3.6.69 \(\int \frac {A+B x}{x^4 (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=166 \[ -\frac {10 b^2 \log (x) (2 A b-a B)}{a^7}+\frac {10 b^2 (2 A b-a B) \log (a+b x)}{a^7}-\frac {2 b^2 (5 A b-3 a B)}{a^6 (a+b x)}-\frac {2 b (5 A b-2 a B)}{a^6 x}-\frac {b^2 (4 A b-3 a B)}{2 a^5 (a+b x)^2}+\frac {4 A b-a B}{2 a^5 x^2}-\frac {b^2 (A b-a B)}{3 a^4 (a+b x)^3}-\frac {A}{3 a^4 x^3} \]

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Rubi [A]  time = 0.17, antiderivative size = 166, 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} \begin {gather*} -\frac {2 b^2 (5 A b-3 a B)}{a^6 (a+b x)}-\frac {b^2 (4 A b-3 a B)}{2 a^5 (a+b x)^2}-\frac {b^2 (A b-a B)}{3 a^4 (a+b x)^3}-\frac {10 b^2 \log (x) (2 A b-a B)}{a^7}+\frac {10 b^2 (2 A b-a B) \log (a+b x)}{a^7}+\frac {4 A b-a B}{2 a^5 x^2}-\frac {2 b (5 A b-2 a B)}{a^6 x}-\frac {A}{3 a^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

-A/(3*a^4*x^3) + (4*A*b - a*B)/(2*a^5*x^2) - (2*b*(5*A*b - 2*a*B))/(a^6*x) - (b^2*(A*b - a*B))/(3*a^4*(a + b*x
)^3) - (b^2*(4*A*b - 3*a*B))/(2*a^5*(a + b*x)^2) - (2*b^2*(5*A*b - 3*a*B))/(a^6*(a + b*x)) - (10*b^2*(2*A*b -
a*B)*Log[x])/a^7 + (10*b^2*(2*A*b - a*B)*Log[a + b*x])/a^7

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^4 \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {A+B x}{x^4 (a+b x)^4} \, dx\\ &=\int \left (\frac {A}{a^4 x^4}+\frac {-4 A b+a B}{a^5 x^3}-\frac {2 b (-5 A b+2 a B)}{a^6 x^2}+\frac {10 b^2 (-2 A b+a B)}{a^7 x}-\frac {b^3 (-A b+a B)}{a^4 (a+b x)^4}-\frac {b^3 (-4 A b+3 a B)}{a^5 (a+b x)^3}-\frac {2 b^3 (-5 A b+3 a B)}{a^6 (a+b x)^2}-\frac {10 b^3 (-2 A b+a B)}{a^7 (a+b x)}\right ) \, dx\\ &=-\frac {A}{3 a^4 x^3}+\frac {4 A b-a B}{2 a^5 x^2}-\frac {2 b (5 A b-2 a B)}{a^6 x}-\frac {b^2 (A b-a B)}{3 a^4 (a+b x)^3}-\frac {b^2 (4 A b-3 a B)}{2 a^5 (a+b x)^2}-\frac {2 b^2 (5 A b-3 a B)}{a^6 (a+b x)}-\frac {10 b^2 (2 A b-a B) \log (x)}{a^7}+\frac {10 b^2 (2 A b-a B) \log (a+b x)}{a^7}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 148, normalized size = 0.89 \begin {gather*} \frac {\frac {a \left (-\left (a^5 (2 A+3 B x)\right )+3 a^4 b x (2 A+5 B x)+10 a^3 b^2 x^2 (11 B x-3 A)+10 a^2 b^3 x^3 (15 B x-22 A)+60 a b^4 x^4 (B x-5 A)-120 A b^5 x^5\right )}{x^3 (a+b x)^3}-60 b^2 \log (x) (2 A b-a B)+60 b^2 (2 A b-a B) \log (a+b x)}{6 a^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

((a*(-120*A*b^5*x^5 + 60*a*b^4*x^4*(-5*A + B*x) - a^5*(2*A + 3*B*x) + 3*a^4*b*x*(2*A + 5*B*x) + 10*a^3*b^2*x^2
*(-3*A + 11*B*x) + 10*a^2*b^3*x^3*(-22*A + 15*B*x)))/(x^3*(a + b*x)^3) - 60*b^2*(2*A*b - a*B)*Log[x] + 60*b^2*
(2*A*b - a*B)*Log[a + b*x])/(6*a^7)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

