\(\int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx\) [6759]

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 e^{16}}{5 \log (-1-x)} \]

[Out]

4/5*exp(16)/ln(-1-x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2437, 2339, 30} \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 e^{16}}{5 \log (-x-1)} \]

[In]

Int[(-4*E^16)/((5 + 5*x)*Log[-1 - x]^2),x]

[Out]

(4*E^16)/(5*Log[-1 - 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (4 e^{16}\right ) \int \frac {1}{(5+5 x) \log ^2(-1-x)} \, dx\right ) \\ & = \left (4 e^{16}\right ) \text {Subst}\left (\int -\frac {1}{5 x \log ^2(x)} \, dx,x,-1-x\right ) \\ & = -\left (\frac {1}{5} \left (4 e^{16}\right ) \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-1-x\right )\right ) \\ & = -\left (\frac {1}{5} \left (4 e^{16}\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-1-x)\right )\right ) \\ & = \frac {4 e^{16}}{5 \log (-1-x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 e^{16}}{5 \log (-1-x)} \]

[In]

Integrate[(-4*E^16)/((5 + 5*x)*Log[-1 - x]^2),x]

[Out]

(4*E^16)/(5*Log[-1 - x])

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {4 \,{\mathrm e}^{16}}{5 \ln \left (-1-x \right )}\)	\(13\)
default	\(\frac {4 \,{\mathrm e}^{16}}{5 \ln \left (-1-x \right )}\)	\(13\)
norman	\(\frac {4 \,{\mathrm e}^{16}}{5 \ln \left (-1-x \right )}\)	\(13\)
risch	\(\frac {4 \,{\mathrm e}^{16}}{5 \ln \left (-1-x \right )}\)	\(13\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{16}}{5 \ln \left (-1-x \right )}\)	\(13\)

[In]

int(-4*exp(16)/(5*x+5)/ln(-1-x)^2,x,method=_RETURNVERBOSE)

[Out]

4/5*exp(16)/ln(-1-x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \]

[In]

integrate(-4*exp(16)/(5*x+5)/log(-1-x)^2,x, algorithm="fricas")

[Out]

4/5*e^16/log(-x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 e^{16}}{5 \log {\left (- x - 1 \right )}} \]

[In]

integrate(-4*exp(16)/(5*x+5)/ln(-1-x)**2,x)

[Out]

4*exp(16)/(5*log(-x - 1))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \]

[In]

integrate(-4*exp(16)/(5*x+5)/log(-1-x)^2,x, algorithm="maxima")

[Out]

4/5*e^16/log(-x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \]

[In]

integrate(-4*exp(16)/(5*x+5)/log(-1-x)^2,x, algorithm="giac")

[Out]

4/5*e^16/log(-x - 1)

Mupad [B] (verification not implemented)

Time = 11.40 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{16}}{(5+5 x) \log ^2(-1-x)} \, dx=\frac {4\,{\mathrm {e}}^{16}}{5\,\ln \left (-x-1\right )} \]

[In]

int(-(4*exp(16))/(log(- x - 1)^2*(5*x + 5)),x)

[Out]

(4*exp(16))/(5*log(- x - 1))