3.2403 \(\int \frac{5-x}{(3+2 x)^2 (2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=94 \[ -\frac{3 (47 x+37)}{10 (2 x+3) \left (3 x^2+5 x+2\right )^2}+\frac{10848 x+9293}{50 (2 x+3) \left (3 x^2+5 x+2\right )}+\frac{12946}{125 (2 x+3)}-175 \log (x+1)+\frac{4912}{625} \log (2 x+3)+\frac{104463}{625} \log (3 x+2) \]

[Out]

12946/(125*(3 + 2*x)) - (3*(37 + 47*x))/(10*(3 + 2*x)*(2 + 5*x + 3*x^2)^2) + (9293 + 10848*x)/(50*(3 + 2*x)*(2
 + 5*x + 3*x^2)) - 175*Log[1 + x] + (4912*Log[3 + 2*x])/625 + (104463*Log[2 + 3*x])/625

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Rubi [A]  time = 0.0599182, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {822, 800} \[ -\frac{3 (47 x+37)}{10 (2 x+3) \left (3 x^2+5 x+2\right )^2}+\frac{10848 x+9293}{50 (2 x+3) \left (3 x^2+5 x+2\right )}+\frac{12946}{125 (2 x+3)}-175 \log (x+1)+\frac{4912}{625} \log (2 x+3)+\frac{104463}{625} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

12946/(125*(3 + 2*x)) - (3*(37 + 47*x))/(10*(3 + 2*x)*(2 + 5*x + 3*x^2)^2) + (9293 + 10848*x)/(50*(3 + 2*x)*(2
 + 5*x + 3*x^2)) - 175*Log[1 + x] + (4912*Log[3 + 2*x])/625 + (104463*Log[2 + 3*x])/625

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 )^3} \, dx &=-\frac{3 (37+47 x)}{10 (3+2 x) \left (2+5 x+3 x^2\right )^2}-\frac{1}{10} \int \frac{1439+1128 x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{3 (37+47 x)}{10 (3+2 x) \left (2+5 x+3 x^2\right )^2}+\frac{9293+10848 x}{50 (3+2 x) \left (2+5 x+3 x^2\right )}+\frac{1}{50} \int \frac{52142+43392 x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{3 (37+47 x)}{10 (3+2 x) \left (2+5 x+3 x^2\right )^2}+\frac{9293+10848 x}{50 (3+2 x) \left (2+5 x+3 x^2\right )}+\frac{1}{50} \int \left (-\frac{8750}{1+x}-\frac{51784}{5 (3+2 x)^2}+\frac{19648}{25 (3+2 x)}+\frac{626778}{25 (2+3 x)}\right ) \, dx\\ &=\frac{12946}{125 (3+2 x)}-\frac{3 (37+47 x)}{10 (3+2 x) \left (2+5 x+3 x^2\right )^2}+\frac{9293+10848 x}{50 (3+2 x) \left (2+5 x+3 x^2\right )}-175 \log (1+x)+\frac{4912}{625} \log (3+2 x)+\frac{104463}{625} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0459667, size = 78, normalized size = 0.83 \[ \frac{1}{625} \left (-\frac{75 (201 x+151)}{2 \left (3 x^2+5 x+2\right )^2}+\frac{5 (39462 x+33697)}{6 x^2+10 x+4}-\frac{1040}{2 x+3}+104463 \log (-6 x-4)-109375 \log (-2 (x+1))+4912 \log (2 x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1040/(3 + 2*x) - (75*(151 + 201*x))/(2*(2 + 5*x + 3*x^2)^2) + (5*(33697 + 39462*x))/(4 + 10*x + 6*x^2) + 104
463*Log[-4 - 6*x] - 109375*Log[-2*(1 + x)] + 4912*Log[3 + 2*x])/625

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Maple [A]  time = 0.013, size = 65, normalized size = 0.7 \begin{align*} 3\, \left ( 1+x \right ) ^{-2}+29\, \left ( 1+x \right ) ^{-1}-175\,\ln \left ( 1+x \right ) -{\frac{208}{375+250\,x}}+{\frac{4912\,\ln \left ( 3+2\,x \right ) }{625}}-{\frac{459}{50\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{8856}{250+375\,x}}+{\frac{104463\,\ln \left ( 2+3\,x \right ) }{625}} \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)^3,x)

