3.2392 \(\int \frac {5-x}{(3+2 x) (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac {3 (47 x+37)}{5 \left (3 x^2+5 x+2\right )}+23 \log (x+1)+\frac {52}{25} \log (2 x+3)-\frac {627}{25} \log (3 x+2) \]

[Out]

-3/5*(37+47*x)/(3*x^2+5*x+2)+23*ln(1+x)+52/25*ln(3+2*x)-627/25*ln(2+3*x)

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {822, 800} \[ -\frac {3 (47 x+37)}{5 \left (3 x^2+5 x+2\right )}+23 \log (x+1)+\frac {52}{25} \log (2 x+3)-\frac {627}{25} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) + 23*Log[1 + x] + (52*Log[3 + 2*x])/25 - (627*Log[2 + 3*x])/25

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 822

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

Rubi steps

\begin {align*} \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {3 (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-\frac {1}{5} \int \frac {397+282 x}{(3+2 x) \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac {3 (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-\frac {1}{5} \int \left (-\frac {115}{1+x}-\frac {104}{5 (3+2 x)}+\frac {1881}{5 (2+3 x)}\right ) \, dx\\ &=-\frac {3 (37+47 x)}{5 \left (2+5 x+3 x^2\right )}+23 \log (1+x)+\frac {52}{25} \log (3+2 x)-\frac {627}{25} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 1.00 \[ \frac {1}{25} \left (-\frac {15 (47 x+37)}{3 x^2+5 x+2}-627 \log (-6 x-4)+575 \log (-2 (x+1))+52 \log (2 x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-15*(37 + 47*x))/(2 + 5*x + 3*x^2) - 627*Log[-4 - 6*x] + 575*Log[-2*(1 + x)] + 52*Log[3 + 2*x])/25

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fricas [A]  time = 0.83, size = 71, normalized size = 1.48 \[ -\frac {627 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 52 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (2 \, x + 3\right ) - 575 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 705 \, x + 555}{25 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(627*(3*x^2 + 5*x + 2)*log(3*x + 2) - 52*(3*x^2 + 5*x + 2)*log(2*x + 3) - 575*(3*x^2 + 5*x + 2)*log(x +
1) + 705*x + 555)/(3*x^2 + 5*x + 2)

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giac [A]  time = 0.15, size = 45, normalized size = 0.94 \[ -\frac {3 \, {\left (47 \, x + 37\right )}}{5 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )}} - \frac {627}{25} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {52}{25} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) + 23 \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*(47*x + 37)/((3*x + 2)*(x + 1)) - 627/25*log(abs(3*x + 2)) + 52/25*log(abs(2*x + 3)) + 23*log(abs(x + 1))

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maple [A]  time = 0.05, size = 40, normalized size = 0.83 \[ -\frac {627 \ln \left (3 x +2\right )}{25}+\frac {52 \ln \left (2 x +3\right )}{25}+23 \ln \left (x +1\right )-\frac {51}{5 \left (3 x +2\right )}-\frac {6}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-51/5/(3*x+2)-627/25*ln(3*x+2)+52/25*ln(2*x+3)-6/(x+1)+23*ln(x+1)

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maxima [A]  time = 0.46, size = 42, normalized size = 0.88 \[ -\frac {3 \, {\left (47 \, x + 37\right )}}{5 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} - \frac {627}{25} \, \log \left (3 \, x + 2\right ) + \frac {52}{25} \, \log \left (2 \, x + 3\right ) + 23 \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*(47*x + 37)/(3*x^2 + 5*x + 2) - 627/25*log(3*x + 2) + 52/25*log(2*x + 3) + 23*log(x + 1)

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mupad [B]  time = 0.04, size = 36, normalized size = 0.75 \[ 23\,\ln \left (x+1\right )-\frac {627\,\ln \left (x+\frac {2}{3}\right )}{25}+\frac {52\,\ln \left (x+\frac {3}{2}\right )}{25}-\frac {\frac {47\,x}{5}+\frac {37}{5}}{x^2+\frac {5\,x}{3}+\frac {2}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

23*log(x + 1) - (627*log(x + 2/3))/25 + (52*log(x + 3/2))/25 - ((47*x)/5 + 37/5)/((5*x)/3 + x^2 + 2/3)

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sympy [A]  time = 0.18, size = 41, normalized size = 0.85 \[ - \frac {141 x + 111}{15 x^{2} + 25 x + 10} - \frac {627 \log {\left (x + \frac {2}{3} \right )}}{25} + 23 \log {\left (x + 1 \right )} + \frac {52 \log {\left (x + \frac {3}{2} \right )}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(141*x + 111)/(15*x**2 + 25*x + 10) - 627*log(x + 2/3)/25 + 23*log(x + 1) + 52*log(x + 3/2)/25

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