3.2402 \(\int \frac{5-x}{(3+2 x) (2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=69 \[ -\frac{3 (47 x+37)}{10 \left (3 x^2+5 x+2\right )^2}+\frac{11442 x+9587}{50 \left (3 x^2+5 x+2\right )}-233 \log (x+1)+\frac{208}{125} \log (2 x+3)+\frac{28917}{125} \log (3 x+2) \]

[Out]

(-3*(37 + 47*x))/(10*(2 + 5*x + 3*x^2)^2) + (9587 + 11442*x)/(50*(2 + 5*x + 3*x^2)) - 233*Log[1 + x] + (208*Lo
g[3 + 2*x])/125 + (28917*Log[2 + 3*x])/125

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Rubi [A]  time = 0.0528511, antiderivative size = 69, 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} \[ -\frac{3 (47 x+37)}{10 \left (3 x^2+5 x+2\right )^2}+\frac{11442 x+9587}{50 \left (3 x^2+5 x+2\right )}-233 \log (x+1)+\frac{208}{125} \log (2 x+3)+\frac{28917}{125} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(37 + 47*x))/(10*(2 + 5*x + 3*x^2)^2) + (9587 + 11442*x)/(50*(2 + 5*x + 3*x^2)) - 233*Log[1 + x] + (208*Lo
g[3 + 2*x])/125 + (28917*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) \left (2+5 x+3 x^2\right )^3} \, dx &=-\frac{3 (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}-\frac{1}{10} \int \frac{1217+846 x}{(3+2 x) \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{3 (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{9587+11442 x}{50 \left (2+5 x+3 x^2\right )}+\frac{1}{50} \int \frac{34534+22884 x}{(3+2 x) \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{3 (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{9587+11442 x}{50 \left (2+5 x+3 x^2\right )}+\frac{1}{50} \int \left (-\frac{11650}{1+x}+\frac{832}{5 (3+2 x)}+\frac{173502}{5 (2+3 x)}\right ) \, dx\\ &=-\frac{3 (37+47 x)}{10 \left (2+5 x+3 x^2\right )^2}+\frac{9587+11442 x}{50 \left (2+5 x+3 x^2\right )}-233 \log (1+x)+\frac{208}{125} \log (3+2 x)+\frac{28917}{125} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0432263, size = 68, normalized size = 0.99 \[ \frac{1}{125} \left (-\frac{75 (47 x+37)}{2 \left (3 x^2+5 x+2\right )^2}+\frac{57210 x+47935}{6 x^2+10 x+4}+28917 \log (-6 x-4)-29125 \log (-2 (x+1))+208 \log (2 x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-75*(37 + 47*x))/(2*(2 + 5*x + 3*x^2)^2) + (47935 + 57210*x)/(4 + 10*x + 6*x^2) + 28917*Log[-4 - 6*x] - 2912
5*Log[-2*(1 + x)] + 208*Log[3 + 2*x])/125

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Maple [A]  time = 0.009, size = 56, normalized size = 0.8 \begin{align*} 3\, \left ( 1+x \right ) ^{-2}+41\, \left ( 1+x \right ) ^{-1}-233\,\ln \left ( 1+x \right ) +{\frac{208\,\ln \left ( 3+2\,x \right ) }{125}}-{\frac{153}{10\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{2646}{50+75\,x}}+{\frac{28917\,\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)/(3*x^2+5*x+2)^3,x)

[Out]

3/(1+x)^2+41/(1+x)-233*ln(1+x)+208/125*ln(3+2*x)-153/10/(2+3*x)^2+2646/25/(2+3*x)+28917/125*ln(2+3*x)

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Maxima [A]  time = 1.18862, size = 84, normalized size = 1.22 \begin{align*} \frac{34326 \, x^{3} + 85971 \, x^{2} + 70114 \, x + 18619}{50 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + \frac{28917}{125} \, \log \left (3 \, x + 2\right ) + \frac{208}{125} \, \log \left (2 \, x + 3\right ) - 233 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(34326*x^3 + 85971*x^2 + 70114*x + 18619)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 28917/125*log(3*x + 2) +
208/125*log(2*x + 3) - 233*log(x + 1)

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Fricas [A]  time = 1.23134, size = 350, normalized size = 5.07 \begin{align*} \frac{171630 \, x^{3} + 429855 \, x^{2} + 57834 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 416 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (2 \, x + 3\right ) - 58250 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 350570 \, x + 93095}{250 \,{\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+2*x)/(3*x^2+5*x+2)^3,x, algorithm="fricas")

[Out]

1/250*(171630*x^3 + 429855*x^2 + 57834*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x + 2) + 416*(9*x^4 + 30*x^3
+ 37*x^2 + 20*x + 4)*log(2*x + 3) - 58250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(x + 1) + 350570*x + 93095)
/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)

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Sympy [A]  time = 0.223326, size = 61, normalized size = 0.88 \begin{align*} \frac{34326 x^{3} + 85971 x^{2} + 70114 x + 18619}{450 x^{4} + 1500 x^{3} + 1850 x^{2} + 1000 x + 200} + \frac{28917 \log{\left (x + \frac{2}{3} \right )}}{125} - 233 \log{\left (x + 1 \right )} + \frac{208 \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)/(3*x**2+5*x+2)**3,x)

[Out]

(34326*x**3 + 85971*x**2 + 70114*x + 18619)/(450*x**4 + 1500*x**3 + 1850*x**2 + 1000*x + 200) + 28917*log(x +
2/3)/125 - 233*log(x + 1) + 208*log(x + 3/2)/125

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Giac [A]  time = 1.13199, size = 74, normalized size = 1.07 \begin{align*} \frac{34326 \, x^{3} + 85971 \, x^{2} + 70114 \, x + 18619}{50 \,{\left (3 \, x + 2\right )}^{2}{\left (x + 1\right )}^{2}} + \frac{28917}{125} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac{208}{125} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - 233 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(34326*x^3 + 85971*x^2 + 70114*x + 18619)/((3*x + 2)^2*(x + 1)^2) + 28917/125*log(abs(3*x + 2)) + 208/125
*log(abs(2*x + 3)) - 233*log(abs(x + 1))