### 3.1532 $$\int \frac{d+e x}{(a^2+2 a b x+b^2 x^2)^3} \, dx$$

Optimal. Leaf size=38 $-\frac{b d-a e}{5 b^2 (a+b x)^5}-\frac{e}{4 b^2 (a+b x)^4}$

[Out]

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

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Rubi [A]  time = 0.024086, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {27, 43} $-\frac{b d-a e}{5 b^2 (a+b x)^5}-\frac{e}{4 b^2 (a+b x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{d+e x}{(a+b x)^6} \, dx\\ &=\int \left (\frac{b d-a e}{b (a+b x)^6}+\frac{e}{b (a+b x)^5}\right ) \, dx\\ &=-\frac{b d-a e}{5 b^2 (a+b x)^5}-\frac{e}{4 b^2 (a+b x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0094128, size = 27, normalized size = 0.71 $-\frac{a e+4 b d+5 b e x}{20 b^2 (a+b x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

-(4*b*d + a*e + 5*b*e*x)/(20*b^2*(a + b*x)^5)

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Maple [A]  time = 0.045, size = 35, normalized size = 0.9 \begin{align*} -{\frac{-ae+bd}{5\,{b}^{2} \left ( bx+a \right ) ^{5}}}-{\frac{e}{4\,{b}^{2} \left ( bx+a \right ) ^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.14417, size = 97, normalized size = 2.55 \begin{align*} -\frac{5 \, b e x + 4 \, b d + a e}{20 \,{\left (b^{7} x^{5} + 5 \, a b^{6} x^{4} + 10 \, a^{2} b^{5} x^{3} + 10 \, a^{3} b^{4} x^{2} + 5 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

-1/20*(5*b*e*x + 4*b*d + a*e)/(b^7*x^5 + 5*a*b^6*x^4 + 10*a^2*b^5*x^3 + 10*a^3*b^4*x^2 + 5*a^4*b^3*x + a^5*b^2
)

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Fricas [B]  time = 1.50123, size = 153, normalized size = 4.03 \begin{align*} -\frac{5 \, b e x + 4 \, b d + a e}{20 \,{\left (b^{7} x^{5} + 5 \, a b^{6} x^{4} + 10 \, a^{2} b^{5} x^{3} + 10 \, a^{3} b^{4} x^{2} + 5 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

-1/20*(5*b*e*x + 4*b*d + a*e)/(b^7*x^5 + 5*a*b^6*x^4 + 10*a^2*b^5*x^3 + 10*a^3*b^4*x^2 + 5*a^4*b^3*x + a^5*b^2
)

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Sympy [B]  time = 0.772541, size = 76, normalized size = 2. \begin{align*} - \frac{a e + 4 b d + 5 b e x}{20 a^{5} b^{2} + 100 a^{4} b^{3} x + 200 a^{3} b^{4} x^{2} + 200 a^{2} b^{5} x^{3} + 100 a b^{6} x^{4} + 20 b^{7} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

-(a*e + 4*b*d + 5*b*e*x)/(20*a**5*b**2 + 100*a**4*b**3*x + 200*a**3*b**4*x**2 + 200*a**2*b**5*x**3 + 100*a*b**
6*x**4 + 20*b**7*x**5)

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Giac [A]  time = 1.13497, size = 36, normalized size = 0.95 \begin{align*} -\frac{5 \, b x e + 4 \, b d + a e}{20 \,{\left (b x + a\right )}^{5} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

-1/20*(5*b*x*e + 4*b*d + a*e)/((b*x + a)^5*b^2)