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

Optimal. Leaf size=15 \[ \frac {4 e^{16}}{5 \log (-1-x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2390, 2302, 30} \begin {gather*} \frac {4 e^{16}}{5 \log (-x-1)} \end {gather*}

Antiderivative was successfully verified.

[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 2302

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 2390

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 {gather*} \begin {aligned} \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 ) \operatorname {Subst}\left (\int -\frac {1}{5 x \log ^2(x)} \, dx,x,-1-x\right )\\ &=-\left (\frac {1}{5} \left (4 e^{16}\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-1-x\right )\right )\\ &=-\left (\frac {1}{5} \left (4 e^{16}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-1-x)\right )\right )\\ &=\frac {4 e^{16}}{5 \log (-1-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {4 e^{16}}{5 \log (-1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E^16)/(5*Log[-1 - x])

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fricas [A]  time = 0.60, size = 12, normalized size = 0.80 \begin {gather*} \frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*e^16/log(-x - 1)

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giac [A]  time = 0.12, size = 12, normalized size = 0.80 \begin {gather*} \frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*e^16/log(-x - 1)

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maple [A]  time = 0.08, size = 13, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 12, normalized size = 0.80 \begin {gather*} \frac {4 \, e^{16}}{5 \, \log \left (-x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*e^16/log(-x - 1)

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mupad [B]  time = 0.31, size = 12, normalized size = 0.80 \begin {gather*} \frac {4\,{\mathrm {e}}^{16}}{5\,\ln \left (-x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(16))/(5*log(- x - 1))

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sympy [A]  time = 0.09, size = 12, normalized size = 0.80 \begin {gather*} \frac {4 e^{16}}{5 \log {\left (- x - 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*exp(16)/(5*log(-x - 1))

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