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

Optimal. Leaf size=59 \[ -\frac {A b-2 a B}{2 b^3 (a+b x)^2}+\frac {a (A b-a B)}{3 b^3 (a+b x)^3}-\frac {B}{b^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 77} \[ -\frac {A b-2 a B}{2 b^3 (a+b x)^2}+\frac {a (A b-a B)}{3 b^3 (a+b x)^3}-\frac {B}{b^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*b - a*B))/(3*b^3*(a + b*x)^3) - (A*b - 2*a*B)/(2*b^3*(a + b*x)^2) - B/(b^3*(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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.71 \[ -\frac {2 a^2 B+a b (A+6 B x)+3 b^2 x (A+2 B x)}{6 b^3 (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 71, normalized size = 1.20 \[ -\frac {6 \, B b^{2} x^{2} + 2 \, B a^{2} + A a b + 3 \, {\left (2 \, B a b + A b^{2}\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*B*b^2*x^2 + 2*B*a^2 + A*a*b + 3*(2*B*a*b + A*b^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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giac [A]  time = 0.19, size = 45, normalized size = 0.76 \[ -\frac {6 \, B b^{2} x^{2} + 6 \, B a b x + 3 \, A b^{2} x + 2 \, B a^{2} + A a b}{6 \, {\left (b x + a\right )}^{3} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 56, normalized size = 0.95 \[ -\frac {B}{\left (b x +a \right ) b^{3}}+\frac {\left (A b -B a \right ) a}{3 \left (b x +a \right )^{3} b^{3}}-\frac {A b -2 B a}{2 \left (b x +a \right )^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 71, normalized size = 1.20 \[ -\frac {6 \, B b^{2} x^{2} + 2 \, B a^{2} + A a b + 3 \, {\left (2 \, B a b + A b^{2}\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*B*b^2*x^2 + 2*B*a^2 + A*a*b + 3*(2*B*a*b + A*b^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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mupad [B]  time = 0.04, size = 68, normalized size = 1.15 \[ -\frac {\frac {B\,x^2}{b}+\frac {a\,\left (A\,b+2\,B\,a\right )}{6\,b^3}+\frac {x\,\left (A\,b+2\,B\,a\right )}{2\,b^2}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.45, size = 75, normalized size = 1.27 \[ \frac {- A a b - 2 B a^{2} - 6 B b^{2} x^{2} + x \left (- 3 A b^{2} - 6 B a b\right )}{6 a^{3} b^{3} + 18 a^{2} b^{4} x + 18 a b^{5} x^{2} + 6 b^{6} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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