### 3.1510 $$\int \frac{1}{a^2+2 a b x+b^2 x^2} \, dx$$

Optimal. Leaf size=12 $-\frac{1}{b (a+b x)}$

[Out]

-(1/(b*(a + b*x)))

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Rubi [A]  time = 0.0022466, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {27, 32} $-\frac{1}{b (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(-1),x]

[Out]

-(1/(b*(a + b*x)))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{1}{(a+b x)^2} \, dx\\ &=-\frac{1}{b (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0029372, size = 12, normalized size = 1. $-\frac{1}{b (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(-1),x]

[Out]

-(1/(b*(a + b*x)))

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Maple [A]  time = 0.041, size = 13, normalized size = 1.1 \begin{align*} -{\frac{1}{b \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

-1/b/(b*x+a)

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Maxima [A]  time = 1.14039, size = 18, normalized size = 1.5 \begin{align*} -\frac{1}{b^{2} x + a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

-1/(b^2*x + a*b)

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Fricas [A]  time = 1.70483, size = 24, normalized size = 2. \begin{align*} -\frac{1}{b^{2} x + a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

-1/(b^2*x + a*b)

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Sympy [A]  time = 0.305257, size = 10, normalized size = 0.83 \begin{align*} - \frac{1}{a b + b^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

-1/(a*b + b**2*x)

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Giac [A]  time = 1.16519, size = 16, normalized size = 1.33 \begin{align*} -\frac{1}{{\left (b x + a\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

-1/((b*x + a)*b)