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

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

[Out]

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

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Rubi [A]  time = 0.0492757, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {27, 77} \[ \frac{A}{a^3 (a+b x)}+\frac{A}{2 a^2 (a+b x)^2}-\frac{A \log (a+b x)}{a^4}+\frac{A \log (x)}{a^4}+\frac{A b-a B}{3 a b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0462261, size = 65, normalized size = 0.9 \[ \frac{\frac{a \left (11 a^2 A b-2 a^3 B+15 a A b^2 x+6 A b^3 x^2\right )}{b (a+b x)^3}-6 A \log (a+b x)+6 A \log (x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(11*a^2*A*b - 2*a^3*B + 15*a*A*b^2*x + 6*A*b^3*x^2))/(b*(a + b*x)^3) + 6*A*Log[x] - 6*A*Log[a + b*x])/(6*a
^4)

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Maple [A]  time = 0.008, size = 72, normalized size = 1. \begin{align*}{\frac{A\ln \left ( x \right ) }{{a}^{4}}}+{\frac{A}{3\,a \left ( bx+a \right ) ^{3}}}-{\frac{B}{3\,b \left ( bx+a \right ) ^{3}}}-{\frac{A\ln \left ( bx+a \right ) }{{a}^{4}}}+{\frac{A}{{a}^{3} \left ( bx+a \right ) }}+{\frac{A}{2\,{a}^{2} \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0186, size = 123, normalized size = 1.71 \begin{align*} \frac{6 \, A b^{3} x^{2} + 15 \, A a b^{2} x - 2 \, B a^{3} + 11 \, A a^{2} b}{6 \,{\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} - \frac{A \log \left (b x + a\right )}{a^{4}} + \frac{A \log \left (x\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*A*b^3*x^2 + 15*A*a*b^2*x - 2*B*a^3 + 11*A*a^2*b)/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b) -
A*log(b*x + a)/a^4 + A*log(x)/a^4

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Fricas [B]  time = 1.3773, size = 336, normalized size = 4.67 \begin{align*} \frac{6 \, A a b^{3} x^{2} + 15 \, A a^{2} b^{2} x - 2 \, B a^{4} + 11 \, A a^{3} b - 6 \,{\left (A b^{4} x^{3} + 3 \, A a b^{3} x^{2} + 3 \, A a^{2} b^{2} x + A a^{3} b\right )} \log \left (b x + a\right ) + 6 \,{\left (A b^{4} x^{3} + 3 \, A a b^{3} x^{2} + 3 \, A a^{2} b^{2} x + A a^{3} b\right )} \log \left (x\right )}{6 \,{\left (a^{4} b^{4} x^{3} + 3 \, a^{5} b^{3} x^{2} + 3 \, a^{6} b^{2} x + a^{7} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*A*a*b^3*x^2 + 15*A*a^2*b^2*x - 2*B*a^4 + 11*A*a^3*b - 6*(A*b^4*x^3 + 3*A*a*b^3*x^2 + 3*A*a^2*b^2*x + A*
a^3*b)*log(b*x + a) + 6*(A*b^4*x^3 + 3*A*a*b^3*x^2 + 3*A*a^2*b^2*x + A*a^3*b)*log(x))/(a^4*b^4*x^3 + 3*a^5*b^3
*x^2 + 3*a^6*b^2*x + a^7*b)

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Sympy [A]  time = 0.784203, size = 90, normalized size = 1.25 \begin{align*} \frac{A \left (\log{\left (x \right )} - \log{\left (\frac{a}{b} + x \right )}\right )}{a^{4}} + \frac{11 A a^{2} b + 15 A a b^{2} x + 6 A b^{3} x^{2} - 2 B a^{3}}{6 a^{6} b + 18 a^{5} b^{2} x + 18 a^{4} b^{3} x^{2} + 6 a^{3} b^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(log(x) - log(a/b + x))/a**4 + (11*A*a**2*b + 15*A*a*b**2*x + 6*A*b**3*x**2 - 2*B*a**3)/(6*a**6*b + 18*a**5*
b**2*x + 18*a**4*b**3*x**2 + 6*a**3*b**4*x**3)

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Giac [A]  time = 1.1663, size = 96, normalized size = 1.33 \begin{align*} -\frac{A \log \left ({\left | b x + a \right |}\right )}{a^{4}} + \frac{A \log \left ({\left | x \right |}\right )}{a^{4}} + \frac{6 \, A a b^{3} x^{2} + 15 \, A a^{2} b^{2} x - 2 \, B a^{4} + 11 \, A a^{3} b}{6 \,{\left (b x + a\right )}^{3} a^{4} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*log(abs(b*x + a))/a^4 + A*log(abs(x))/a^4 + 1/6*(6*A*a*b^3*x^2 + 15*A*a^2*b^2*x - 2*B*a^4 + 11*A*a^3*b)/((b
*x + a)^3*a^4*b)