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

Optimal. Leaf size=69 \[ -\frac {(35 x+29) (2 x+3)^3}{2 \left (3 x^2+5 x+2\right )^2}+\frac {141 (8 x+7) (2 x+3)}{2 \left (3 x^2+5 x+2\right )}-705 \log (x+1)+705 \log (3 x+2) \]

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {804, 722, 616, 31} \begin {gather*} -\frac {(35 x+29) (2 x+3)^3}{2 \left (3 x^2+5 x+2\right )^2}+\frac {141 (8 x+7) (2 x+3)}{2 \left (3 x^2+5 x+2\right )}-705 \log (x+1)+705 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((3 + 2*x)^3*(29 + 35*x))/(2*(2 + 5*x + 3*x^2)^2) + (141*(3 + 2*x)*(7 + 8*x))/(2*(2 + 5*x + 3*x^2)) - 705*Log
[1 + x] + 705*Log[2 + 3*x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 722

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

Rule 804

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

Rubi steps

\begin {align*} \int \frac {(5-x) (3+2 x)^3}{\left (2+5 x+3 x^2\right )^3} \, dx &=-\frac {(3+2 x)^3 (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}-\frac {141}{2} \int \frac {(3+2 x)^2}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac {(3+2 x)^3 (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {141 (3+2 x) (7+8 x)}{2 \left (2+5 x+3 x^2\right )}+705 \int \frac {1}{2+5 x+3 x^2} \, dx\\ &=-\frac {(3+2 x)^3 (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {141 (3+2 x) (7+8 x)}{2 \left (2+5 x+3 x^2\right )}+2115 \int \frac {1}{2+3 x} \, dx-2115 \int \frac {1}{3+3 x} \, dx\\ &=-\frac {(3+2 x)^3 (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {141 (3+2 x) (7+8 x)}{2 \left (2+5 x+3 x^2\right )}-705 \log (1+x)+705 \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 59, normalized size = 0.86 \begin {gather*} -\frac {2611 x+2449}{54 \left (3 x^2+5 x+2\right )^2}+\frac {38118 x+31673}{54 \left (3 x^2+5 x+2\right )}+705 \log (-6 x-4)-705 \log (-2 (x+1)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/54*(2449 + 2611*x)/(2 + 5*x + 3*x^2)^2 + (31673 + 38118*x)/(54*(2 + 5*x + 3*x^2)) + 705*Log[-4 - 6*x] - 705
*Log[-2*(1 + x)]

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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 )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 93, normalized size = 1.35 \begin {gather*} \frac {38118 \, x^{3} + 95203 \, x^{2} + 12690 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 12690 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 77330 \, x + 20299}{18 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

1/18*(38118*x^3 + 95203*x^2 + 12690*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x + 2) - 12690*(9*x^4 + 30*x^3
+ 37*x^2 + 20*x + 4)*log(x + 1) + 77330*x + 20299)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)

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giac [A]  time = 0.18, size = 46, normalized size = 0.67 \begin {gather*} \frac {38118 \, x^{3} + 95203 \, x^{2} + 77330 \, x + 20299}{18 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{2}} + 705 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 705 \, \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)^3,x, algorithm="giac")

[Out]

1/18*(38118*x^3 + 95203*x^2 + 77330*x + 20299)/(3*x^2 + 5*x + 2)^2 + 705*log(abs(3*x + 2)) - 705*log(abs(x + 1
))

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maple [A]  time = 0.05, size = 48, normalized size = 0.70 \begin {gather*} 705 \ln \left (3 x +2\right )-705 \ln \left (x +1\right )-\frac {2125}{18 \left (3 x +2\right )^{2}}+\frac {3950}{9 \left (3 x +2\right )}+\frac {3}{\left (x +1\right )^{2}}+\frac {89}{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)^3,x)

[Out]

-2125/18/(3*x+2)^2+3950/9/(3*x+2)+705*ln(3*x+2)+3/(x+1)^2+89/(x+1)-705*ln(x+1)

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maxima [A]  time = 0.55, size = 54, normalized size = 0.78 \begin {gather*} \frac {38118 \, x^{3} + 95203 \, x^{2} + 77330 \, x + 20299}{18 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + 705 \, \log \left (3 \, x + 2\right ) - 705 \, \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)^3,x, algorithm="maxima")

[Out]

1/18*(38118*x^3 + 95203*x^2 + 77330*x + 20299)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 705*log(3*x + 2) - 705*l
og(x + 1)

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mupad [B]  time = 0.05, size = 45, normalized size = 0.65 \begin {gather*} \frac {\frac {6353\,x^3}{27}+\frac {95203\,x^2}{162}+\frac {38665\,x}{81}+\frac {20299}{162}}{x^4+\frac {10\,x^3}{3}+\frac {37\,x^2}{9}+\frac {20\,x}{9}+\frac {4}{9}}-1410\,\mathrm {atanh}\left (6\,x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((38665*x)/81 + (95203*x^2)/162 + (6353*x^3)/27 + 20299/162)/((20*x)/9 + (37*x^2)/9 + (10*x^3)/3 + x^4 + 4/9)
- 1410*atanh(6*x + 5)

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sympy [A]  time = 0.17, size = 51, normalized size = 0.74 \begin {gather*} - \frac {- 38118 x^{3} - 95203 x^{2} - 77330 x - 20299}{162 x^{4} + 540 x^{3} + 666 x^{2} + 360 x + 72} + 705 \log {\left (x + \frac {2}{3} \right )} - 705 \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)**3,x)

[Out]

-(-38118*x**3 - 95203*x**2 - 77330*x - 20299)/(162*x**4 + 540*x**3 + 666*x**2 + 360*x + 72) + 705*log(x + 2/3)
- 705*log(x + 1)

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