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

Optimal. Leaf size=111 \[ -\frac {\log (x) (4 A b-a B)}{a^5}+\frac {(4 A b-a B) \log (a+b x)}{a^5}-\frac {3 A b-a B}{a^4 (a+b x)}-\frac {A}{a^4 x}-\frac {2 A b-a B}{2 a^3 (a+b x)^2}-\frac {A b-a B}{3 a^2 (a+b x)^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, 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} \[ -\frac {3 A b-a B}{a^4 (a+b x)}-\frac {2 A b-a B}{2 a^3 (a+b x)^2}-\frac {A b-a B}{3 a^2 (a+b x)^3}-\frac {\log (x) (4 A b-a B)}{a^5}+\frac {(4 A b-a B) \log (a+b x)}{a^5}-\frac {A}{a^4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^4*x)) - (A*b - a*B)/(3*a^2*(a + b*x)^3) - (2*A*b - a*B)/(2*a^3*(a + b*x)^2) - (3*A*b - a*B)/(a^4*(a + b
*x)) - ((4*A*b - a*B)*Log[x])/a^5 + ((4*A*b - a*B)*Log[a + b*x])/a^5

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^2 \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {A+B x}{x^2 (a+b x)^4} \, dx\\ &=\int \left (\frac {A}{a^4 x^2}+\frac {-4 A b+a B}{a^5 x}-\frac {b (-A b+a B)}{a^2 (a+b x)^4}-\frac {b (-2 A b+a B)}{a^3 (a+b x)^3}-\frac {b (-3 A b+a B)}{a^4 (a+b x)^2}-\frac {b (-4 A b+a B)}{a^5 (a+b x)}\right ) \, dx\\ &=-\frac {A}{a^4 x}-\frac {A b-a B}{3 a^2 (a+b x)^3}-\frac {2 A b-a B}{2 a^3 (a+b x)^2}-\frac {3 A b-a B}{a^4 (a+b x)}-\frac {(4 A b-a B) \log (x)}{a^5}+\frac {(4 A b-a B) \log (a+b x)}{a^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 102, normalized size = 0.92 \[ \frac {\frac {2 a^3 (a B-A b)}{(a+b x)^3}+\frac {3 a^2 (a B-2 A b)}{(a+b x)^2}+\frac {6 a (a B-3 A b)}{a+b x}+6 \log (x) (a B-4 A b)+6 (4 A b-a B) \log (a+b x)-\frac {6 a A}{x}}{6 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 267, normalized size = 2.41 \[ -\frac {6 \, A a^{4} - 6 \, {\left (B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} - 15 \, {\left (B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} - 11 \, {\left (B a^{4} - 4 \, A a^{3} b\right )} x + 6 \, {\left ({\left (B a b^{3} - 4 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 4 \, A a^{3} b\right )} x\right )} \log \left (b x + a\right ) - 6 \, {\left ({\left (B a b^{3} - 4 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 4 \, A a^{3} b\right )} x\right )} \log \relax (x)}{6 \, {\left (a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{3} + 3 \, a^{7} b x^{2} + a^{8} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*A*a^4 - 6*(B*a^2*b^2 - 4*A*a*b^3)*x^3 - 15*(B*a^3*b - 4*A*a^2*b^2)*x^2 - 11*(B*a^4 - 4*A*a^3*b)*x + 6*
((B*a*b^3 - 4*A*b^4)*x^4 + 3*(B*a^2*b^2 - 4*A*a*b^3)*x^3 + 3*(B*a^3*b - 4*A*a^2*b^2)*x^2 + (B*a^4 - 4*A*a^3*b)
*x)*log(b*x + a) - 6*((B*a*b^3 - 4*A*b^4)*x^4 + 3*(B*a^2*b^2 - 4*A*a*b^3)*x^3 + 3*(B*a^3*b - 4*A*a^2*b^2)*x^2
+ (B*a^4 - 4*A*a^3*b)*x)*log(x))/(a^5*b^3*x^4 + 3*a^6*b^2*x^3 + 3*a^7*b*x^2 + a^8*x)

