3.205 \(\int \frac{(1+2 x)^2}{(3+5 x+2 x^2)^5} \, dx\)

Optimal. Leaf size=102 \[ \frac{620 (4 x+5)}{2 x^2+5 x+3}-\frac{155 (4 x+5)}{3 \left (2 x^2+5 x+3\right )^2}+\frac{62 x+73}{3 \left (2 x^2+5 x+3\right )^3}+\frac{(2 x+1) (6 x+7)}{4 \left (2 x^2+5 x+3\right )^4}+2480 \log (x+1)-2480 \log (2 x+3) \]

[Out]

((1 + 2*x)*(7 + 6*x))/(4*(3 + 5*x + 2*x^2)^4) + (73 + 62*x)/(3*(3 + 5*x + 2*x^2)^3) - (155*(5 + 4*x))/(3*(3 +
5*x + 2*x^2)^2) + (620*(5 + 4*x))/(3 + 5*x + 2*x^2) + 2480*Log[1 + x] - 2480*Log[3 + 2*x]

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Rubi [A]  time = 0.0371751, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {738, 638, 614, 616, 31} \[ \frac{620 (4 x+5)}{2 x^2+5 x+3}-\frac{155 (4 x+5)}{3 \left (2 x^2+5 x+3\right )^2}+\frac{62 x+73}{3 \left (2 x^2+5 x+3\right )^3}+\frac{(2 x+1) (6 x+7)}{4 \left (2 x^2+5 x+3\right )^4}+2480 \log (x+1)-2480 \log (2 x+3) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)^2/(3 + 5*x + 2*x^2)^5,x]

[Out]

((1 + 2*x)*(7 + 6*x))/(4*(3 + 5*x + 2*x^2)^4) + (73 + 62*x)/(3*(3 + 5*x + 2*x^2)^3) - (155*(5 + 4*x))/(3*(3 +
5*x + 2*x^2)^2) + (620*(5 + 4*x))/(3 + 5*x + 2*x^2) + 2480*Log[1 + x] - 2480*Log[3 + 2*x]

Rule 738

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[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 638

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

Rule 614

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

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 31

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

Rubi steps

\begin{align*} \int \frac{(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}-\frac{1}{4} \int \frac{-28-72 x}{\left (3+5 x+2 x^2\right )^4} \, dx\\ &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac{73+62 x}{3 \left (3+5 x+2 x^2\right )^3}+\frac{310}{3} \int \frac{1}{\left (3+5 x+2 x^2\right )^3} \, dx\\ &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac{73+62 x}{3 \left (3+5 x+2 x^2\right )^3}-\frac{155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}-620 \int \frac{1}{\left (3+5 x+2 x^2\right )^2} \, dx\\ &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac{73+62 x}{3 \left (3+5 x+2 x^2\right )^3}-\frac{155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}+\frac{620 (5+4 x)}{3+5 x+2 x^2}+2480 \int \frac{1}{3+5 x+2 x^2} \, dx\\ &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac{73+62 x}{3 \left (3+5 x+2 x^2\right )^3}-\frac{155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}+\frac{620 (5+4 x)}{3+5 x+2 x^2}+4960 \int \frac{1}{2+2 x} \, dx-4960 \int \frac{1}{3+2 x} \, dx\\ &=\frac{(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac{73+62 x}{3 \left (3+5 x+2 x^2\right )^3}-\frac{155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}+\frac{620 (5+4 x)}{3+5 x+2 x^2}+2480 \log (1+x)-2480 \log (3+2 x)\\ \end{align*}

Mathematica [A]  time = 0.0513033, size = 99, normalized size = 0.97 \[ \frac{620 (4 x+5)}{2 x^2+5 x+3}-\frac{155 (4 x+5)}{3 \left (2 x^2+5 x+3\right )^2}+\frac{31 (4 x+5)}{6 \left (2 x^2+5 x+3\right )^3}-\frac{10 x+11}{4 \left (2 x^2+5 x+3\right )^4}+2480 \log (2 (x+1))-2480 \log (2 x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)^2/(3 + 5*x + 2*x^2)^5,x]

[Out]

