3.83 \(\int \frac {A+B x^2}{x^3 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=76 \[ \frac {(2 A b-a B) \log \left (a+b x^2\right )}{2 a^3}-\frac {\log (x) (2 A b-a B)}{a^3}-\frac {A b-a B}{2 a^2 \left (a+b x^2\right )}-\frac {A}{2 a^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 76, 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} \[ -\frac {A b-a B}{2 a^2 \left (a+b x^2\right )}+\frac {(2 A b-a B) \log \left (a+b x^2\right )}{2 a^3}-\frac {\log (x) (2 A b-a B)}{a^3}-\frac {A}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^3*(a + b*x^2)^2),x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.84 \[ \frac {\frac {a (a B-A b)}{a+b x^2}+(2 A b-a B) \log \left (a+b x^2\right )+2 \log (x) (a B-2 A b)-\frac {a A}{x^2}}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^3*(a + b*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 117, normalized size = 1.54 \[ -\frac {A a^{2} - {\left (B a^{2} - 2 \, A a b\right )} x^{2} + {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{4} + {\left (B a^{2} - 2 \, A a b\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{4} + {\left (B a^{2} - 2 \, A a b\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 82, normalized size = 1.08 \[ \frac {{\left (B a - 2 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{3}} + \frac {B a x^{2} - 2 \, A b x^{2} - A a}{2 \, {\left (b x^{4} + a x^{2}\right )} a^{2}} - \frac {{\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 86, normalized size = 1.13 \[ -\frac {A b}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {2 A b \ln \relax (x )}{a^{3}}+\frac {A b \ln \left (b \,x^{2}+a \right )}{a^{3}}+\frac {B}{2 \left (b \,x^{2}+a \right ) a}+\frac {B \ln \relax (x )}{a^{2}}-\frac {B \ln \left (b \,x^{2}+a \right )}{2 a^{2}}-\frac {A}{2 a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^3/(b*x^2+a)^2,x)

[Out]

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

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maxima [A]  time = 1.06, size = 76, normalized size = 1.00 \[ \frac {{\left (B a - 2 \, A b\right )} x^{2} - A a}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} - \frac {{\left (B a - 2 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {{\left (B a - 2 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.14, size = 78, normalized size = 1.03 \[ \frac {\ln \left (b\,x^2+a\right )\,\left (2\,A\,b-B\,a\right )}{2\,a^3}-\frac {\frac {A}{2\,a}+\frac {x^2\,\left (2\,A\,b-B\,a\right )}{2\,a^2}}{b\,x^4+a\,x^2}-\frac {\ln \relax (x)\,\left (2\,A\,b-B\,a\right )}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^3*(a + b*x^2)^2),x)

[Out]

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

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sympy [A]  time = 0.88, size = 70, normalized size = 0.92 \[ \frac {- A a + x^{2} \left (- 2 A b + B a\right )}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac {\left (- 2 A b + B a\right ) \log {\relax (x )}}{a^{3}} - \frac {\left (- 2 A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**3/(b*x**2+a)**2,x)

[Out]

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

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