3.85 \(\int \frac {A+B x^2}{x^5 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=97 \[ -\frac {b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}+\frac {b \log (x) (3 A b-2 a B)}{a^4}+\frac {b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {A}{4 a^2 x^4} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 77} \[ \frac {b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}+\frac {b \log (x) (3 A b-2 a B)}{a^4}-\frac {A}{4 a^2 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^2),x]

[Out]

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

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]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^3 (a+b x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {A}{a^2 x^3}+\frac {-2 A b+a B}{a^3 x^2}-\frac {b (-3 A b+2 a B)}{a^4 x}+\frac {b^2 (-A b+a B)}{a^3 (a+b x)^2}+\frac {b^2 (-3 A b+2 a B)}{a^4 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {A}{4 a^2 x^4}+\frac {2 A b-a B}{2 a^3 x^2}+\frac {b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac {b (3 A b-2 a B) \log (x)}{a^4}-\frac {b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 85, normalized size = 0.88 \[ -\frac {\frac {a^2 A}{x^4}+\frac {2 a b (a B-A b)}{a+b x^2}+\frac {2 a (a B-2 A b)}{x^2}+2 b (3 A b-2 a B) \log \left (a+b x^2\right )-4 b \log (x) (3 A b-2 a B)}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.45, size = 154, normalized size = 1.59 \[ -\frac {2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4} + A a^{3} + {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} x^{2} - 2 \, {\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \relax (x)}{4 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.39, size = 150, normalized size = 1.55 \[ -\frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} + \frac {{\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b} - \frac {2 \, B a b^{2} x^{2} - 3 \, A b^{3} x^{2} + 3 \, B a^{2} b - 4 \, A a b^{2}}{2 \, {\left (b x^{2} + a\right )} a^{4}} + \frac {6 \, B a b x^{4} - 9 \, A b^{2} x^{4} - 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} - A a^{2}}{4 \, a^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 114, normalized size = 1.18 \[ \frac {A \,b^{2}}{2 \left (b \,x^{2}+a \right ) a^{3}}+\frac {3 A \,b^{2} \ln \relax (x )}{a^{4}}-\frac {3 A \,b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{4}}-\frac {B b}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {2 B b \ln \relax (x )}{a^{3}}+\frac {B b \ln \left (b \,x^{2}+a \right )}{a^{3}}+\frac {A b}{a^{3} x^{2}}-\frac {B}{2 a^{2} x^{2}}-\frac {A}{4 a^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(b*x^2+a)^2,x)

[Out]

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

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maxima [A]  time = 1.13, size = 106, normalized size = 1.09 \[ -\frac {2 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + A a^{2} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}}{4 \, {\left (a^{3} b x^{6} + a^{4} x^{4}\right )}} + \frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.14, size = 100, normalized size = 1.03 \[ \frac {\frac {x^2\,\left (3\,A\,b-2\,B\,a\right )}{4\,a^2}-\frac {A}{4\,a}+\frac {b\,x^4\,\left (3\,A\,b-2\,B\,a\right )}{2\,a^3}}{b\,x^6+a\,x^4}-\frac {\ln \left (b\,x^2+a\right )\,\left (3\,A\,b^2-2\,B\,a\,b\right )}{2\,a^4}+\frac {\ln \relax (x)\,\left (3\,A\,b^2-2\,B\,a\,b\right )}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^5*(a + b*x^2)^2),x)

[Out]

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

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sympy [A]  time = 1.02, size = 100, normalized size = 1.03 \[ \frac {- A a^{2} + x^{4} \left (6 A b^{2} - 4 B a b\right ) + x^{2} \left (3 A a b - 2 B a^{2}\right )}{4 a^{4} x^{4} + 4 a^{3} b x^{6}} - \frac {b \left (- 3 A b + 2 B a\right ) \log {\relax (x )}}{a^{4}} + \frac {b \left (- 3 A b + 2 B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**2,x)

[Out]

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

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