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

Optimal. Leaf size=34 \[ \frac{A \log (x)}{a}-\frac{(A b-a B) \log \left (a+b x^2\right )}{2 a b} \]

[Out]

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

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Rubi [A]  time = 0.0335423, antiderivative size = 34, 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, 72} \[ \frac{A \log (x)}{a}-\frac{(A b-a B) \log \left (a+b x^2\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x*(a + b*x^2)),x]

[Out]

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

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0121514, size = 34, normalized size = 1. \[ \frac{(a B-A b) \log \left (a+b x^2\right )}{2 a b}+\frac{A \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 37, normalized size = 1.1 \begin{align*}{\frac{A\ln \left ( x \right ) }{a}}-{\frac{\ln \left ( b{x}^{2}+a \right ) A}{2\,a}}+{\frac{\ln \left ( b{x}^{2}+a \right ) B}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00515, size = 47, normalized size = 1.38 \begin{align*} \frac{A \log \left (x^{2}\right )}{2 \, a} + \frac{{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.21652, size = 74, normalized size = 2.18 \begin{align*} \frac{2 \, A b \log \left (x\right ) +{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.654914, size = 26, normalized size = 0.76 \begin{align*} \frac{A \log{\left (x \right )}}{a} + \frac{\left (- A b + B a\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14665, size = 49, normalized size = 1.44 \begin{align*} \frac{A \log \left (x^{2}\right )}{2 \, a} + \frac{{\left (B a - A b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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