3.16.85 \(\int \frac {-48+17 x^2+e^x (48+96 x+103 x^2+34 x^3)}{-48 x-60 x^2-17 x^3+e^x (48 x+60 x^2+17 x^3)} \, dx\)

Optimal. Leaf size=34 \[ \log \left (\frac {\left (2-2 e^x\right )^2}{4+x+\frac {3-\frac {x}{4}+\frac {x^2}{16}}{x}}\right ) \]

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Rubi [A]  time = 0.73, antiderivative size = 26, normalized size of antiderivative = 0.76, number of steps used = 10, number of rules used = 8, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.131, Rules used = {6741, 6728, 2282, 36, 31, 29, 1628, 628} \begin {gather*} -\log \left (17 x^2+60 x+48\right )+2 \log \left (1-e^x\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48 + 17*x^2 + E^x*(48 + 96*x + 103*x^2 + 34*x^3))/(-48*x - 60*x^2 - 17*x^3 + E^x*(48*x + 60*x^2 + 17*x^3
)),x]

[Out]

2*Log[1 - E^x] + Log[x] - Log[48 + 60*x + 17*x^2]

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 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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {48-17 x^2-e^x \left (48+96 x+103 x^2+34 x^3\right )}{\left (1-e^x\right ) x \left (48+60 x+17 x^2\right )} \, dx\\ &=\int \left (\frac {2}{-1+e^x}+\frac {48+96 x+103 x^2+34 x^3}{x \left (48+60 x+17 x^2\right )}\right ) \, dx\\ &=2 \int \frac {1}{-1+e^x} \, dx+\int \frac {48+96 x+103 x^2+34 x^3}{x \left (48+60 x+17 x^2\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{(-1+x) x} \, dx,x,e^x\right )+\int \left (2+\frac {1}{x}-\frac {2 (30+17 x)}{48+60 x+17 x^2}\right ) \, dx\\ &=2 x+\log (x)-2 \int \frac {30+17 x}{48+60 x+17 x^2} \, dx+2 \operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,e^x\right )-2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )\\ &=2 \log \left (1-e^x\right )+\log (x)-\log \left (48+60 x+17 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 0.76 \begin {gather*} 2 \log \left (1-e^x\right )+\log (x)-\log \left (48+60 x+17 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48 + 17*x^2 + E^x*(48 + 96*x + 103*x^2 + 34*x^3))/(-48*x - 60*x^2 - 17*x^3 + E^x*(48*x + 60*x^2 +
17*x^3)),x]

[Out]

2*Log[1 - E^x] + Log[x] - Log[48 + 60*x + 17*x^2]

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fricas [A]  time = 0.88, size = 23, normalized size = 0.68 \begin {gather*} -\log \left (17 \, x^{2} + 60 \, x + 48\right ) + \log \relax (x) + 2 \, \log \left (e^{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((34*x^3+103*x^2+96*x+48)*exp(x)+17*x^2-48)/((17*x^3+60*x^2+48*x)*exp(x)-17*x^3-60*x^2-48*x),x, algo
rithm="fricas")

[Out]

-log(17*x^2 + 60*x + 48) + log(x) + 2*log(e^x - 1)

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giac [A]  time = 0.21, size = 23, normalized size = 0.68 \begin {gather*} -\log \left (17 \, x^{2} + 60 \, x + 48\right ) + \log \relax (x) + 2 \, \log \left (e^{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((34*x^3+103*x^2+96*x+48)*exp(x)+17*x^2-48)/((17*x^3+60*x^2+48*x)*exp(x)-17*x^3-60*x^2-48*x),x, algo
rithm="giac")

[Out]

-log(17*x^2 + 60*x + 48) + log(x) + 2*log(e^x - 1)

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maple [A]  time = 0.04, size = 24, normalized size = 0.71

method	result	size
norman	\(2 \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (17 x^{2}+60 x +48\right )+\ln \relax (x )\)	\(24\)
risch	\(2 \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (17 x^{2}+60 x +48\right )+\ln \relax (x )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((34*x^3+103*x^2+96*x+48)*exp(x)+17*x^2-48)/((17*x^3+60*x^2+48*x)*exp(x)-17*x^3-60*x^2-48*x),x,method=_RET
URNVERBOSE)

[Out]

2*ln(exp(x)-1)-ln(17*x^2+60*x+48)+ln(x)

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maxima [A]  time = 0.48, size = 23, normalized size = 0.68 \begin {gather*} -\log \left (17 \, x^{2} + 60 \, x + 48\right ) + \log \relax (x) + 2 \, \log \left (e^{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((34*x^3+103*x^2+96*x+48)*exp(x)+17*x^2-48)/((17*x^3+60*x^2+48*x)*exp(x)-17*x^3-60*x^2-48*x),x, algo
rithm="maxima")

[Out]

-log(17*x^2 + 60*x + 48) + log(x) + 2*log(e^x - 1)

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mupad [B]  time = 0.42, size = 23, normalized size = 0.68 \begin {gather*} 2\,\ln \left ({\mathrm {e}}^x-1\right )-\ln \left (17\,x^2+60\,x+48\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(17*x^2 + exp(x)*(96*x + 103*x^2 + 34*x^3 + 48) - 48)/(48*x + 60*x^2 + 17*x^3 - exp(x)*(48*x + 60*x^2 + 1
7*x^3)),x)

[Out]

2*log(exp(x) - 1) - log(60*x + 17*x^2 + 48) + log(x)

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sympy [A]  time = 0.16, size = 22, normalized size = 0.65 \begin {gather*} \log {\relax (x )} + 2 \log {\left (e^{x} - 1 \right )} - \log {\left (17 x^{2} + 60 x + 48 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((34*x**3+103*x**2+96*x+48)*exp(x)+17*x**2-48)/((17*x**3+60*x**2+48*x)*exp(x)-17*x**3-60*x**2-48*x),
x)

[Out]

log(x) + 2*log(exp(x) - 1) - log(17*x**2 + 60*x + 48)

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