IntegrateAlgebraic[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)^2), x]

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fricas [B]  time = 0.41, size = 333, normalized size = 2.01 \begin {gather*} -\frac {2 \, A a^{6} - 60 \, {\left (B a^{2} b^{4} - 2 \, A a b^{5}\right )} x^{5} - 150 \, {\left (B a^{3} b^{3} - 2 \, A a^{2} b^{4}\right )} x^{4} - 110 \, {\left (B a^{4} b^{2} - 2 \, A a^{3} b^{3}\right )} x^{3} - 15 \, {\left (B a^{5} b - 2 \, A a^{4} b^{2}\right )} x^{2} + 3 \, {\left (B a^{6} - 2 \, A a^{5} b\right )} x + 60 \, {\left ({\left (B a b^{5} - 2 \, A b^{6}\right )} x^{6} + 3 \, {\left (B a^{2} b^{4} - 2 \, A a b^{5}\right )} x^{5} + 3 \, {\left (B a^{3} b^{3} - 2 \, A a^{2} b^{4}\right )} x^{4} + {\left (B a^{4} b^{2} - 2 \, A a^{3} b^{3}\right )} x^{3}\right )} \log \left (b x + a\right ) - 60 \, {\left ({\left (B a b^{5} - 2 \, A b^{6}\right )} x^{6} + 3 \, {\left (B a^{2} b^{4} - 2 \, A a b^{5}\right )} x^{5} + 3 \, {\left (B a^{3} b^{3} - 2 \, A a^{2} b^{4}\right )} x^{4} + {\left (B a^{4} b^{2} - 2 \, A a^{3} b^{3}\right )} x^{3}\right )} \log \relax (x)}{6 \, {\left (a^{7} b^{3} x^{6} + 3 \, a^{8} b^{2} x^{5} + 3 \, a^{9} b x^{4} + a^{10} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

-1/6*(2*A*a^6 - 60*(B*a^2*b^4 - 2*A*a*b^5)*x^5 - 150*(B*a^3*b^3 - 2*A*a^2*b^4)*x^4 - 110*(B*a^4*b^2 - 2*A*a^3*
b^3)*x^3 - 15*(B*a^5*b - 2*A*a^4*b^2)*x^2 + 3*(B*a^6 - 2*A*a^5*b)*x + 60*((B*a*b^5 - 2*A*b^6)*x^6 + 3*(B*a^2*b
^4 - 2*A*a*b^5)*x^5 + 3*(B*a^3*b^3 - 2*A*a^2*b^4)*x^4 + (B*a^4*b^2 - 2*A*a^3*b^3)*x^3)*log(b*x + a) - 60*((B*a
*b^5 - 2*A*b^6)*x^6 + 3*(B*a^2*b^4 - 2*A*a*b^5)*x^5 + 3*(B*a^3*b^3 - 2*A*a^2*b^4)*x^4 + (B*a^4*b^2 - 2*A*a^3*b
^3)*x^3)*log(x))/(a^7*b^3*x^6 + 3*a^8*b^2*x^5 + 3*a^9*b*x^4 + a^10*x^3)

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giac [A]  time = 0.15, size = 175, normalized size = 1.05 \begin {gather*} \frac {10 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | x \right |}\right )}{a^{7}} - \frac {10 \, {\left (B a b^{3} - 2 \, A b^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{7} b} + \frac {60 \, B a b^{4} x^{5} - 120 \, A b^{5} x^{5} + 150 \, B a^{2} b^{3} x^{4} - 300 \, A a b^{4} x^{4} + 110 \, B a^{3} b^{2} x^{3} - 220 \, A a^{2} b^{3} x^{3} + 15 \, B a^{4} b x^{2} - 30 \, A a^{3} b^{2} x^{2} - 3 \, B a^{5} x + 6 \, A a^{4} b x - 2 \, A a^{5}}{6 \, {\left (b x^{2} + a x\right )}^{3} a^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

