3.25.1 \(\int \frac {5-x}{(2+5 x+3 x^2)^3} \, dx\) [2401]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 = {652, 628, 630, 31} \begin {gather*} \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) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - x)/(2 + 5*x + 3*x^2)^3,x]

[Out]

-1/2*(29 + 35*x)/(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 31

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

Rule 628

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 630

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 652

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), 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 )^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*}

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Mathematica [A]
time = 0.01, size = 57, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.03, size = 48, normalized size = 0.84

method	result	size
risch	\(\frac {945 x^{3}+\frac {4725}{2} x^{2}+1925 x +\frac {1021}{2}}{\left (3 x^{2}+5 x +2\right )^{2}}-315 \ln \left (1+x \right )+315 \ln \left (2+3 x \right )\)	\(45\)
default	\(\frac {3}{\left (1+x \right )^{2}}+\frac {53}{1+x}-315 \ln \left (1+x \right )-\frac {51}{2 \left (2+3 x \right )^{2}}+\frac {156}{2+3 x}+315 \ln \left (2+3 x \right )\)	\(48\)
norman	\(\frac {-\frac {1255}{2} x -\frac {9189}{8} x^{4}-\frac {11535}{4} x^{3}-\frac {18877}{8} x^{2}}{\left (3 x^{2}+5 x +2\right )^{2}}-315 \ln \left (1+x \right )+315 \ln \left (2+3 x \right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.26, size = 54, normalized size = 0.95 \begin {gather*} \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 {gather*}

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 = 2.75, size = 93, normalized size = 1.63 \begin {gather*} \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 {gather*}

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 = 0.06, size = 51, normalized size = 0.89 \begin {gather*} - \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 {gather*}

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*lo
g(x + 1)

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Giac [A]
time = 1.63, size = 46, normalized size = 0.81 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 0.05, size = 45, normalized size = 0.79 \begin {gather*} \frac {105\,x^3+\frac {525\,x^2}{2}+\frac {1925\,x}{9}+\frac {1021}{18}}{x^4+\frac {10\,x^3}{3}+\frac {37\,x^2}{9}+\frac {20\,x}{9}+\frac {4}{9}}-630\,\mathrm {atanh}\left (6\,x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 5)/(5*x + 3*x^2 + 2)^3,x)

[Out]

((1925*x)/9 + (525*x^2)/2 + 105*x^3 + 1021/18)/((20*x)/9 + (37*x^2)/9 + (10*x^3)/3 + x^4 + 4/9) - 630*atanh(6*
x + 5)

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