3.22.50 \(\int \frac {5-x}{(2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac {35 x+29}{3 x^2+5 x+2}+35 \log (x+1)-35 \log (3 x+2) \]

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {638, 616, 31} \begin {gather*} -\frac {35 x+29}{3 x^2+5 x+2}+35 \log (x+1)-35 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - x)/(2 + 5*x + 3*x^2)^2,x]

[Out]

-((29 + 35*x)/(2 + 5*x + 3*x^2)) + 35*Log[1 + x] - 35*Log[2 + 3*x]

Rule 31

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

Rule 616

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

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]

Rubi steps

\begin {align*} \int \frac {5-x}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {29+35 x}{2+5 x+3 x^2}-35 \int \frac {1}{2+5 x+3 x^2} \, dx\\ &=-\frac {29+35 x}{2+5 x+3 x^2}-105 \int \frac {1}{2+3 x} \, dx+105 \int \frac {1}{3+3 x} \, dx\\ &=-\frac {29+35 x}{2+5 x+3 x^2}+35 \log (1+x)-35 \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.97 \begin {gather*} \frac {-35 x-29}{3 x^2+5 x+2}+35 \log (x+1)-35 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)/(2 + 5*x + 3*x^2)^2,x]

[Out]

(-29 - 35*x)/(2 + 5*x + 3*x^2) + 35*Log[1 + x] - 35*Log[2 + 3*x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5-x}{\left (2+5 x+3 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(5 - x)/(2 + 5*x + 3*x^2)^2,x]

[Out]

IntegrateAlgebraic[(5 - x)/(2 + 5*x + 3*x^2)^2, x]

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fricas [A]  time = 0.38, size = 53, normalized size = 1.56 \begin {gather*} -\frac {35 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 35 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 35 \, x + 29}{3 \, x^{2} + 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2)^2,x, algorithm="fricas")

[Out]

-(35*(3*x^2 + 5*x + 2)*log(3*x + 2) - 35*(3*x^2 + 5*x + 2)*log(x + 1) + 35*x + 29)/(3*x^2 + 5*x + 2)

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giac [A]  time = 0.15, size = 36, normalized size = 1.06 \begin {gather*} -\frac {35 \, x + 29}{3 \, x^{2} + 5 \, x + 2} - 35 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + 35 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-(35*x + 29)/(3*x^2 + 5*x + 2) - 35*log(abs(3*x + 2)) + 35*log(abs(x + 1))

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maple [A]  time = 0.05, size = 32, normalized size = 0.94 \begin {gather*} -35 \ln \left (3 x +2\right )+35 \ln \left (x +1\right )-\frac {17}{3 x +2}-\frac {6}{x +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)/(3*x^2+5*x+2)^2,x)

[Out]

-17/(3*x+2)-35*ln(3*x+2)-6/(x+1)+35*ln(x+1)

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maxima [A]  time = 0.53, size = 34, normalized size = 1.00 \begin {gather*} -\frac {35 \, x + 29}{3 \, x^{2} + 5 \, x + 2} - 35 \, \log \left (3 \, x + 2\right ) + 35 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-(35*x + 29)/(3*x^2 + 5*x + 2) - 35*log(3*x + 2) + 35*log(x + 1)

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mupad [B]  time = 2.26, size = 26, normalized size = 0.76 \begin {gather*} 70\,\mathrm {atanh}\left (6\,x+5\right )-\frac {\frac {35\,x}{3}+\frac {29}{3}}{x^2+\frac {5\,x}{3}+\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 5)/(5*x + 3*x^2 + 2)^2,x)

[Out]

70*atanh(6*x + 5) - ((35*x)/3 + 29/3)/((5*x)/3 + x^2 + 2/3)

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sympy [A]  time = 0.13, size = 29, normalized size = 0.85 \begin {gather*} - \frac {35 x + 29}{3 x^{2} + 5 x + 2} - 35 \log {\left (x + \frac {2}{3} \right )} + 35 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x**2+5*x+2)**2,x)

[Out]

-(35*x + 29)/(3*x**2 + 5*x + 2) - 35*log(x + 2/3) + 35*log(x + 1)

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