3.91.2 \(\int \frac {16+e^{8 x} (-1+8 x)}{-16 x+e^{8 x} x} \, dx\)

Optimal. Leaf size=12 \[ \log \left (\frac {-16+e^{8 x}}{x}\right ) \]

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Rubi [B]  time = 0.23, antiderivative size = 49, normalized size of antiderivative = 4.08, number of steps used = 20, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6742, 2282, 36, 31, 29, 800, 628, 43} \begin {gather*} \log \left (2-e^{2 x}\right )+\log \left (e^{2 x}+2\right )+\log \left (-2 e^x+e^{2 x}+2\right )+\log \left (2 e^x+e^{2 x}+2\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + E^(8*x)*(-1 + 8*x))/(-16*x + E^(8*x)*x),x]

[Out]

Log[2 - E^(2*x)] + Log[2 + E^(2*x)] + Log[2 - 2*E^x + E^(2*x)] + Log[2 + 2*E^x + E^(2*x)] - Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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]]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4}{-2+e^{2 x}}-\frac {4}{2+e^{2 x}}+\frac {2 \left (-2+e^x\right )}{2-2 e^x+e^{2 x}}-\frac {2 \left (2+e^x\right )}{2+2 e^x+e^{2 x}}+\frac {-1+8 x}{x}\right ) \, dx\\ &=2 \int \frac {-2+e^x}{2-2 e^x+e^{2 x}} \, dx-2 \int \frac {2+e^x}{2+2 e^x+e^{2 x}} \, dx+4 \int \frac {1}{-2+e^{2 x}} \, dx-4 \int \frac {1}{2+e^{2 x}} \, dx+\int \frac {-1+8 x}{x} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{(-2+x) x} \, dx,x,e^{2 x}\right )-2 \operatorname {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^{2 x}\right )+2 \operatorname {Subst}\left (\int \frac {-2+x}{x \left (2-2 x+x^2\right )} \, dx,x,e^x\right )-2 \operatorname {Subst}\left (\int \frac {2+x}{x \left (2+2 x+x^2\right )} \, dx,x,e^x\right )+\int \left (8-\frac {1}{x}\right ) \, dx\\ &=8 x-\log (x)+2 \operatorname {Subst}\left (\int \left (-\frac {1}{x}+\frac {-1+x}{2-2 x+x^2}\right ) \, dx,x,e^x\right )-2 \operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {-1-x}{2+2 x+x^2}\right ) \, dx,x,e^x\right )+\operatorname {Subst}\left (\int \frac {1}{-2+x} \, dx,x,e^{2 x}\right )-2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^{2 x}\right )\\ &=\log \left (2-e^{2 x}\right )+\log \left (2+e^{2 x}\right )-\log (x)+2 \operatorname {Subst}\left (\int \frac {-1+x}{2-2 x+x^2} \, dx,x,e^x\right )-2 \operatorname {Subst}\left (\int \frac {-1-x}{2+2 x+x^2} \, dx,x,e^x\right )\\ &=\log \left (2-e^{2 x}\right )+\log \left (2+e^{2 x}\right )+\log \left (2-2 e^x+e^{2 x}\right )+\log \left (2+2 e^x+e^{2 x}\right )-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 15, normalized size = 1.25 \begin {gather*} \log \left (16-e^{8 x}\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + E^(8*x)*(-1 + 8*x))/(-16*x + E^(8*x)*x),x]

[Out]

Log[16 - E^(8*x)] - Log[x]

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fricas [A]  time = 0.51, size = 12, normalized size = 1.00 \begin {gather*} -\log \relax (x) + \log \left (e^{\left (8 \, x\right )} - 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-1)*exp(8*x)+16)/(x*exp(8*x)-16*x),x, algorithm="fricas")

[Out]

-log(x) + log(e^(8*x) - 16)

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giac [A]  time = 0.16, size = 12, normalized size = 1.00 \begin {gather*} -\log \relax (x) + \log \left (e^{\left (8 \, x\right )} - 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-1)*exp(8*x)+16)/(x*exp(8*x)-16*x),x, algorithm="giac")

[Out]

-log(x) + log(e^(8*x) - 16)

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maple [A]  time = 0.14, size = 13, normalized size = 1.08

method	result	size
norman	\(-\ln \relax (x )+\ln \left ({\mathrm e}^{8 x}-16\right )\)	\(13\)
risch	\(-\ln \relax (x )+\ln \left ({\mathrm e}^{8 x}-16\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x-1)*exp(8*x)+16)/(x*exp(8*x)-16*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(exp(8*x)-16)

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maxima [B]  time = 0.48, size = 41, normalized size = 3.42 \begin {gather*} -\log \relax (x) + \log \left (e^{\left (2 \, x\right )} + 2 \, e^{x} + 2\right ) + \log \left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 2\right ) + \log \left (e^{\left (2 \, x\right )} + 2\right ) + \log \left (e^{\left (2 \, x\right )} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-1)*exp(8*x)+16)/(x*exp(8*x)-16*x),x, algorithm="maxima")

[Out]

-log(x) + log(e^(2*x) + 2*e^x + 2) + log(e^(2*x) - 2*e^x + 2) + log(e^(2*x) + 2) + log(e^(2*x) - 2)

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mupad [B]  time = 5.95, size = 12, normalized size = 1.00 \begin {gather*} \ln \left ({\mathrm {e}}^{8\,x}-16\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8*x)*(8*x - 1) + 16)/(16*x - x*exp(8*x)),x)

[Out]

log(exp(8*x) - 16) - log(x)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.83 \begin {gather*} - \log {\relax (x )} + \log {\left (e^{8 x} - 16 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-1)*exp(8*x)+16)/(x*exp(8*x)-16*x),x)

[Out]

-log(x) + log(exp(8*x) - 16)

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