-(11 + 10*x)/(4*(3 + 5*x + 2*x^2)^4) + (31*(5 + 4*x))/(6*(3 + 5*x + 2*x^2)^3) - (155*(5 + 4*x))/(3*(3 + 5*x +
2*x^2)^2) + (620*(5 + 4*x))/(3 + 5*x + 2*x^2) + 2480*Log[2*(1 + x)] - 2480*Log[3 + 2*x]

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Maple [A]  time = 0.01, size = 80, normalized size = 0.8 \begin{align*} 16\, \left ( 3+2\,x \right ) ^{-4}+{\frac{256}{3\, \left ( 3+2\,x \right ) ^{3}}}+328\, \left ( 3+2\,x \right ) ^{-2}+1360\, \left ( 3+2\,x \right ) ^{-1}-2480\,\ln \left ( 3+2\,x \right ) -{\frac{1}{4\, \left ( 1+x \right ) ^{4}}}+{\frac{14}{3\, \left ( 1+x \right ) ^{3}}}-52\, \left ( 1+x \right ) ^{-2}+560\, \left ( 1+x \right ) ^{-1}+2480\,\ln \left ( 1+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)^2/(2*x^2+5*x+3)^5,x)

[Out]

16/(3+2*x)^4+256/3/(3+2*x)^3+328/(3+2*x)^2+1360/(3+2*x)-2480*ln(3+2*x)-1/4/(1+x)^4+14/3/(1+x)^3-52/(1+x)^2+560
/(1+x)+2480*ln(1+x)

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Maxima [A]  time = 0.932153, size = 127, normalized size = 1.25 \begin{align*} \frac{238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} + 5712464 \, x + 977397}{12 \,{\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )}} - 2480 \, \log \left (2 \, x + 3\right ) + 2480 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^2/(2*x^2+5*x+3)^5,x, algorithm="maxima")

[Out]

1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 + 5712464*x + 977397
)/(16*x^8 + 160*x^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81) - 2480*log(2*x + 3) +
2480*log(x + 1)

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Fricas [A]  time = 1.74688, size = 555, normalized size = 5.44 \begin{align*} \frac{238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} - 29760 \,{\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )} \log \left (2 \, x + 3\right ) + 29760 \,{\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )} \log \left (x + 1\right ) + 5712464 \, x + 977397}{12 \,{\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^2/(2*x^2+5*x+3)^5,x, algorithm="fricas")

[Out]

1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 - 29760*(16*x^8 + 16
0*x^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)*log(2*x + 3) + 29760*(16*x^8 + 160*x
^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)*log(x + 1) + 5712464*x + 977397)/(16*x^
8 + 160*x^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)

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Sympy [A]  time = 0.208913, size = 90, normalized size = 0.88 \begin{align*} \frac{238080 x^{7} + 2083200 x^{6} + 7757440 x^{5} + 15934000 x^{4} + 19495776 x^{3} + 14209160 x^{2} + 5712464 x + 977397}{192 x^{8} + 1920 x^{7} + 8352 x^{6} + 20640 x^{5} + 31692 x^{4} + 30960 x^{3} + 18792 x^{2} + 6480 x + 972} + 2480 \log{\left (x + 1 \right )} - 2480 \log{\left (x + \frac{3}{2} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**2/(2*x**2+5*x+3)**5,x)

[Out]

(238080*x**7 + 2083200*x**6 + 7757440*x**5 + 15934000*x**4 + 19495776*x**3 + 14209160*x**2 + 5712464*x + 97739
7)/(192*x**8 + 1920*x**7 + 8352*x**6 + 20640*x**5 + 31692*x**4 + 30960*x**3 + 18792*x**2 + 6480*x + 972) + 248
0*log(x + 1) - 2480*log(x + 3/2)

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Giac [A]  time = 1.06351, size = 89, normalized size = 0.87 \begin{align*} \frac{238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} + 5712464 \, x + 977397}{12 \,{\left (2 \, x^{2} + 5 \, x + 3\right )}^{4}} - 2480 \, \log \left ({\left | 2 \, x + 3 \right |}\right ) + 2480 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^2/(2*x^2+5*x+3)^5,x, algorithm="giac")

[Out]

1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 + 5712464*x + 977397
)/(2*x^2 + 5*x + 3)^4 - 2480*log(abs(2*x + 3)) + 2480*log(abs(x + 1))