3.1705 \(\int \frac{A+B x}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=38 \[ -\frac{A b-a B}{3 b^2 (a+b x)^3}-\frac{B}{2 b^2 (a+b x)^2} \]

[Out]

-(A*b - a*B)/(3*b^2*(a + b*x)^3) - B/(2*b^2*(a + b*x)^2)

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Rubi [A]  time = 0.0233002, 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} \[ -\frac{A b-a B}{3 b^2 (a+b x)^3}-\frac{B}{2 b^2 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(A*b - a*B)/(3*b^2*(a + b*x)^3) - B/(2*b^2*(a + b*x)^2)

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

Mathematica [A]  time = 0.0103597, size = 27, normalized size = 0.71 \[ -\frac{B (a+3 b x)+2 A b}{6 b^2 (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(2*A*b + B*(a + 3*b*x))/(6*b^2*(a + b*x)^3)

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Maple [A]  time = 0.005, size = 35, normalized size = 0.9 \begin{align*} -{\frac{Ab-aB}{3\,{b}^{2} \left ( bx+a \right ) ^{3}}}-{\frac{B}{2\,{b}^{2} \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(A*b-B*a)/b^2/(b*x+a)^3-1/2*B/b^2/(b*x+a)^2

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Maxima [A]  time = 1.18199, size = 68, normalized size = 1.79 \begin{align*} -\frac{3 \, B b x + B a + 2 \, A b}{6 \,{\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47451, size = 105, normalized size = 2.76 \begin{align*} -\frac{3 \, B b x + B a + 2 \, A b}{6 \,{\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.535453, size = 53, normalized size = 1.39 \begin{align*} - \frac{2 A b + B a + 3 B b x}{6 a^{3} b^{2} + 18 a^{2} b^{3} x + 18 a b^{4} x^{2} + 6 b^{5} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

-(2*A*b + B*a + 3*B*b*x)/(6*a**3*b**2 + 18*a**2*b**3*x + 18*a*b**4*x**2 + 6*b**5*x**3)

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Giac [A]  time = 1.1962, size = 34, normalized size = 0.89 \begin{align*} -\frac{3 \, B b x + B a + 2 \, A b}{6 \,{\left (b x + a\right )}^{3} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*B*b*x + B*a + 2*A*b)/((b*x + a)^3*b^2)