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

Optimal. Leaf size=50 \[ \frac{(A b-a B) \log \left (a+b x^2\right )}{2 a^2}-\frac{\log (x) (A b-a B)}{a^2}-\frac{A}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0476906, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 77} \[ \frac{(A b-a B) \log \left (a+b x^2\right )}{2 a^2}-\frac{\log (x) (A b-a B)}{a^2}-\frac{A}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^3*(a + b*x^2)),x]

[Out]

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

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]]

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

Mathematica [A]  time = 0.0203936, size = 49, normalized size = 0.98 \[ \frac{(A b-a B) \log \left (a+b x^2\right )}{2 a^2}+\frac{\log (x) (a B-A b)}{a^2}-\frac{A}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 56, normalized size = 1.1 \begin{align*} -{\frac{A}{2\,a{x}^{2}}}-{\frac{A\ln \left ( x \right ) b}{{a}^{2}}}+{\frac{\ln \left ( x \right ) B}{a}}+{\frac{\ln \left ( b{x}^{2}+a \right ) Ab}{2\,{a}^{2}}}-{\frac{\ln \left ( b{x}^{2}+a \right ) B}{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00942, size = 65, normalized size = 1.3 \begin{align*} -\frac{{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac{{\left (B a - A b\right )} \log \left (x^{2}\right )}{2 \, a^{2}} - \frac{A}{2 \, a x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.18832, size = 111, normalized size = 2.22 \begin{align*} -\frac{{\left (B a - A b\right )} x^{2} \log \left (b x^{2} + a\right ) - 2 \,{\left (B a - A b\right )} x^{2} \log \left (x\right ) + A a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.790597, size = 41, normalized size = 0.82 \begin{align*} - \frac{A}{2 a x^{2}} + \frac{\left (- A b + B a\right ) \log{\left (x \right )}}{a^{2}} - \frac{\left (- A b + B a\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14143, size = 96, normalized size = 1.92 \begin{align*} \frac{{\left (B a - A b\right )} \log \left (x^{2}\right )}{2 \, a^{2}} - \frac{{\left (B a b - A b^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2} b} - \frac{B a x^{2} - A b x^{2} + A a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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