\(\int \frac {A+B x^2}{x (a+b x^2)} \, dx\) [63]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log (x)}{a}-\frac {(A b-a B) \log \left (a+b x^2\right )}{2 a b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 78} \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log (x)}{a}-\frac {(A b-a B) \log \left (a+b x^2\right )}{2 a b} \]

[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 78

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 457

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log (x)}{a}+\frac {(-A b+a B) \log \left (a+b x^2\right )}{2 a b} \]

[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)

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {A \ln \left (x \right )}{a}-\frac {\left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a b}\)	\(33\)
norman	\(\frac {A \ln \left (x \right )}{a}-\frac {\left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a b}\)	\(33\)
risch	\(\frac {A \ln \left (x \right )}{a}-\frac {\ln \left (b \,x^{2}+a \right ) A}{2 a}+\frac {\ln \left (b \,x^{2}+a \right ) B}{2 b}\)	\(37\)
parallelrisch	\(\frac {2 A \ln \left (x \right ) b -A \ln \left (b \,x^{2}+a \right ) b +B \ln \left (b \,x^{2}+a \right ) a}{2 a b}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {2 \, A b \log \left (x\right ) + {\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log {\left (x \right )}}{a} + \frac {\left (- A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a b} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log \left (x^{2}\right )}{2 \, a} + \frac {{\left (B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A \log \left (x^{2}\right )}{2 \, a} + \frac {{\left (B a - A b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a b} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )} \, dx=\frac {A\,\ln \left (x\right )}{a}-\frac {\ln \left (b\,x^2+a\right )\,\left (A\,b-B\,a\right )}{2\,a\,b} \]

[In]

int((A + B*x^2)/(x*(a + b*x^2)),x)

[Out]

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