10*(B*a*b^2 - 2*A*b^3)*log(abs(x))/a^7 - 10*(B*a*b^3 - 2*A*b^4)*log(abs(b*x + a))/(a^7*b) + 1/6*(60*B*a*b^4*x^
5 - 120*A*b^5*x^5 + 150*B*a^2*b^3*x^4 - 300*A*a*b^4*x^4 + 110*B*a^3*b^2*x^3 - 220*A*a^2*b^3*x^3 + 15*B*a^4*b*x
^2 - 30*A*a^3*b^2*x^2 - 3*B*a^5*x + 6*A*a^4*b*x - 2*A*a^5)/((b*x^2 + a*x)^3*a^6)

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maple [A]  time = 0.07, size = 200, normalized size = 1.20 \begin {gather*} -\frac {A \,b^{3}}{3 \left (b x +a \right )^{3} a^{4}}+\frac {B \,b^{2}}{3 \left (b x +a \right )^{3} a^{3}}-\frac {2 A \,b^{3}}{\left (b x +a \right )^{2} a^{5}}+\frac {3 B \,b^{2}}{2 \left (b x +a \right )^{2} a^{4}}-\frac {10 A \,b^{3}}{\left (b x +a \right ) a^{6}}-\frac {20 A \,b^{3} \ln \relax (x )}{a^{7}}+\frac {20 A \,b^{3} \ln \left (b x +a \right )}{a^{7}}+\frac {6 B \,b^{2}}{\left (b x +a \right ) a^{5}}+\frac {10 B \,b^{2} \ln \relax (x )}{a^{6}}-\frac {10 B \,b^{2} \ln \left (b x +a \right )}{a^{6}}-\frac {10 A \,b^{2}}{a^{6} x}+\frac {4 B b}{a^{5} x}+\frac {2 A b}{a^{5} x^{2}}-\frac {B}{2 a^{4} x^{2}}-\frac {A}{3 a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-2*b^3/a^5/(b*x+a)^2*A+3/2*b^2/a^4/(b*x+a)^2*B-1/3*b^3/a^4/(b*x+a)^3*A+1/3*b^2/a^3/(b*x+a)^3*B+20*b^3/a^7*ln(b
*x+a)*A-10*b^2/a^6*ln(b*x+a)*B-10*b^3/a^6/(b*x+a)*A+6*b^2/a^5/(b*x+a)*B-1/3*A/a^4/x^3+2/a^5/x^2*A*b-1/2/a^4/x^
2*B-10*b^2/a^6/x*A+4*b/a^5/x*B-20*b^3/a^7*ln(x)*A+10*b^2/a^6*ln(x)*B

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maxima [A]  time = 0.67, size = 193, normalized size = 1.16 \begin {gather*} -\frac {2 \, A a^{5} - 60 \, {\left (B a b^{4} - 2 \, A b^{5}\right )} x^{5} - 150 \, {\left (B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{4} - 110 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} - 15 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} x}{6 \, {\left (a^{6} b^{3} x^{6} + 3 \, a^{7} b^{2} x^{5} + 3 \, a^{8} b x^{4} + a^{9} x^{3}\right )}} - \frac {10 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} \log \left (b x + a\right )}{a^{7}} + \frac {10 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} \log \relax (x)}{a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

-1/6*(2*A*a^5 - 60*(B*a*b^4 - 2*A*b^5)*x^5 - 150*(B*a^2*b^3 - 2*A*a*b^4)*x^4 - 110*(B*a^3*b^2 - 2*A*a^2*b^3)*x
^3 - 15*(B*a^4*b - 2*A*a^3*b^2)*x^2 + 3*(B*a^5 - 2*A*a^4*b)*x)/(a^6*b^3*x^6 + 3*a^7*b^2*x^5 + 3*a^8*b*x^4 + a^
9*x^3) - 10*(B*a*b^2 - 2*A*b^3)*log(b*x + a)/a^7 + 10*(B*a*b^2 - 2*A*b^3)*log(x)/a^7

