\(\int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx\) [875]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{e^4}}{4 (2+5 x)} \]

[Out]

exp(-ln(20*x+8)+exp(4))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 21, 32} \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{e^4}}{4 (5 x+2)} \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{e^4}}{4 (2+5 x)} \]

[In]

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

[Out]

E^E^4/(4*(2 + 5*x))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \]

[In]

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

[Out]

1/4*e^(e^4)/(5*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {5 e^{e^{4}}}{100 x + 40} \]

[In]

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

[Out]

5*exp(exp(4))/(100*x + 40)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \]

[In]

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

[Out]

1/4*e^(e^4)/(5*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {e^{\left (e^{4}\right )}}{4 \, {\left (5 \, x + 2\right )}} \]

[In]

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

[Out]

1/4*e^(e^4)/(5*x + 2)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int -\frac {5 e^{e^4}}{(2+5 x) (8+20 x)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^4}}{4\,\left (5\,x+2\right )} \]

[In]

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

[Out]

exp(exp(4))/(4*(5*x + 2))