3.12.79 \(\int -\frac {2 e^{16}}{25+10 x+x^2} \, dx\)

Optimal. Leaf size=12 \[ 3+\frac {2 e^{16}}{5+x} \]

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 27, 32} \begin {gather*} \frac {2 e^{16}}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^16)/(25 + 10*x + x^2),x]

[Out]

(2*E^16)/(5 + 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 (2 e^{16}\right ) \int \frac {1}{25+10 x+x^2} \, dx\right )\\ &=-\left (\left (2 e^{16}\right ) \int \frac {1}{(5+x)^2} \, dx\right )\\ &=\frac {2 e^{16}}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.83 \begin {gather*} \frac {2 e^{16}}{5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^16)/(25 + 10*x + x^2),x]

[Out]

(2*E^16)/(5 + x)

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fricas [A]  time = 0.83, size = 9, normalized size = 0.75 \begin {gather*} \frac {2 \, e^{16}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(2)*exp(16)/(x^2+10*x+25)/exp(1)^2,x, algorithm="fricas")

[Out]

2*e^16/(x + 5)

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giac [A]  time = 0.38, size = 9, normalized size = 0.75 \begin {gather*} \frac {2 \, e^{16}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(2)*exp(16)/(x^2+10*x+25)/exp(1)^2,x, algorithm="giac")

[Out]

2*e^16/(x + 5)

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maple [A]  time = 0.20, size = 10, normalized size = 0.83

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{16}}{5+x}\)	\(10\)
meijerg	\(-\frac {2 \,{\mathrm e}^{16} x}{25 \left (1+\frac {x}{5}\right )}\)	\(13\)
gosper	\(\frac {2 \,{\mathrm e}^{18} {\mathrm e}^{-2}}{5+x}\)	\(14\)
default	\(\frac {2 \,{\mathrm e}^{16} {\mathrm e}^{-2} {\mathrm e}^{2}}{5+x}\)	\(16\)
norman	\(\frac {2 \,{\mathrm e}^{16} {\mathrm e}^{-2} {\mathrm e}^{2}}{5+x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*exp(2)*exp(16)/(x^2+10*x+25)/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

2*exp(16)/(5+x)

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maxima [A]  time = 0.53, size = 9, normalized size = 0.75 \begin {gather*} \frac {2 \, e^{16}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(2)*exp(16)/(x^2+10*x+25)/exp(1)^2,x, algorithm="maxima")

[Out]

2*e^16/(x + 5)

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mupad [B]  time = 0.76, size = 9, normalized size = 0.75 \begin {gather*} \frac {2\,{\mathrm {e}}^{16}}{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(16))/(10*x + x^2 + 25),x)

[Out]

(2*exp(16))/(x + 5)

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sympy [A]  time = 0.07, size = 7, normalized size = 0.58 \begin {gather*} \frac {2 e^{16}}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(2)*exp(16)/(x**2+10*x+25)/exp(1)**2,x)

[Out]

2*exp(16)/(x + 5)

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