3.2393 \(\int \frac{5-x}{(3+2 x)^2 (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=66 \[ -\frac{3 (47 x+37)}{5 (2 x+3) \left (3 x^2+5 x+2\right )}-\frac{454}{25 (2 x+3)}+11 \log (x+1)+\frac{812}{125} \log (2 x+3)-\frac{2187}{125} \log (3 x+2) \]

[Out]

-454/(25*(3 + 2*x)) - (3*(37 + 47*x))/(5*(3 + 2*x)*(2 + 5*x + 3*x^2)) + 11*Log[1 + x] + (812*Log[3 + 2*x])/125
 - (2187*Log[2 + 3*x])/125

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Rubi [A]  time = 0.0468649, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {822, 800} \[ -\frac{3 (47 x+37)}{5 (2 x+3) \left (3 x^2+5 x+2\right )}-\frac{454}{25 (2 x+3)}+11 \log (x+1)+\frac{812}{125} \log (2 x+3)-\frac{2187}{125} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-454/(25*(3 + 2*x)) - (3*(37 + 47*x))/(5*(3 + 2*x)*(2 + 5*x + 3*x^2)) + 11*Log[1 + x] + (812*Log[3 + 2*x])/125
 - (2187*Log[2 + 3*x])/125

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])

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]

Rubi steps

\begin{align*} \int \frac{5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^2} \, dx &=-\frac{3 (37+47 x)}{5 (3+2 x) \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \frac{619+564 x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{3 (37+47 x)}{5 (3+2 x) \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \left (-\frac{55}{1+x}-\frac{908}{5 (3+2 x)^2}-\frac{1624}{25 (3+2 x)}+\frac{6561}{25 (2+3 x)}\right ) \, dx\\ &=-\frac{454}{25 (3+2 x)}-\frac{3 (37+47 x)}{5 (3+2 x) \left (2+5 x+3 x^2\right )}+11 \log (1+x)+\frac{812}{125} \log (3+2 x)-\frac{2187}{125} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0280787, size = 57, normalized size = 0.86 \[ \frac{1}{125} \left (-\frac{15 (201 x+151)}{3 x^2+5 x+2}-\frac{260}{2 x+3}-2187 \log (-6 x-4)+1375 \log (-2 (x+1))+812 \log (2 x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-260/(3 + 2*x) - (15*(151 + 201*x))/(2 + 5*x + 3*x^2) - 2187*Log[-4 - 6*x] + 1375*Log[-2*(1 + x)] + 812*Log[3
 + 2*x])/125

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Maple [A]  time = 0.013, size = 49, normalized size = 0.7 \begin{align*} -6\, \left ( 1+x \right ) ^{-1}+11\,\ln \left ( 1+x \right ) -{\frac{52}{75+50\,x}}+{\frac{812\,\ln \left ( 3+2\,x \right ) }{125}}-{\frac{153}{50+75\,x}}-{\frac{2187\,\ln \left ( 2+3\,x \right ) }{125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/(1+x)+11*ln(1+x)-52/25/(3+2*x)+812/125*ln(3+2*x)-153/25/(2+3*x)-2187/125*ln(2+3*x)

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Maxima [A]  time = 1.06862, size = 70, normalized size = 1.06 \begin{align*} -\frac{1362 \, x^{2} + 2975 \, x + 1463}{25 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )}} - \frac{2187}{125} \, \log \left (3 \, x + 2\right ) + \frac{812}{125} \, \log \left (2 \, x + 3\right ) + 11 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(1362*x^2 + 2975*x + 1463)/(6*x^3 + 19*x^2 + 19*x + 6) - 2187/125*log(3*x + 2) + 812/125*log(2*x + 3) +
11*log(x + 1)

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Fricas [A]  time = 1.32404, size = 277, normalized size = 4.2 \begin{align*} -\frac{6810 \, x^{2} + 2187 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) - 812 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (2 \, x + 3\right ) - 1375 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )} \log \left (x + 1\right ) + 14875 \, x + 7315}{125 \,{\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/125*(6810*x^2 + 2187*(6*x^3 + 19*x^2 + 19*x + 6)*log(3*x + 2) - 812*(6*x^3 + 19*x^2 + 19*x + 6)*log(2*x + 3
) - 1375*(6*x^3 + 19*x^2 + 19*x + 6)*log(x + 1) + 14875*x + 7315)/(6*x^3 + 19*x^2 + 19*x + 6)

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Sympy [A]  time = 0.197922, size = 51, normalized size = 0.77 \begin{align*} - \frac{1362 x^{2} + 2975 x + 1463}{150 x^{3} + 475 x^{2} + 475 x + 150} - \frac{2187 \log{\left (x + \frac{2}{3} \right )}}{125} + 11 \log{\left (x + 1 \right )} + \frac{812 \log{\left (x + \frac{3}{2} \right )}}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1362*x**2 + 2975*x + 1463)/(150*x**3 + 475*x**2 + 475*x + 150) - 2187*log(x + 2/3)/125 + 11*log(x + 1) + 812
*log(x + 3/2)/125

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Giac [A]  time = 1.13398, size = 104, normalized size = 1.58 \begin{align*} -\frac{52}{25 \,{\left (2 \, x + 3\right )}} + \frac{6 \,{\left (\frac{1403}{2 \, x + 3} - 903\right )}}{125 \,{\left (\frac{5}{2 \, x + 3} - 3\right )}{\left (\frac{1}{2 \, x + 3} - 1\right )}} + 11 \, \log \left ({\left | -\frac{1}{2 \, x + 3} + 1 \right |}\right ) - \frac{2187}{125} \, \log \left ({\left | -\frac{5}{2 \, x + 3} + 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-52/25/(2*x + 3) + 6/125*(1403/(2*x + 3) - 903)/((5/(2*x + 3) - 3)*(1/(2*x + 3) - 1)) + 11*log(abs(-1/(2*x + 3
) + 1)) - 2187/125*log(abs(-5/(2*x + 3) + 3))