3.69.14 $\int \frac{-e^{-x}+8e^{8x}}{-5+e^{-x}+e^{8x}}dx$

Optimal. Leaf size=17 $$\log (5-e^{-x}-e^{8x})$$

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Rubi [A]  time = 0.14, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.100, Rules used = {2282, 6742, 1587} $$\begin{array}{c}\log (-5e^x+e^{9x}+1)-x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + 8*E^(8*x))/(-5 + E^(-x) + E^(8*x)),x]

[Out]

-x + Log[1 - 5*E^x + E^(9*x)]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\mathrm{Subst}(\int \frac{-1+8x^9}{x(1-5x+x^9)}dx,x,e^x)\\ & =\mathrm{Subst}(\int (-\frac{1}{x}+\frac{-5+9x^8}{1-5x+x^9})dx,x,e^x)\\ & =-x+\mathrm{Subst}(\int \frac{-5+9x^8}{1-5x+x^9}dx,x,e^x)\\ & =-x+\log (1-5e^x+e^{9x})\end{array}\end{array}$$

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Mathematica [A]  time = 0.06, size = 17, normalized size = 1.00 $$\begin{array}{c}-x+\log (1-5e^x+e^{9x})\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

-x + Log[1 - 5*E^x + E^(9*x)]

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fricas [A]  time = 0.60, size = 17, normalized size = 1.00 $$\begin{array}{c}8x+\log (-5e^{(-8x)}+e^{(-9x)}+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x + log(-5*e^(-8*x) + e^(-9*x) + 1)

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giac [A]  time = 0.24, size = 16, normalized size = 0.94 $$\begin{array}{c}-x+\log \left(|e^{\left(9x\right)}-5e^x+1|\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + log(abs(e^(9*x) - 5*e^x + 1))

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maple [A]  time = 0.04, size = 12, normalized size = 0.71

method	result	size
derivativedivides	$\ln ({\mathrm{e}}^{8x}+{\mathrm{e}}^{-x}-5)$	$12$
default	$\ln ({\mathrm{e}}^{8x}+{\mathrm{e}}^{-x}-5)$	$12$
risch	$8x+\ln ({\mathrm{e}}^{-9x}-5{\mathrm{e}}^{-8x}+1)$	$18$
norman	$8x+\ln ({\mathrm{e}}^{-9x}-5{\mathrm{e}}^{-8x}+1)$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(8*x)-exp(-x))/(exp(8*x)+exp(-x)-5),x,method=_RETURNVERBOSE)

[Out]

ln(exp(8*x)+exp(-x)-5)

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maxima [A]  time = 0.35, size = 11, normalized size = 0.65 $$\begin{array}{c}\log (e^{\left(8x\right)}+e^{(-x)}-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(8*x) + e^(-x) - 5)

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mupad [B]  time = 0.09, size = 15, normalized size = 0.88 $$\begin{array}{c}\ln ({\mathrm{e}}^{9x}-5{\mathrm{e}}^x+1)-x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x) - 8*exp(8*x))/(exp(-x) + exp(8*x) - 5),x)

[Out]

log(exp(9*x) - 5*exp(x) + 1) - x

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sympy [A]  time = 0.13, size = 19, normalized size = 1.12 $$\begin{array}{c}8x+\log (1-5e^{-8x}+e^{-9x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(8*x)-exp(-x))/(exp(8*x)+exp(-x)-5),x)

[Out]

8*x + log(1 - 5*exp(-8*x) + exp(-9*x))

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