3.5.10 \(\int \frac {-16-268 x-781 x^2-728 x^3-260 x^4+e^2 (24 x+26 x^2-8 x^3)}{-16 x-188 x^2-656 x^3-592 x^4-64 x^5+64 x^6} \, dx\)

Optimal. Leaf size=35 \[ -\frac {3-e^2-x}{-5+(3+4 x)^2}+\log \left (\frac {4 x}{4-x}\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 37, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 4, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2074, 638, 618, 206} \begin {gather*} -\frac {-x-e^2+3}{4 \left (4 x^2+6 x+1\right )}-\log (4-x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 268*x - 781*x^2 - 728*x^3 - 260*x^4 + E^2*(24*x + 26*x^2 - 8*x^3))/(-16*x - 188*x^2 - 656*x^3 - 592
*x^4 - 64*x^5 + 64*x^6),x]

[Out]

-1/4*(3 - E^2 - x)/(1 + 6*x + 4*x^2) - Log[4 - x] + Log[x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{4-x}+\frac {1}{x}+\frac {10-3 e^2+\left (15-4 e^2\right ) x}{2 \left (1+6 x+4 x^2\right )^2}-\frac {1}{4 \left (1+6 x+4 x^2\right )}\right ) \, dx\\ &=-\log (4-x)+\log (x)-\frac {1}{4} \int \frac {1}{1+6 x+4 x^2} \, dx+\frac {1}{2} \int \frac {10-3 e^2+\left (15-4 e^2\right ) x}{\left (1+6 x+4 x^2\right )^2} \, dx\\ &=-\frac {3-e^2-x}{4 \left (1+6 x+4 x^2\right )}-\log (4-x)+\log (x)+\frac {1}{4} \int \frac {1}{1+6 x+4 x^2} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,6+8 x\right )\\ &=-\frac {3-e^2-x}{4 \left (1+6 x+4 x^2\right )}+\frac {\tanh ^{-1}\left (\frac {3+4 x}{\sqrt {5}}\right )}{4 \sqrt {5}}-\log (4-x)+\log (x)-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,6+8 x\right )\\ &=-\frac {3-e^2-x}{4 \left (1+6 x+4 x^2\right )}-\log (4-x)+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 1.03 \begin {gather*} \frac {1}{4} \left (\frac {-3+e^2+x}{1+6 x+4 x^2}-4 \log (4-x)+4 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 268*x - 781*x^2 - 728*x^3 - 260*x^4 + E^2*(24*x + 26*x^2 - 8*x^3))/(-16*x - 188*x^2 - 656*x^3
- 592*x^4 - 64*x^5 + 64*x^6),x]

[Out]

((-3 + E^2 + x)/(1 + 6*x + 4*x^2) - 4*Log[4 - x] + 4*Log[x])/4

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fricas [A]  time = 0.48, size = 53, normalized size = 1.51 \begin {gather*} -\frac {4 \, {\left (4 \, x^{2} + 6 \, x + 1\right )} \log \left (x - 4\right ) - 4 \, {\left (4 \, x^{2} + 6 \, x + 1\right )} \log \relax (x) - x - e^{2} + 3}{4 \, {\left (4 \, x^{2} + 6 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+26*x^2+24*x)*exp(2)-260*x^4-728*x^3-781*x^2-268*x-16)/(64*x^6-64*x^5-592*x^4-656*x^3-188*x^
2-16*x),x, algorithm="fricas")

[Out]

-1/4*(4*(4*x^2 + 6*x + 1)*log(x - 4) - 4*(4*x^2 + 6*x + 1)*log(x) - x - e^2 + 3)/(4*x^2 + 6*x + 1)

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giac [A]  time = 0.29, size = 30, normalized size = 0.86 \begin {gather*} \frac {x + e^{2} - 3}{4 \, {\left (4 \, x^{2} + 6 \, x + 1\right )}} - \log \left ({\left | x - 4 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+26*x^2+24*x)*exp(2)-260*x^4-728*x^3-781*x^2-268*x-16)/(64*x^6-64*x^5-592*x^4-656*x^3-188*x^
2-16*x),x, algorithm="giac")

[Out]

1/4*(x + e^2 - 3)/(4*x^2 + 6*x + 1) - log(abs(x - 4)) + log(abs(x))

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maple [A]  time = 0.12, size = 30, normalized size = 0.86

method	result	size
risch	\(\frac {\frac {x}{16}+\frac {{\mathrm e}^{2}}{16}-\frac {3}{16}}{x^{2}+\frac {3}{2} x +\frac {1}{4}}-\ln \left (x -4\right )+\ln \relax (x )\)	\(30\)
default	\(-\frac {-\frac {x}{4}+\frac {3}{4}-\frac {{\mathrm e}^{2}}{4}}{4 \left (x^{2}+\frac {3}{2} x +\frac {1}{4}\right )}+\ln \relax (x )-\ln \left (x -4\right )\)	\(31\)
norman	\(\frac {\left (\frac {19}{4}-\frac {3 \,{\mathrm e}^{2}}{2}\right ) x +\left (-{\mathrm e}^{2}+3\right ) x^{2}}{4 x^{2}+6 x +1}-\ln \left (x -4\right )+\ln \relax (x )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^3+26*x^2+24*x)*exp(2)-260*x^4-728*x^3-781*x^2-268*x-16)/(64*x^6-64*x^5-592*x^4-656*x^3-188*x^2-16*x
),x,method=_RETURNVERBOSE)

[Out]

(1/16*x+1/16*exp(2)-3/16)/(x^2+3/2*x+1/4)-ln(x-4)+ln(x)

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maxima [A]  time = 0.38, size = 28, normalized size = 0.80 \begin {gather*} \frac {x + e^{2} - 3}{4 \, {\left (4 \, x^{2} + 6 \, x + 1\right )}} - \log \left (x - 4\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+26*x^2+24*x)*exp(2)-260*x^4-728*x^3-781*x^2-268*x-16)/(64*x^6-64*x^5-592*x^4-656*x^3-188*x^
2-16*x),x, algorithm="maxima")

[Out]

1/4*(x + e^2 - 3)/(4*x^2 + 6*x + 1) - log(x - 4) + log(x)

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mupad [B]  time = 0.48, size = 27, normalized size = 0.77 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {x}{2}-1\right )+\frac {x+{\mathrm {e}}^2-3}{16\,x^2+24\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((268*x - exp(2)*(24*x + 26*x^2 - 8*x^3) + 781*x^2 + 728*x^3 + 260*x^4 + 16)/(16*x + 188*x^2 + 656*x^3 + 59
2*x^4 + 64*x^5 - 64*x^6),x)

[Out]

2*atanh(x/2 - 1) + (x + exp(2) - 3)/(24*x + 16*x^2 + 4)

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sympy [A]  time = 1.44, size = 24, normalized size = 0.69 \begin {gather*} - \frac {- x - e^{2} + 3}{16 x^{2} + 24 x + 4} + \log {\relax (x )} - \log {\left (x - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**3+26*x**2+24*x)*exp(2)-260*x**4-728*x**3-781*x**2-268*x-16)/(64*x**6-64*x**5-592*x**4-656*x*
*3-188*x**2-16*x),x)

[Out]

-(-x - exp(2) + 3)/(16*x**2 + 24*x + 4) + log(x) - log(x - 4)

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