\(\int \frac {1}{x (a+b x^2) (c+d x^2)^2} \, dx\) [247]

Optimal result

Integrand size = 22, antiderivative size = 100 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {d}{2 c (b c-a d) \left (c+d x^2\right )}+\frac {\log (x)}{a c^2}-\frac {b^2 \log \left (a+b x^2\right )}{2 a (b c-a d)^2}+\frac {d (2 b c-a d) \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 100, 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} \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {b^2 \log \left (a+b x^2\right )}{2 a (b c-a d)^2}+\frac {d (2 b c-a d) \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^2}-\frac {d}{2 c \left (c+d x^2\right ) (b c-a d)}+\frac {\log (x)}{a c^2} \]

[In]

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

[Out]

-1/2*d/(c*(b*c - a*d)*(c + d*x^2)) + Log[x]/(a*c^2) - (b^2*Log[a + b*x^2])/(2*a*(b*c - a*d)^2) + (d*(2*b*c - a
*d)*Log[c + d*x^2])/(2*c^2*(b*c - a*d)^2)

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {1}{2} \left (-\frac {d}{c (b c-a d) \left (c+d x^2\right )}+\frac {2 \log (x)}{a c^2}-\frac {b^2 \log \left (a+b x^2\right )}{a (b c-a d)^2}+\frac {d (2 b c-a d) \log \left (c+d x^2\right )}{c^2 (b c-a d)^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\ln \left (x \right )}{a \,c^{2}}-\frac {b^{2} \ln \left (b \,x^{2}+a \right )}{2 a \left (a d -b c \right )^{2}}-\frac {d^{2} \left (\frac {\left (a d -2 b c \right ) \ln \left (d \,x^{2}+c \right )}{d}-\frac {\left (a d -b c \right ) c}{d \left (d \,x^{2}+c \right )}\right )}{2 \left (a d -b c \right )^{2} c^{2}}\)	\(99\)
norman	\(-\frac {d^{2} x^{2}}{2 c^{2} \left (a d -b c \right ) \left (d \,x^{2}+c \right )}+\frac {\ln \left (x \right )}{a \,c^{2}}-\frac {b^{2} \ln \left (b \,x^{2}+a \right )}{2 a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {d \left (a d -2 b c \right ) \ln \left (d \,x^{2}+c \right )}{2 c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(125\)
risch	\(\frac {d}{2 \left (a d -b c \right ) c \left (d \,x^{2}+c \right )}+\frac {\ln \left (x \right )}{a \,c^{2}}-\frac {d^{2} \ln \left (-d \,x^{2}-c \right ) a}{2 c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d \ln \left (-d \,x^{2}-c \right ) b}{c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {b^{2} \ln \left (b \,x^{2}+a \right )}{2 a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(158\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2} a^{2} d^{3}-4 \ln \left (x \right ) x^{2} a b c \,d^{2}+2 \ln \left (x \right ) x^{2} b^{2} c^{2} d -\ln \left (b \,x^{2}+a \right ) x^{2} b^{2} c^{2} d -\ln \left (d \,x^{2}+c \right ) x^{2} a^{2} d^{3}+2 \ln \left (d \,x^{2}+c \right ) x^{2} a b c \,d^{2}-x^{2} a^{2} d^{3}+x^{2} a b c \,d^{2}+2 \ln \left (x \right ) a^{2} c \,d^{2}-4 \ln \left (x \right ) a b \,c^{2} d +2 \ln \left (x \right ) b^{2} c^{3}-\ln \left (b \,x^{2}+a \right ) b^{2} c^{3}-\ln \left (d \,x^{2}+c \right ) a^{2} c \,d^{2}+2 \ln \left (d \,x^{2}+c \right ) a b \,c^{2} d}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) a \left (d \,x^{2}+c \right ) c^{2}}\)	\(241\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (94) = 188\).

Time = 0.69 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.19 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {a b c^{2} d - a^{2} c d^{2} + {\left (b^{2} c^{2} d x^{2} + b^{2} c^{3}\right )} \log \left (b x^{2} + a\right ) - {\left (2 \, a b c^{2} d - a^{2} c d^{2} + {\left (2 \, a b c d^{2} - a^{2} d^{3}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) - 2 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} d^{2} + {\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x^{2}\right )}} \]

[In]

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

[Out]

-1/2*(a*b*c^2*d - a^2*c*d^2 + (b^2*c^2*d*x^2 + b^2*c^3)*log(b*x^2 + a) - (2*a*b*c^2*d - a^2*c*d^2 + (2*a*b*c*d
^2 - a^2*d^3)*x^2)*log(d*x^2 + c) - 2*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)
*x^2)*log(x))/(a*b^2*c^5 - 2*a^2*b*c^4*d + a^3*c^3*d^2 + (a*b^2*c^4*d - 2*a^2*b*c^3*d^2 + a^3*c^2*d^3)*x^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {b^{2} \log \left (b x^{2} + a\right )}{2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )}} - \frac {d}{2 \, {\left (b c^{3} - a c^{2} d + {\left (b c^{2} d - a c d^{2}\right )} x^{2}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

-1/2*b^2*log(b*x^2 + a)/(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2) + 1/2*(2*b*c*d - a*d^2)*log(d*x^2 + c)/(b^2*c^4 -
2*a*b*c^3*d + a^2*c^2*d^2) - 1/2*d/(b*c^3 - a*c^2*d + (b*c^2*d - a*c*d^2)*x^2) + 1/2*log(x^2)/(a*c^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.85 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {b^{3} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )}} + \frac {{\left (2 \, b c d^{2} - a d^{3}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )}} - \frac {2 \, b c d^{2} x^{2} - a d^{3} x^{2} + 3 \, b c^{2} d - 2 \, a c d^{2}}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} {\left (d x^{2} + c\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

-1/2*b^3*log(abs(b*x^2 + a))/(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2) + 1/2*(2*b*c*d^2 - a*d^3)*log(abs(d*x^2 +
c))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3) - 1/2*(2*b*c*d^2*x^2 - a*d^3*x^2 + 3*b*c^2*d - 2*a*c*d^2)/((b^2
*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*(d*x^2 + c)) + 1/2*log(x^2)/(a*c^2)

Mupad [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {\ln \left (x\right )}{a\,c^2}-\frac {\ln \left (d\,x^2+c\right )\,\left (a\,d^2-2\,b\,c\,d\right )}{2\,a^2\,c^2\,d^2-4\,a\,b\,c^3\,d+2\,b^2\,c^4}-\frac {b^2\,\ln \left (b\,x^2+a\right )}{2\,a^3\,d^2-4\,a^2\,b\,c\,d+2\,a\,b^2\,c^2}+\frac {d}{2\,c\,\left (d\,x^2+c\right )\,\left (a\,d-b\,c\right )} \]

[In]

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

[Out]

log(x)/(a*c^2) - (log(c + d*x^2)*(a*d^2 - 2*b*c*d))/(2*b^2*c^4 + 2*a^2*c^2*d^2 - 4*a*b*c^3*d) - (b^2*log(a + b
*x^2))/(2*a^3*d^2 + 2*a*b^2*c^2 - 4*a^2*b*c*d) + d/(2*c*(c + d*x^2)*(a*d - b*c))