\(\int \frac {1}{x (a+b x^2) (c+d x^2)} \, dx\) [234]

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {\log (x)}{a c}-\frac {b \log \left (a+b x^2\right )}{2 a (b c-a d)}+\frac {d \log \left (c+d x^2\right )}{2 c (b c-a d)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 84} \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {b \log \left (a+b x^2\right )}{2 a (b c-a d)}+\frac {d \log \left (c+d x^2\right )}{2 c (b c-a d)}+\frac {\log (x)}{a c} \]

[In]

Int[1/(x*(a + b*x^2)*(c + d*x^2)),x]

[Out]

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

Rule 84

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {2 b c \log (x)-2 a d \log (x)-b c \log \left (a+b x^2\right )+a d \log \left (c+d x^2\right )}{2 a b c^2-2 a^2 c d} \]

[In]

Integrate[1/(x*(a + b*x^2)*(c + d*x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {2 \ln \left (x \right ) a d -2 c \ln \left (x \right ) b +\ln \left (b \,x^{2}+a \right ) b c -d \ln \left (d \,x^{2}+c \right ) a}{2 a c \left (a d -b c \right )}\)	\(55\)
default	\(\frac {\ln \left (x \right )}{a c}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) a}-\frac {d \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) c}\)	\(59\)
norman	\(\frac {\ln \left (x \right )}{a c}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) a}-\frac {d \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) c}\)	\(59\)
risch	\(\frac {\ln \left (x \right )}{a c}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) a}-\frac {d \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) c}\)	\(59\)

[In]

int(1/x/(b*x^2+a)/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {b c \log \left (b x^{2} + a\right ) - a d \log \left (d x^{2} + c\right ) - 2 \, {\left (b c - a d\right )} \log \left (x\right )}{2 \, {\left (a b c^{2} - a^{2} c d\right )}} \]

[In]

integrate(1/x/(b*x^2+a)/(d*x^2+c),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/x/(b*x**2+a)/(d*x**2+c),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {b \log \left (b x^{2} + a\right )}{2 \, {\left (a b c - a^{2} d\right )}} + \frac {d \log \left (d x^{2} + c\right )}{2 \, {\left (b c^{2} - a c d\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a c} \]

[In]

integrate(1/x/(b*x^2+a)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=-\frac {b^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a b^{2} c - a^{2} b d\right )}} + \frac {d^{2} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b c^{2} d - a c d^{2}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a c} \]

[In]

integrate(1/x/(b*x^2+a)/(d*x^2+c),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx=\frac {b\,\ln \left (b\,x^2+a\right )}{2\,a^2\,d-2\,a\,b\,c}+\frac {d\,\ln \left (d\,x^2+c\right )}{2\,b\,c^2-2\,a\,c\,d}+\frac {\ln \left (x\right )}{a\,c} \]

[In]

int(1/(x*(a + b*x^2)*(c + d*x^2)),x)

[Out]

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