\(\int \frac {c+d x^2}{x (a+b x^2)} \, dx\) [204]

Optimal result

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

[Out]

c*ln(x)/a-1/2*(-a*d+b*c)*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} \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c \log (x)}{a}-\frac {(b c-a d) \log \left (a+b x^2\right )}{2 a b} \]

[In]

Int[(c + d*x^2)/(x*(a + b*x^2)),x]

[Out]

(c*Log[x])/a - ((b*c - a*d)*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 {c+d x}{x (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {c}{a x}+\frac {-b c+a d}{a (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {c \log (x)}{a}-\frac {(b c-a d) \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 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c \log (x)}{a}+\frac {(-b c+a d) \log \left (a+b x^2\right )}{2 a b} \]

[In]

Integrate[(c + d*x^2)/(x*(a + b*x^2)),x]

[Out]

(c*Log[x])/a + ((-(b*c) + a*d)*Log[a + b*x^2])/(2*a*b)

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {c \ln \left (x \right )}{a}+\frac {\left (a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 a b}\)	\(33\)
norman	\(\frac {c \ln \left (x \right )}{a}+\frac {\left (a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 a b}\)	\(33\)
parallelrisch	\(\frac {2 c \ln \left (x \right ) b +\ln \left (b \,x^{2}+a \right ) a d -\ln \left (b \,x^{2}+a \right ) b c}{2 a b}\)	\(39\)
risch	\(\frac {c \ln \left (x \right )}{a}+\frac {\ln \left (-b \,x^{2}-a \right ) d}{2 b}-\frac {\ln \left (-b \,x^{2}-a \right ) c}{2 a}\)	\(43\)

[In]

int((d*x^2+c)/x/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

c*ln(x)/a+1/2*(a*d-b*c)/a/b*ln(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {2 \, b c \log \left (x\right ) - {\left (b c - a d\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[In]

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

[Out]

1/2*(2*b*c*log(x) - (b*c - a*d)*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 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c \log {\left (x \right )}}{a} + \frac {\left (a d - b c\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a b} \]

[In]

integrate((d*x**2+c)/x/(b*x**2+a),x)

[Out]

c*log(x)/a + (a*d - b*c)*log(a/b + x**2)/(2*a*b)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c \log \left (x^{2}\right )}{2 \, a} - \frac {{\left (b c - a d\right )} \log \left (b x^{2} + a\right )}{2 \, a b} \]

[In]

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

[Out]

1/2*c*log(x^2)/a - 1/2*(b*c - a*d)*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 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c \log \left (x^{2}\right )}{2 \, a} - \frac {{\left (b c - a d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a b} \]

[In]

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

[Out]

1/2*c*log(x^2)/a - 1/2*(b*c - a*d)*log(abs(b*x^2 + a))/(a*b)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x^2}{x \left (a+b x^2\right )} \, dx=\frac {c\,\ln \left (x\right )}{a}+\frac {\ln \left (b\,x^2+a\right )\,\left (a\,d-b\,c\right )}{2\,a\,b} \]

[In]

int((c + d*x^2)/(x*(a + b*x^2)),x)

[Out]

(c*log(x))/a + (log(a + b*x^2)*(a*d - b*c))/(2*a*b)