[Out]

3/(1+x)^2+29/(1+x)-175*ln(1+x)-208/125/(3+2*x)+4912/625*ln(3+2*x)-459/50/(2+3*x)^2+8856/125/(2+3*x)+104463/625
*ln(2+3*x)

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Maxima [A]  time = 1.05139, size = 97, normalized size = 1.03 \begin{align*} \frac{233028 \, x^{4} + 939480 \, x^{3} + 1368599 \, x^{2} + 855120 \, x + 193723}{250 \,{\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )}} + \frac{104463}{625} \, \log \left (3 \, x + 2\right ) + \frac{4912}{625} \, \log \left (2 \, x + 3\right ) - 175 \, \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)^3,x, algorithm="maxima")

[Out]

1/250*(233028*x^4 + 939480*x^3 + 1368599*x^2 + 855120*x + 193723)/(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x
+ 12) + 104463/625*log(3*x + 2) + 4912/625*log(2*x + 3) - 175*log(x + 1)

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Fricas [A]  time = 1.33629, size = 450, normalized size = 4.79 \begin{align*} \frac{1165140 \, x^{4} + 4697400 \, x^{3} + 6842995 \, x^{2} + 208926 \,{\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (3 \, x + 2\right ) + 9824 \,{\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (2 \, x + 3\right ) - 218750 \,{\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (x + 1\right ) + 4275600 \, x + 968615}{1250 \,{\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\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)^3,x, algorithm="fricas")

[Out]

1/1250*(1165140*x^4 + 4697400*x^3 + 6842995*x^2 + 208926*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log
(3*x + 2) + 9824*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log(2*x + 3) - 218750*(18*x^5 + 87*x^4 + 16
4*x^3 + 151*x^2 + 68*x + 12)*log(x + 1) + 4275600*x + 968615)/(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12
)

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Sympy [A]  time = 0.239875, size = 71, normalized size = 0.76 \begin{align*} \frac{233028 x^{4} + 939480 x^{3} + 1368599 x^{2} + 855120 x + 193723}{4500 x^{5} + 21750 x^{4} + 41000 x^{3} + 37750 x^{2} + 17000 x + 3000} + \frac{104463 \log{\left (x + \frac{2}{3} \right )}}{625} - 175 \log{\left (x + 1 \right )} + \frac{4912 \log{\left (x + \frac{3}{2} \right )}}{625} \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)**3,x)

[Out]

(233028*x**4 + 939480*x**3 + 1368599*x**2 + 855120*x + 193723)/(4500*x**5 + 21750*x**4 + 41000*x**3 + 37750*x*
*2 + 17000*x + 3000) + 104463*log(x + 2/3)/625 - 175*log(x + 1) + 4912*log(x + 3/2)/625

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Giac [A]  time = 1.13641, size = 128, normalized size = 1.36 \begin{align*} -\frac{208}{125 \,{\left (2 \, x + 3\right )}} - \frac{2 \,{\left (\frac{168231}{2 \, x + 3} - \frac{211036}{{\left (2 \, x + 3\right )}^{2}} + \frac{82447}{{\left (2 \, x + 3\right )}^{3}} - 42642\right )}}{125 \,{\left (\frac{5}{2 \, x + 3} - 3\right )}^{2}{\left (\frac{1}{2 \, x + 3} - 1\right )}^{2}} - 175 \, \log \left ({\left | -\frac{1}{2 \, x + 3} + 1 \right |}\right ) + \frac{104463}{625} \, \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)^3,x, algorithm="giac")

[Out]

-208/125/(2*x + 3) - 2/125*(168231/(2*x + 3) - 211036/(2*x + 3)^2 + 82447/(2*x + 3)^3 - 42642)/((5/(2*x + 3) -
 3)^2*(1/(2*x + 3) - 1)^2) - 175*log(abs(-1/(2*x + 3) + 1)) + 104463/625*log(abs(-5/(2*x + 3) + 3))