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giac [A]  time = 0.17, size = 122, normalized size = 1.10 \[ \frac {{\left (B a - 4 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{5}} - \frac {{\left (B a b - 4 \, A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b} - \frac {6 \, A a^{4} - 6 \, {\left (B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} - 15 \, {\left (B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} - 11 \, {\left (B a^{4} - 4 \, A a^{3} b\right )} x}{6 \, {\left (b x + a\right )}^{3} a^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 132, normalized size = 1.19 \[ -\frac {A b}{3 \left (b x +a \right )^{3} a^{2}}+\frac {B}{3 \left (b x +a \right )^{3} a}-\frac {A b}{\left (b x +a \right )^{2} a^{3}}+\frac {B}{2 \left (b x +a \right )^{2} a^{2}}-\frac {3 A b}{\left (b x +a \right ) a^{4}}-\frac {4 A b \ln \relax (x )}{a^{5}}+\frac {4 A b \ln \left (b x +a \right )}{a^{5}}+\frac {B}{\left (b x +a \right ) a^{3}}+\frac {B \ln \relax (x )}{a^{4}}-\frac {B \ln \left (b x +a \right )}{a^{4}}-\frac {A}{a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 134, normalized size = 1.21 \[ -\frac {6 \, A a^{3} - 6 \, {\left (B a b^{2} - 4 \, A b^{3}\right )} x^{3} - 15 \, {\left (B a^{2} b - 4 \, A a b^{2}\right )} x^{2} - 11 \, {\left (B a^{3} - 4 \, A a^{2} b\right )} x}{6 \, {\left (a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{3} + 3 \, a^{6} b x^{2} + a^{7} x\right )}} - \frac {{\left (B a - 4 \, A b\right )} \log \left (b x + a\right )}{a^{5}} + \frac {{\left (B a - 4 \, A b\right )} \log \relax (x)}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 118, normalized size = 1.06 \[ \frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )\,\left (4\,A\,b-B\,a\right )}{a^5}-\frac {\frac {A}{a}+\frac {11\,x\,\left (4\,A\,b-B\,a\right )}{6\,a^2}+\frac {b^2\,x^3\,\left (4\,A\,b-B\,a\right )}{a^4}+\frac {5\,b\,x^2\,\left (4\,A\,b-B\,a\right )}{2\,a^3}}{a^3\,x+3\,a^2\,b\,x^2+3\,a\,b^2\,x^3+b^3\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.76, size = 204, normalized size = 1.84 \[ \frac {- 6 A a^{3} + x^{3} \left (- 24 A b^{3} + 6 B a b^{2}\right ) + x^{2} \left (- 60 A a b^{2} + 15 B a^{2} b\right ) + x \left (- 44 A a^{2} b + 11 B a^{3}\right )}{6 a^{7} x + 18 a^{6} b x^{2} + 18 a^{5} b^{2} x^{3} + 6 a^{4} b^{3} x^{4}} + \frac {\left (- 4 A b + B a\right ) \log {\left (x + \frac {- 4 A a b + B a^{2} - a \left (- 4 A b + B a\right )}{- 8 A b^{2} + 2 B a b} \right )}}{a^{5}} - \frac {\left (- 4 A b + B a\right ) \log {\left (x + \frac {- 4 A a b + B a^{2} + a \left (- 4 A b + B a\right )}{- 8 A b^{2} + 2 B a b} \right )}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-6*A*a**3 + x**3*(-24*A*b**3 + 6*B*a*b**2) + x**2*(-60*A*a*b**2 + 15*B*a**2*b) + x*(-44*A*a**2*b + 11*B*a**3)
)/(6*a**7*x + 18*a**6*b*x**2 + 18*a**5*b**2*x**3 + 6*a**4*b**3*x**4) + (-4*A*b + B*a)*log(x + (-4*A*a*b + B*a*
*2 - a*(-4*A*b + B*a))/(-8*A*b**2 + 2*B*a*b))/a**5 - (-4*A*b + B*a)*log(x + (-4*A*a*b + B*a**2 + a*(-4*A*b + B
*a))/(-8*A*b**2 + 2*B*a*b))/a**5

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