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

Optimal. Leaf size=69 \[ -\frac {b (A b-a B) \log \left (a+b x^2\right )}{2 a^3}+\frac {b \log (x) (A b-a B)}{a^3}+\frac {A b-a B}{2 a^2 x^2}-\frac {A}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 69, 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} \[ \frac {A b-a B}{2 a^2 x^2}-\frac {b (A b-a B) \log \left (a+b x^2\right )}{2 a^3}+\frac {b \log (x) (A b-a B)}{a^3}-\frac {A}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 70, normalized size = 1.01 \[ \frac {4 b x^4 \log (x) (A b-a B)-a \left (a A+2 a B x^2-2 A b x^2\right )+2 b x^4 (a B-A b) \log \left (a+b x^2\right )}{4 a^3 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 73, normalized size = 1.06 \[ \frac {2 \, {\left (B a b - A b^{2}\right )} x^{4} \log \left (b x^{2} + a\right ) - 4 \, {\left (B a b - A b^{2}\right )} x^{4} \log \relax (x) - A a^{2} - 2 \, {\left (B a^{2} - A a b\right )} x^{2}}{4 \, a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 100, normalized size = 1.45 \[ -\frac {{\left (B a b - A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3}} + \frac {{\left (B a b^{2} - A b^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} b} + \frac {3 \, B a b x^{4} - 3 \, A b^{2} x^{4} - 2 \, B a^{2} x^{2} + 2 \, A a b x^{2} - A a^{2}}{4 \, a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 81, normalized size = 1.17 \[ \frac {A \,b^{2} \ln \relax (x )}{a^{3}}-\frac {A \,b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{3}}-\frac {B b \ln \relax (x )}{a^{2}}+\frac {B b \ln \left (b \,x^{2}+a \right )}{2 a^{2}}+\frac {A b}{2 a^{2} x^{2}}-\frac {B}{2 a \,x^{2}}-\frac {A}{4 a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.12, size = 70, normalized size = 1.01 \[ \frac {{\left (B a b - A b^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} - \frac {{\left (B a b - A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac {2 \, {\left (B a - A b\right )} x^{2} + A a}{4 \, a^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 70, normalized size = 1.01 \[ \frac {\ln \relax (x)\,\left (A\,b^2-B\,a\,b\right )}{a^3}-\frac {\ln \left (b\,x^2+a\right )\,\left (A\,b^2-B\,a\,b\right )}{2\,a^3}-\frac {\frac {A}{4\,a}-\frac {x^2\,\left (A\,b-B\,a\right )}{2\,a^2}}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^5*(a + b*x^2)),x)

[Out]

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

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sympy [A]  time = 0.83, size = 61, normalized size = 0.88 \[ \frac {- A a + x^{2} \left (2 A b - 2 B a\right )}{4 a^{2} x^{4}} - \frac {b \left (- A b + B a\right ) \log {\relax (x )}}{a^{3}} + \frac {b \left (- A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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