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

Optimal. Leaf size=101 \[ \frac {(3 A b-a B) \log \left (a+b x^2\right )}{2 a^4}-\frac {\log (x) (3 A b-a B)}{a^4}-\frac {2 A b-a B}{2 a^3 \left (a+b x^2\right )}-\frac {A}{2 a^3 x^2}-\frac {A b-a B}{4 a^2 \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 101, 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} \[ -\frac {2 A b-a B}{2 a^3 \left (a+b x^2\right )}-\frac {A b-a B}{4 a^2 \left (a+b x^2\right )^2}+\frac {(3 A b-a B) \log \left (a+b x^2\right )}{2 a^4}-\frac {\log (x) (3 A b-a B)}{a^4}-\frac {A}{2 a^3 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.86 \[ \frac {\frac {a^2 (a B-A b)}{\left (a+b x^2\right )^2}+\frac {2 a (a B-2 A b)}{a+b x^2}+2 (3 A b-a B) \log \left (a+b x^2\right )+4 \log (x) (a B-3 A b)-\frac {2 a A}{x^2}}{4 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 197, normalized size = 1.95 \[ \frac {2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{4} - 2 \, A a^{3} + 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} x^{2} - 2 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{6} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{4} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{6} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{4} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x^{2}\right )} \log \relax (x)}{4 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 138, normalized size = 1.37 \[ \frac {{\left (B a - 3 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{4}} - \frac {{\left (B a b - 3 \, A b^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b} + \frac {3 \, B a b^{2} x^{4} - 9 \, A b^{3} x^{4} + 8 \, B a^{2} b x^{2} - 22 \, A a b^{2} x^{2} + 6 \, B a^{3} - 14 \, A a^{2} b}{4 \, {\left (b x^{2} + a\right )}^{2} a^{4}} - \frac {B a x^{2} - 3 \, A b x^{2} + A a}{2 \, a^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*a - 3*A*b)*log(x^2)/a^4 - 1/2*(B*a*b - 3*A*b^2)*log(abs(b*x^2 + a))/(a^4*b) + 1/4*(3*B*a*b^2*x^4 - 9*A*
b^3*x^4 + 8*B*a^2*b*x^2 - 22*A*a*b^2*x^2 + 6*B*a^3 - 14*A*a^2*b)/((b*x^2 + a)^2*a^4) - 1/2*(B*a*x^2 - 3*A*b*x^
2 + A*a)/(a^4*x^2)

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maple [A]  time = 0.02, size = 118, normalized size = 1.17 \[ -\frac {A b}{4 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {B}{4 \left (b \,x^{2}+a \right )^{2} a}-\frac {A b}{\left (b \,x^{2}+a \right ) a^{3}}-\frac {3 A b \ln \relax (x )}{a^{4}}+\frac {3 A b \ln \left (b \,x^{2}+a \right )}{2 a^{4}}+\frac {B}{2 \left (b \,x^{2}+a \right ) a^{2}}+\frac {B \ln \relax (x )}{a^{3}}-\frac {B \ln \left (b \,x^{2}+a \right )}{2 a^{3}}-\frac {A}{2 a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.00, size = 109, normalized size = 1.08 \[ \frac {2 \, {\left (B a b - 3 \, A b^{2}\right )} x^{4} - 2 \, A a^{2} + 3 \, {\left (B a^{2} - 3 \, A a b\right )} x^{2}}{4 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} - \frac {{\left (B a - 3 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {{\left (B a - 3 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 107, normalized size = 1.06 \[ \frac {\ln \left (b\,x^2+a\right )\,\left (3\,A\,b-B\,a\right )}{2\,a^4}-\frac {\frac {A}{2\,a}+\frac {3\,x^2\,\left (3\,A\,b-B\,a\right )}{4\,a^2}+\frac {b\,x^4\,\left (3\,A\,b-B\,a\right )}{2\,a^3}}{a^2\,x^2+2\,a\,b\,x^4+b^2\,x^6}-\frac {\ln \relax (x)\,\left (3\,A\,b-B\,a\right )}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.05, size = 107, normalized size = 1.06 \[ \frac {- 2 A a^{2} + x^{4} \left (- 6 A b^{2} + 2 B a b\right ) + x^{2} \left (- 9 A a b + 3 B a^{2}\right )}{4 a^{5} x^{2} + 8 a^{4} b x^{4} + 4 a^{3} b^{2} x^{6}} + \frac {\left (- 3 A b + B a\right ) \log {\relax (x )}}{a^{4}} - \frac {\left (- 3 A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*A*a**2 + x**4*(-6*A*b**2 + 2*B*a*b) + x**2*(-9*A*a*b + 3*B*a**2))/(4*a**5*x**2 + 8*a**4*b*x**4 + 4*a**3*b*
*2*x**6) + (-3*A*b + B*a)*log(x)/a**4 - (-3*A*b + B*a)*log(a/b + x**2)/(2*a**4)

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