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

Optimal. Leaf size=77 \[ -\frac {3 (47 x+37)}{5 (2 x+3)^2 \left (3 x^2+5 x+2\right )}-\frac {2618}{125 (2 x+3)}-\frac {428}{25 (2 x+3)^2}-\log (x+1)+\frac {8104}{625} \log (2 x+3)-\frac {7479}{625} \log (3 x+2) \]

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Rubi [A]  time = 0.05, antiderivative size = 77, 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} \begin {gather*} -\frac {3 (47 x+37)}{5 (2 x+3)^2 \left (3 x^2+5 x+2\right )}-\frac {2618}{125 (2 x+3)}-\frac {428}{25 (2 x+3)^2}-\log (x+1)+\frac {8104}{625} \log (2 x+3)-\frac {7479}{625} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-428/(25*(3 + 2*x)^2) - 2618/(125*(3 + 2*x)) - (3*(37 + 47*x))/(5*(3 + 2*x)^2*(2 + 5*x + 3*x^2)) - Log[1 + x]
+ (8104*Log[3 + 2*x])/625 - (7479*Log[2 + 3*x])/625

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)^3 \left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {3 (37+47 x)}{5 (3+2 x)^2 \left (2+5 x+3 x^2\right )}-\frac {1}{5} \int \frac {841+846 x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac {3 (37+47 x)}{5 (3+2 x)^2 \left (2+5 x+3 x^2\right )}-\frac {1}{5} \int \left (\frac {5}{1+x}-\frac {1712}{5 (3+2 x)^3}-\frac {5236}{25 (3+2 x)^2}-\frac {16208}{125 (3+2 x)}+\frac {22437}{125 (2+3 x)}\right ) \, dx\\ &=-\frac {428}{25 (3+2 x)^2}-\frac {2618}{125 (3+2 x)}-\frac {3 (37+47 x)}{5 (3+2 x)^2 \left (2+5 x+3 x^2\right )}-\log (1+x)+\frac {8104}{625} \log (3+2 x)-\frac {7479}{625} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.86 \begin {gather*} \frac {1}{625} \left (-\frac {15 (903 x+653)}{3 x^2+5 x+2}-\frac {4060}{2 x+3}-\frac {650}{(2 x+3)^2}-7479 \log (-6 x-4)-625 \log (-2 (x+1))+8104 \log (2 x+3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-650/(3 + 2*x)^2 - 4060/(3 + 2*x) - (15*(653 + 903*x))/(2 + 5*x + 3*x^2) - 7479*Log[-4 - 6*x] - 625*Log[-2*(1
+ x)] + 8104*Log[3 + 2*x])/625

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 121, normalized size = 1.57 \begin {gather*} -\frac {78540 \, x^{3} + 280810 \, x^{2} + 7479 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (3 \, x + 2\right ) - 8104 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (2 \, x + 3\right ) + 625 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (x + 1\right ) + 319835 \, x + 113815}{625 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/625*(78540*x^3 + 280810*x^2 + 7479*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log(3*x + 2) - 8104*(12*x^4 + 56*
x^3 + 95*x^2 + 69*x + 18)*log(2*x + 3) + 625*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log(x + 1) + 319835*x + 11
3815)/(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)

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giac [A]  time = 0.15, size = 62, normalized size = 0.81 \begin {gather*} -\frac {15708 \, x^{3} + 56162 \, x^{2} + 63967 \, x + 22763}{125 \, {\left (3 \, x + 2\right )} {\left (2 \, x + 3\right )}^{2} {\left (x + 1\right )}} - \frac {7479}{625} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {8104}{625} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/125*(15708*x^3 + 56162*x^2 + 63967*x + 22763)/((3*x + 2)*(2*x + 3)^2*(x + 1)) - 7479/625*log(abs(3*x + 2))
+ 8104/625*log(abs(2*x + 3)) - log(abs(x + 1))

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maple [A]  time = 0.06, size = 58, normalized size = 0.75 \begin {gather*} -\frac {7479 \ln \left (3 x +2\right )}{625}+\frac {8104 \ln \left (2 x +3\right )}{625}-\ln \left (x +1\right )-\frac {459}{125 \left (3 x +2\right )}-\frac {26}{25 \left (2 x +3\right )^{2}}-\frac {812}{125 \left (2 x +3\right )}-\frac {6}{x +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-459/125/(3*x+2)-7479/625*ln(3*x+2)-26/25/(2*x+3)^2-812/125/(2*x+3)+8104/625*ln(2*x+3)-6/(x+1)-ln(x+1)

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maxima [A]  time = 0.62, size = 62, normalized size = 0.81 \begin {gather*} -\frac {15708 \, x^{3} + 56162 \, x^{2} + 63967 \, x + 22763}{125 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} - \frac {7479}{625} \, \log \left (3 \, x + 2\right ) + \frac {8104}{625} \, \log \left (2 \, x + 3\right ) - \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/125*(15708*x^3 + 56162*x^2 + 63967*x + 22763)/(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18) - 7479/625*log(3*x + 2
) + 8104/625*log(2*x + 3) - log(x + 1)

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mupad [B]  time = 2.31, size = 56, normalized size = 0.73 \begin {gather*} \frac {8104\,\ln \left (x+\frac {3}{2}\right )}{625}-\frac {7479\,\ln \left (x+\frac {2}{3}\right )}{625}-\ln \left (x+1\right )-\frac {\frac {1309\,x^3}{125}+\frac {28081\,x^2}{750}+\frac {63967\,x}{1500}+\frac {22763}{1500}}{x^4+\frac {14\,x^3}{3}+\frac {95\,x^2}{12}+\frac {23\,x}{4}+\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8104*log(x + 3/2))/625 - (7479*log(x + 2/3))/625 - log(x + 1) - ((63967*x)/1500 + (28081*x^2)/750 + (1309*x^3
)/125 + 22763/1500)/((23*x)/4 + (95*x^2)/12 + (14*x^3)/3 + x^4 + 3/2)

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sympy [A]  time = 0.21, size = 60, normalized size = 0.78 \begin {gather*} - \frac {15708 x^{3} + 56162 x^{2} + 63967 x + 22763}{1500 x^{4} + 7000 x^{3} + 11875 x^{2} + 8625 x + 2250} - \frac {7479 \log {\left (x + \frac {2}{3} \right )}}{625} - \log {\left (x + 1 \right )} + \frac {8104 \log {\left (x + \frac {3}{2} \right )}}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(15708*x**3 + 56162*x**2 + 63967*x + 22763)/(1500*x**4 + 7000*x**3 + 11875*x**2 + 8625*x + 2250) - 7479*log(x
+ 2/3)/625 - log(x + 1) + 8104*log(x + 3/2)/625

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