3.9.75 \(\int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {e^{e^4}}{4 (2+5 x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 21, 32} \begin {gather*} \frac {e^{e^4}}{4 (5 x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*E^E^4)/((2 + 5*x)*(8 + 20*x)),x]

[Out]

E^E^4/(4*(2 + 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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 (5 e^{e^4}\right ) \int \frac {1}{(2+5 x) (8+20 x)} \, dx\right )\\ &=-\left (\frac {1}{4} \left (5 e^{e^4}\right ) \int \frac {1}{(2+5 x)^2} \, dx\right )\\ &=\frac {e^{e^4}}{4 (2+5 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {e^{e^4}}{4 (2+5 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^E^4)/((2 + 5*x)*(8 + 20*x)),x]

[Out]

E^E^4/(4*(2 + 5*x))

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fricas [A]  time = 0.66, size = 12, normalized size = 0.75 \begin {gather*} \frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(-log(20*x+8)+exp(4))/(5*x+2),x, algorithm="fricas")

[Out]

1/4*e^(e^4)/(5*x + 2)

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giac [A]  time = 0.43, size = 12, normalized size = 0.75 \begin {gather*} \frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(-log(20*x+8)+exp(4))/(5*x+2),x, algorithm="giac")

[Out]

1/4*e^(e^4)/(5*x + 2)

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maple [A]  time = 0.19, size = 11, normalized size = 0.69

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{4}}}{20 x +8}\)	\(11\)
gosper	\({\mathrm e}^{-\ln \left (20 x +8\right )+{\mathrm e}^{4}}\)	\(13\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{4}}}{20 x +8}\)	\(13\)
norman	\(-\frac {5 \,{\mathrm e}^{{\mathrm e}^{4}} x}{8 \left (5 x +2\right )}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*exp(-ln(20*x+8)+exp(4))/(5*x+2),x,method=_RETURNVERBOSE)

[Out]

1/20*exp(exp(4))/(x+2/5)

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maxima [A]  time = 0.35, size = 12, normalized size = 0.75 \begin {gather*} \frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(-log(20*x+8)+exp(4))/(5*x+2),x, algorithm="maxima")

[Out]

1/4*e^(e^4)/(5*x + 2)

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mupad [B]  time = 0.07, size = 11, normalized size = 0.69 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^4}}{4\,\left (5\,x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(exp(4) - log(20*x + 8)))/(5*x + 2),x)

[Out]

exp(exp(4))/(4*(5*x + 2))

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sympy [A]  time = 0.08, size = 10, normalized size = 0.62 \begin {gather*} \frac {5 e^{e^{4}}}{100 x + 40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(-ln(20*x+8)+exp(4))/(5*x+2),x)

[Out]

5*exp(exp(4))/(100*x + 40)

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