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

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]

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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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, 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]

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

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fricas [A]  time = 0.89, size = 50, normalized size = 1.32 \[ -\frac {3 \, B b x + B a + 2 \, A b}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \]

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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giac [A]  time = 0.15, size = 25, normalized size = 0.66 \[ -\frac {3 \, B b x + B a + 2 \, A b}{6 \, {\left (b x + a\right )}^{3} b^{2}} \]

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)

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maple [A]  time = 0.06, size = 35, normalized size = 0.92 \[ -\frac {B}{2 \left (b x +a \right )^{2} b^{2}}-\frac {A b -B a}{3 \left (b x +a \right )^{3} b^{2}} \]

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 = 0.51, size = 50, normalized size = 1.32 \[ -\frac {3 \, B b x + B a + 2 \, A b}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \]

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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mupad [B]  time = 0.03, size = 52, normalized size = 1.37 \[ -\frac {\frac {2\,A\,b+B\,a}{6\,b^2}+\frac {B\,x}{2\,b}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.35, size = 53, normalized size = 1.39 \[ \frac {- 2 A b - B a - 3 B b x}{6 a^{3} b^{2} + 18 a^{2} b^{3} x + 18 a b^{4} x^{2} + 6 b^{5} x^{3}} \]

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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