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

Optimal. Leaf size=72 \[ -\frac{a^2 B}{2 b^4 (a+b x)^2}+\frac{x^3 (A b-a B)}{3 a b (a+b x)^3}+\frac{2 a B}{b^4 (a+b x)}+\frac{B \log (a+b x)}{b^4} \]

[Out]

((A*b - a*B)*x^3)/(3*a*b*(a + b*x)^3) - (a^2*B)/(2*b^4*(a + b*x)^2) + (2*a*B)/(b^4*(a + b*x)) + (B*Log[a + b*x
])/b^4

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Rubi [A]  time = 0.0329438, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {27, 78, 43} \[ -\frac{a^2 B}{2 b^4 (a+b x)^2}+\frac{x^3 (A b-a B)}{3 a b (a+b x)^3}+\frac{2 a B}{b^4 (a+b x)}+\frac{B \log (a+b x)}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^3)/(3*a*b*(a + b*x)^3) - (a^2*B)/(2*b^4*(a + b*x)^2) + (2*a*B)/(b^4*(a + b*x)) + (B*Log[a + b*x
])/b^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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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{x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{x^2 (A+B x)}{(a+b x)^4} \, dx\\ &=\frac{(A b-a B) x^3}{3 a b (a+b x)^3}+\frac{B \int \frac{x^2}{(a+b x)^3} \, dx}{b}\\ &=\frac{(A b-a B) x^3}{3 a b (a+b x)^3}+\frac{B \int \left (\frac{a^2}{b^2 (a+b x)^3}-\frac{2 a}{b^2 (a+b x)^2}+\frac{1}{b^2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{(A b-a B) x^3}{3 a b (a+b x)^3}-\frac{a^2 B}{2 b^4 (a+b x)^2}+\frac{2 a B}{b^4 (a+b x)}+\frac{B \log (a+b x)}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.0255028, size = 73, normalized size = 1.01 \[ \frac{a^2 (27 b B x-2 A b)+11 a^3 B-6 a b^2 x (A-3 B x)+6 B (a+b x)^3 \log (a+b x)-6 A b^3 x^2}{6 b^4 (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*a^3*B - 6*A*b^3*x^2 - 6*a*b^2*x*(A - 3*B*x) + a^2*(-2*A*b + 27*b*B*x) + 6*B*(a + b*x)^3*Log[a + b*x])/(6*b
^4*(a + b*x)^3)

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Maple [A]  time = 0.006, size = 101, normalized size = 1.4 \begin{align*} -{\frac{A}{{b}^{3} \left ( bx+a \right ) }}+3\,{\frac{aB}{{b}^{4} \left ( bx+a \right ) }}+{\frac{B\ln \left ( bx+a \right ) }{{b}^{4}}}+{\frac{aA}{{b}^{3} \left ( bx+a \right ) ^{2}}}-{\frac{3\,B{a}^{2}}{2\,{b}^{4} \left ( bx+a \right ) ^{2}}}-{\frac{A{a}^{2}}{3\,{b}^{3} \left ( bx+a \right ) ^{3}}}+{\frac{B{a}^{3}}{3\,{b}^{4} \left ( bx+a \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00749, size = 135, normalized size = 1.88 \begin{align*} \frac{11 \, B a^{3} - 2 \, A a^{2} b + 6 \,{\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 3 \,{\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} x}{6 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac{B \log \left (b x + a\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(11*B*a^3 - 2*A*a^2*b + 6*(3*B*a*b^2 - A*b^3)*x^2 + 3*(9*B*a^2*b - 2*A*a*b^2)*x)/(b^7*x^3 + 3*a*b^6*x^2 +
3*a^2*b^5*x + a^3*b^4) + B*log(b*x + a)/b^4

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Fricas [A]  time = 1.39955, size = 271, normalized size = 3.76 \begin{align*} \frac{11 \, B a^{3} - 2 \, A a^{2} b + 6 \,{\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 3 \,{\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} x + 6 \,{\left (B b^{3} x^{3} + 3 \, B a b^{2} x^{2} + 3 \, B a^{2} b x + B a^{3}\right )} \log \left (b x + a\right )}{6 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(11*B*a^3 - 2*A*a^2*b + 6*(3*B*a*b^2 - A*b^3)*x^2 + 3*(9*B*a^2*b - 2*A*a*b^2)*x + 6*(B*b^3*x^3 + 3*B*a*b^2
*x^2 + 3*B*a^2*b*x + B*a^3)*log(b*x + a))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4)

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Sympy [A]  time = 0.869151, size = 100, normalized size = 1.39 \begin{align*} \frac{B \log{\left (a + b x \right )}}{b^{4}} + \frac{- 2 A a^{2} b + 11 B a^{3} + x^{2} \left (- 6 A b^{3} + 18 B a b^{2}\right ) + x \left (- 6 A a b^{2} + 27 B a^{2} b\right )}{6 a^{3} b^{4} + 18 a^{2} b^{5} x + 18 a b^{6} x^{2} + 6 b^{7} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(a + b*x)/b**4 + (-2*A*a**2*b + 11*B*a**3 + x**2*(-6*A*b**3 + 18*B*a*b**2) + x*(-6*A*a*b**2 + 27*B*a**2*b
))/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3)

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Giac [A]  time = 1.18682, size = 103, normalized size = 1.43 \begin{align*} \frac{B \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac{6 \,{\left (3 \, B a b - A b^{2}\right )} x^{2} + 3 \,{\left (9 \, B a^{2} - 2 \, A a b\right )} x + \frac{11 \, B a^{3} - 2 \, A a^{2} b}{b}}{6 \,{\left (b x + a\right )}^{3} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(abs(b*x + a))/b^4 + 1/6*(6*(3*B*a*b - A*b^2)*x^2 + 3*(9*B*a^2 - 2*A*a*b)*x + (11*B*a^3 - 2*A*a^2*b)/b)/(
(b*x + a)^3*b^3)