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mupad [B]  time = 1.19, size = 195, normalized size = 1.17 \begin {gather*} \frac {20\,b^2\,\mathrm {atanh}\left (\frac {10\,b^2\,\left (2\,A\,b-B\,a\right )\,\left (a+2\,b\,x\right )}{a\,\left (20\,A\,b^3-10\,B\,a\,b^2\right )}\right )\,\left (2\,A\,b-B\,a\right )}{a^7}-\frac {\frac {A}{3\,a}-\frac {x\,\left (2\,A\,b-B\,a\right )}{2\,a^2}+\frac {55\,b^2\,x^3\,\left (2\,A\,b-B\,a\right )}{3\,a^4}+\frac {25\,b^3\,x^4\,\left (2\,A\,b-B\,a\right )}{a^5}+\frac {10\,b^4\,x^5\,\left (2\,A\,b-B\,a\right )}{a^6}+\frac {5\,b\,x^2\,\left (2\,A\,b-B\,a\right )}{2\,a^3}}{a^3\,x^3+3\,a^2\,b\,x^4+3\,a\,b^2\,x^5+b^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^4*(a^2 + b^2*x^2 + 2*a*b*x)^2),x)

[Out]

(20*b^2*atanh((10*b^2*(2*A*b - B*a)*(a + 2*b*x))/(a*(20*A*b^3 - 10*B*a*b^2)))*(2*A*b - B*a))/a^7 - (A/(3*a) -
(x*(2*A*b - B*a))/(2*a^2) + (55*b^2*x^3*(2*A*b - B*a))/(3*a^4) + (25*b^3*x^4*(2*A*b - B*a))/a^5 + (10*b^4*x^5*
(2*A*b - B*a))/a^6 + (5*b*x^2*(2*A*b - B*a))/(2*a^3))/(a^3*x^3 + b^3*x^6 + 3*a^2*b*x^4 + 3*a*b^2*x^5)

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sympy [A]  time = 0.94, size = 291, normalized size = 1.75 \begin {gather*} \frac {- 2 A a^{5} + x^{5} \left (- 120 A b^{5} + 60 B a b^{4}\right ) + x^{4} \left (- 300 A a b^{4} + 150 B a^{2} b^{3}\right ) + x^{3} \left (- 220 A a^{2} b^{3} + 110 B a^{3} b^{2}\right ) + x^{2} \left (- 30 A a^{3} b^{2} + 15 B a^{4} b\right ) + x \left (6 A a^{4} b - 3 B a^{5}\right )}{6 a^{9} x^{3} + 18 a^{8} b x^{4} + 18 a^{7} b^{2} x^{5} + 6 a^{6} b^{3} x^{6}} + \frac {10 b^{2} \left (- 2 A b + B a\right ) \log {\left (x + \frac {- 20 A a b^{3} + 10 B a^{2} b^{2} - 10 a b^{2} \left (- 2 A b + B a\right )}{- 40 A b^{4} + 20 B a b^{3}} \right )}}{a^{7}} - \frac {10 b^{2} \left (- 2 A b + B a\right ) \log {\left (x + \frac {- 20 A a b^{3} + 10 B a^{2} b^{2} + 10 a b^{2} \left (- 2 A b + B a\right )}{- 40 A b^{4} + 20 B a b^{3}} \right )}}{a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**4/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

(-2*A*a**5 + x**5*(-120*A*b**5 + 60*B*a*b**4) + x**4*(-300*A*a*b**4 + 150*B*a**2*b**3) + x**3*(-220*A*a**2*b**
3 + 110*B*a**3*b**2) + x**2*(-30*A*a**3*b**2 + 15*B*a**4*b) + x*(6*A*a**4*b - 3*B*a**5))/(6*a**9*x**3 + 18*a**
8*b*x**4 + 18*a**7*b**2*x**5 + 6*a**6*b**3*x**6) + 10*b**2*(-2*A*b + B*a)*log(x + (-20*A*a*b**3 + 10*B*a**2*b*
*2 - 10*a*b**2*(-2*A*b + B*a))/(-40*A*b**4 + 20*B*a*b**3))/a**7 - 10*b**2*(-2*A*b + B*a)*log(x + (-20*A*a*b**3
+ 10*B*a**2*b**2 + 10*a*b**2*(-2*A*b + B*a))/(-40*A*b**4 + 20*B*a*b**3))/a**7

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