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

Optimal. Leaf size=14 \[ \frac {5}{-1+e^{x (1+\log (16))}} \]

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Rubi [A]  time = 0.07, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 2282, 32} \begin {gather*} -\frac {5}{1-(16 e)^x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(1 - (16*E)^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 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]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.50 \begin {gather*} \frac {5 (1+\log (16))}{\left (-1+(16 e)^x\right ) \log (16 e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(1 + Log[16]))/((-1 + (16*E)^x)*Log[16*E])

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fricas [A]  time = 0.44, size = 14, normalized size = 1.00 \begin {gather*} \frac {5}{e^{\left (4 \, x \log \relax (2) + x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(2)-5)*exp(4*x*log(2)+x)/(exp(4*x*log(2)+x)^2-2*exp(4*x*log(2)+x)+1),x, algorithm="fricas")

[Out]

5/(e^(4*x*log(2) + x) - 1)

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giac [A]  time = 0.16, size = 15, normalized size = 1.07 \begin {gather*} \frac {5}{e^{\left (x {\left (4 \, \log \relax (2) + 1\right )}\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(2)-5)*exp(4*x*log(2)+x)/(exp(4*x*log(2)+x)^2-2*exp(4*x*log(2)+x)+1),x, algorithm="giac")

[Out]

5/(e^(x*(4*log(2) + 1)) - 1)

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maple [A]  time = 0.04, size = 15, normalized size = 1.07

method	result	size
norman	\(\frac {5}{{\mathrm e}^{4 x \ln \relax (2)+x}-1}\)	\(15\)
derivativedivides	\(-\frac {5 \left (-4 \ln \relax (2)-1\right )}{\left (1+4 \ln \relax (2)\right ) \left ({\mathrm e}^{4 x \ln \relax (2)+x}-1\right )}\)	\(29\)
default	\(-\frac {-20 \ln \relax (2)-5}{\left (1+4 \ln \relax (2)\right ) \left ({\mathrm e}^{4 x \ln \relax (2)+x}-1\right )}\)	\(29\)
risch	\(\frac {20 \ln \relax (2)}{\left (1+4 \ln \relax (2)\right ) \left (16^{x} {\mathrm e}^{x}-1\right )}+\frac {5}{\left (1+4 \ln \relax (2)\right ) \left (16^{x} {\mathrm e}^{x}-1\right )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*ln(2)-5)*exp(4*x*ln(2)+x)/(exp(4*x*ln(2)+x)^2-2*exp(4*x*ln(2)+x)+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.61, size = 14, normalized size = 1.00 \begin {gather*} \frac {5}{e^{\left (4 \, x \log \relax (2) + x\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(2)-5)*exp(4*x*log(2)+x)/(exp(4*x*log(2)+x)^2-2*exp(4*x*log(2)+x)+1),x, algorithm="maxima")

[Out]

5/(e^(4*x*log(2) + x) - 1)

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mupad [B]  time = 4.64, size = 144, normalized size = 10.29 \begin {gather*} \frac {10\,\mathrm {atan}\left (\frac {\ln \left (256\right )+2}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}-\frac {2^{4\,x}\,{\mathrm {e}}^x\,\left (\ln \left (256\right )+2\right )}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}\right )}{\sqrt {4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}+\frac {40\,\ln \relax (2)\,\mathrm {atan}\left (\frac {\ln \left (256\right )-2\,2^{4\,x}\,{\mathrm {e}}^x-2\,2^{4\,x}\,{\mathrm {e}}^x\,\ln \left (16\right )+2}{\sqrt {8\,\ln \left (16\right )-4\,\ln \left (256\right )+4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}}\right )}{\sqrt {8\,\ln \left (16\right )-4\,\ln \left (256\right )+4\,{\ln \left (16\right )}^2-{\ln \left (256\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + 4*x*log(2))*(20*log(2) + 5))/(exp(2*x + 8*x*log(2)) - 2*exp(x + 4*x*log(2)) + 1),x)

[Out]

(10*atan((log(256) + 2)/(4*log(16)^2 - log(256)^2)^(1/2) - (2^(4*x)*exp(x)*(log(256) + 2))/(4*log(16)^2 - log(
256)^2)^(1/2)))/(4*log(16)^2 - log(256)^2)^(1/2) + (40*log(2)*atan((log(256) - 2*2^(4*x)*exp(x) - 2*2^(4*x)*ex
p(x)*log(16) + 2)/(8*log(16) - 4*log(256) + 4*log(16)^2 - log(256)^2)^(1/2)))/(8*log(16) - 4*log(256) + 4*log(
16)^2 - log(256)^2)^(1/2)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.86 \begin {gather*} \frac {5}{e^{x + 4 x \log {\relax (2 )}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*ln(2)-5)*exp(4*x*ln(2)+x)/(exp(4*x*ln(2)+x)**2-2*exp(4*x*ln(2)+x)+1),x)

[Out]

5/(exp(x + 4*x*log(2)) - 1)

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