

E^(1 - 65*x)^(-1)

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 2209

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=65 \int \frac {e^{-\frac {1}{-1+65 x}}}{1-130 x+4225 x^2} \, dx\\ &=65 \int \frac {e^{-\frac {1}{-1+65 x}}}{(-1+65 x)^2} \, dx\\ &=e^{\frac {1}{1-65 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 0.69 \begin {gather*} e^{\frac {1}{1-65 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[65/(E^(-1 + 65*x)^(-1)*(1 - 130*x + 4225*x^2)),x]

[Out]

E^(1 - 65*x)^(-1)

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fricas [A]  time = 1.62, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (-\frac {1}{65 \, x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(65*exp(-1/(65*x-1))/(4225*x^2-130*x+1),x, algorithm="fricas")

[Out]

e^(-1/(65*x - 1))

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giac [A]  time = 0.14, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (-\frac {1}{65 \, x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(65*exp(-1/(65*x-1))/(4225*x^2-130*x+1),x, algorithm="giac")

[Out]

e^(-1/(65*x - 1))

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maple [A]  time = 0.29, size = 11, normalized size = 0.85

method	result	size
gosper	\({\mathrm e}^{-\frac {1}{65 x -1}}\)	\(11\)
derivativedivides	\({\mathrm e}^{-\frac {1}{65 x -1}}\)	\(11\)
default	\({\mathrm e}^{-\frac {1}{65 x -1}}\)	\(11\)
risch	\({\mathrm e}^{-\frac {1}{65 x -1}}\)	\(11\)
norman	\(\frac {65 x \,{\mathrm e}^{-\frac {1}{65 x -1}}-{\mathrm e}^{-\frac {1}{65 x -1}}}{65 x -1}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(65*exp(-1/(65*x-1))/(4225*x^2-130*x+1),x,method=_RETURNVERBOSE)

[Out]

exp(-1/(65*x-1))

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maxima [A]  time = 0.37, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (-\frac {1}{65 \, x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(65*exp(-1/(65*x-1))/(4225*x^2-130*x+1),x, algorithm="maxima")

[Out]

e^(-1/(65*x - 1))

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mupad [B]  time = 3.43, size = 10, normalized size = 0.77 \begin {gather*} {\mathrm {e}}^{-\frac {1}{65\,x-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((65*exp(-1/(65*x - 1)))/(4225*x^2 - 130*x + 1),x)

[Out]

exp(-1/(65*x - 1))

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sympy [A]  time = 0.12, size = 8, normalized size = 0.62 \begin {gather*} e^{- \frac {1}{65 x - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(65*exp(-1/(65*x-1))/(4225*x**2-130*x+1),x)

[Out]

exp(-1/(65*x - 1))

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