3.2401 \(\int \frac{5-x}{(2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=57 \[ \frac{105 (6 x+5)}{2 \left (3 x^2+5 x+2\right )}-\frac{35 x+29}{2 \left (3 x^2+5 x+2\right )^2}-315 \log (x+1)+315 \log (3 x+2) \]

[Out]

-(29 + 35*x)/(2*(2 + 5*x + 3*x^2)^2) + (105*(5 + 6*x))/(2*(2 + 5*x + 3*x^2)) - 315*Log[1 + x] + 315*Log[2 + 3*
x]

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Rubi [A]  time = 0.015495, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {638, 614, 616, 31} \[ \frac{105 (6 x+5)}{2 \left (3 x^2+5 x+2\right )}-\frac{35 x+29}{2 \left (3 x^2+5 x+2\right )^2}-315 \log (x+1)+315 \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[(5 - x)/(2 + 5*x + 3*x^2)^3,x]

[Out]

-(29 + 35*x)/(2*(2 + 5*x + 3*x^2)^2) + (105*(5 + 6*x))/(2*(2 + 5*x + 3*x^2)) - 315*Log[1 + x] + 315*Log[2 + 3*
x]

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 614

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

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 31

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

Rubi steps

\begin{align*} \int \frac{5-x}{\left (2+5 x+3 x^2\right )^3} \, dx &=-\frac{29+35 x}{2 \left (2+5 x+3 x^2\right )^2}-\frac{105}{2} \int \frac{1}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{29+35 x}{2 \left (2+5 x+3 x^2\right )^2}+\frac{105 (5+6 x)}{2 \left (2+5 x+3 x^2\right )}+315 \int \frac{1}{2+5 x+3 x^2} \, dx\\ &=-\frac{29+35 x}{2 \left (2+5 x+3 x^2\right )^2}+\frac{105 (5+6 x)}{2 \left (2+5 x+3 x^2\right )}+945 \int \frac{1}{2+3 x} \, dx-945 \int \frac{1}{3+3 x} \, dx\\ &=-\frac{29+35 x}{2 \left (2+5 x+3 x^2\right )^2}+\frac{105 (5+6 x)}{2 \left (2+5 x+3 x^2\right )}-315 \log (1+x)+315 \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0145249, size = 57, normalized size = 1. \[ \frac{-35 x-29}{2 \left (3 x^2+5 x+2\right )^2}+\frac{105 (6 x+5)}{2 \left (3 x^2+5 x+2\right )}-315 \log (x+1)+315 \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-29 - 35*x)/(2*(2 + 5*x + 3*x^2)^2) + (105*(5 + 6*x))/(2*(2 + 5*x + 3*x^2)) - 315*Log[1 + x] + 315*Log[2 + 3*
x]

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Maple [A]  time = 0.007, size = 48, normalized size = 0.8 \begin{align*} 3\, \left ( 1+x \right ) ^{-2}+53\, \left ( 1+x \right ) ^{-1}-315\,\ln \left ( 1+x \right ) -{\frac{51}{2\, \left ( 2+3\,x \right ) ^{2}}}+156\, \left ( 2+3\,x \right ) ^{-1}+315\,\ln \left ( 2+3\,x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(1+x)^2+53/(1+x)-315*ln(1+x)-51/2/(2+3*x)^2+156/(2+3*x)+315*ln(2+3*x)

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Maxima [A]  time = 1.00078, size = 73, normalized size = 1.28 \begin{align*} \frac{1890 \, x^{3} + 4725 \, x^{2} + 3850 \, x + 1021}{2 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + 315 \, \log \left (3 \, x + 2\right ) - 315 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1890*x^3 + 4725*x^2 + 3850*x + 1021)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 315*log(3*x + 2) - 315*log(x
+ 1)

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Fricas [A]  time = 1.25189, size = 257, normalized size = 4.51 \begin{align*} \frac{1890 \, x^{3} + 4725 \, x^{2} + 630 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 630 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 3850 \, x + 1021}{2 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1890*x^3 + 4725*x^2 + 630*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x + 2) - 630*(9*x^4 + 30*x^3 + 37*x^
2 + 20*x + 4)*log(x + 1) + 3850*x + 1021)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)

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Sympy [A]  time = 0.1704, size = 49, normalized size = 0.86 \begin{align*} \frac{1890 x^{3} + 4725 x^{2} + 3850 x + 1021}{18 x^{4} + 60 x^{3} + 74 x^{2} + 40 x + 8} + 315 \log{\left (x + \frac{2}{3} \right )} - 315 \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1890*x**3 + 4725*x**2 + 3850*x + 1021)/(18*x**4 + 60*x**3 + 74*x**2 + 40*x + 8) + 315*log(x + 2/3) - 315*log(
x + 1)

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Giac [A]  time = 1.18099, size = 62, normalized size = 1.09 \begin{align*} \frac{1890 \, x^{3} + 4725 \, x^{2} + 3850 \, x + 1021}{2 \,{\left (3 \, x^{2} + 5 \, x + 2\right )}^{2}} + 315 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 315 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1890*x^3 + 4725*x^2 + 3850*x + 1021)/(3*x^2 + 5*x + 2)^2 + 315*log(abs(3*x + 2)) - 315*log(abs(x + 1))