3.3.6 \(\int -\frac {15 e^2}{1550-620 e x+62 e^2 x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {15}{62 \left (-\frac {5}{e}+x\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 27, 32} \begin {gather*} -\frac {15 e}{62 (5-e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*E^2)/(1550 - 620*E*x + 62*E^2*x^2),x]

[Out]

(-15*E)/(62*(5 - E*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 (\left (15 e^2\right ) \int \frac {1}{1550-620 e x+62 e^2 x^2} \, dx\right )\\ &=-\left (\left (15 e^2\right ) \int \frac {1}{62 (-5+e x)^2} \, dx\right )\\ &=-\left (\frac {1}{62} \left (15 e^2\right ) \int \frac {1}{(-5+e x)^2} \, dx\right )\\ &=-\frac {15 e}{62 (5-e x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.92 \begin {gather*} \frac {15 e}{62 (-5+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^2)/(1550 - 620*E*x + 62*E^2*x^2),x]

[Out]

(15*E)/(62*(-5 + E*x))

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fricas [A]  time = 0.68, size = 12, normalized size = 0.92 \begin {gather*} \frac {15 \, e}{62 \, {\left (x e - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*exp(1)^2/(62*x^2*exp(1)^2-620*x*exp(1)+1550),x, algorithm="fricas")

[Out]

15/62*e/(x*e - 5)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*exp(1)^2/(62*x^2*exp(1)^2-620*x*exp(1)+1550),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -15*exp(2)/62*2*1/10/sqrt(-exp(1)^2+exp(
2))*atan((sageVARx*exp(2)-5*exp(1))*1/5/sqrt(-exp(1)^2+exp(2)))

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maple [A]  time = 0.12, size = 13, normalized size = 1.00

method	result	size
gosper	\(\frac {15 \,{\mathrm e}}{62 \left (x \,{\mathrm e}-5\right )}\)	\(13\)
risch	\(\frac {15 \,{\mathrm e}}{62 \left (x \,{\mathrm e}-5\right )}\)	\(13\)
meijerg	\(-\frac {3 \,{\mathrm e}^{2} x}{310 \left (1-\frac {x \,{\mathrm e}}{5}\right )}\)	\(15\)
norman	\(\frac {3 \,{\mathrm e}^{2} x}{62 \left (x \,{\mathrm e}-5\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-15*exp(1)^2/(62*x^2*exp(1)^2-620*x*exp(1)+1550),x,method=_RETURNVERBOSE)

[Out]

15/62*exp(1)/(x*exp(1)-5)

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maxima [A]  time = 0.46, size = 15, normalized size = 1.15 \begin {gather*} \frac {15 \, e^{2}}{62 \, {\left (x e^{2} - 5 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*exp(1)^2/(62*x^2*exp(1)^2-620*x*exp(1)+1550),x, algorithm="maxima")

[Out]

15/62*e^2/(x*e^2 - 5*e)

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mupad [B]  time = 0.08, size = 13, normalized size = 1.00 \begin {gather*} \frac {15\,\mathrm {e}}{62\,\left (x\,\mathrm {e}-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*exp(2))/(62*x^2*exp(2) - 620*x*exp(1) + 1550),x)

[Out]

(15*exp(1))/(62*(x*exp(1) - 5))

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sympy [A]  time = 0.11, size = 15, normalized size = 1.15 \begin {gather*} \frac {15 e^{2}}{62 x e^{2} - 310 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*exp(1)**2/(62*x**2*exp(1)**2-620*x*exp(1)+1550),x)

[Out]

15*exp(2)/(62*x*exp(2) - 310*E)

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