### 3.1519 $$\int \frac{(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^2} \, dx$$

Optimal. Leaf size=28 $-\frac{(d+e x)^3}{3 (a+b x)^3 (b d-a e)}$

[Out]

-(d + e*x)^3/(3*(b*d - a*e)*(a + b*x)^3)

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Rubi [A]  time = 0.0046141, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {27, 37} $-\frac{(d+e x)^3}{3 (a+b x)^3 (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(d + e*x)^3/(3*(b*d - a*e)*(a + b*x)^3)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^2}{(a+b x)^4} \, dx\\ &=-\frac{(d+e x)^3}{3 (b d-a e) (a+b x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0238475, size = 53, normalized size = 1.89 $-\frac{a^2 e^2+a b e (d+3 e x)+b^2 \left (d^2+3 d e x+3 e^2 x^2\right )}{3 b^3 (a+b x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(a^2*e^2 + a*b*e*(d + 3*e*x) + b^2*(d^2 + 3*d*e*x + 3*e^2*x^2))/(3*b^3*(a + b*x)^3)

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Maple [B]  time = 0.046, size = 70, normalized size = 2.5 \begin{align*}{\frac{e \left ( ae-bd \right ) }{{b}^{3} \left ( bx+a \right ) ^{2}}}-{\frac{{a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2}}{3\,{b}^{3} \left ( bx+a \right ) ^{3}}}-{\frac{{e}^{2}}{{b}^{3} \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

e*(a*e-b*d)/b^3/(b*x+a)^2-1/3*(a^2*e^2-2*a*b*d*e+b^2*d^2)/b^3/(b*x+a)^3-e^2/b^3/(b*x+a)

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Maxima [B]  time = 1.12975, size = 113, normalized size = 4.04 \begin{align*} -\frac{3 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + a b d e + a^{2} e^{2} + 3 \,{\left (b^{2} d e + a b e^{2}\right )} x}{3 \,{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

-1/3*(3*b^2*e^2*x^2 + b^2*d^2 + a*b*d*e + a^2*e^2 + 3*(b^2*d*e + a*b*e^2)*x)/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^
4*x + a^3*b^3)

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Fricas [B]  time = 1.74185, size = 170, normalized size = 6.07 \begin{align*} -\frac{3 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + a b d e + a^{2} e^{2} + 3 \,{\left (b^{2} d e + a b e^{2}\right )} x}{3 \,{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

-1/3*(3*b^2*e^2*x^2 + b^2*d^2 + a*b*d*e + a^2*e^2 + 3*(b^2*d*e + a*b*e^2)*x)/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^
4*x + a^3*b^3)

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Sympy [B]  time = 0.958547, size = 88, normalized size = 3.14 \begin{align*} - \frac{a^{2} e^{2} + a b d e + b^{2} d^{2} + 3 b^{2} e^{2} x^{2} + x \left (3 a b e^{2} + 3 b^{2} d e\right )}{3 a^{3} b^{3} + 9 a^{2} b^{4} x + 9 a b^{5} x^{2} + 3 b^{6} x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

-(a**2*e**2 + a*b*d*e + b**2*d**2 + 3*b**2*e**2*x**2 + x*(3*a*b*e**2 + 3*b**2*d*e))/(3*a**3*b**3 + 9*a**2*b**4
*x + 9*a*b**5*x**2 + 3*b**6*x**3)

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Giac [B]  time = 1.21924, size = 78, normalized size = 2.79 \begin{align*} -\frac{3 \, b^{2} x^{2} e^{2} + 3 \, b^{2} d x e + b^{2} d^{2} + 3 \, a b x e^{2} + a b d e + a^{2} e^{2}}{3 \,{\left (b x + a\right )}^{3} b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

-1/3*(3*b^2*x^2*e^2 + 3*b^2*d*x*e + b^2*d^2 + 3*a*b*x*e^2 + a*b*d*e + a^2*e^2)/((b*x + a)^3*b^3)