\(\int \frac {1}{x (a+b x^2)^2 (c+d x^2)} \, dx\) [294]

Optimal result

Integrand size = 22, antiderivative size = 99 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {\log (x)}{a^2 c}-\frac {b (b c-2 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}-\frac {d^2 \log \left (c+d x^2\right )}{2 c (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, 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} \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {b (b c-2 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}+\frac {\log (x)}{a^2 c}-\frac {d^2 \log \left (c+d x^2\right )}{2 c (b c-a d)^2}+\frac {b}{2 a \left (a+b x^2\right ) (b c-a d)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {2 \log (x)-\frac {b c (b c-2 a d) \left (a+b x^2\right ) \log \left (a+b x^2\right )+a \left (b c (-b c+a d)+a d^2 \left (a+b x^2\right ) \log \left (c+d x^2\right )\right )}{(b c-a d)^2 \left (a+b x^2\right )}}{2 a^2 c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\ln \left (x \right )}{a^{2} c}+\frac {b^{2} \left (\frac {\left (2 a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}\right )}{2 \left (a d -b c \right )^{2} a^{2}}-\frac {d^{2} \ln \left (d \,x^{2}+c \right )}{2 c \left (a d -b c \right )^{2}}\)	\(100\)
norman	\(\frac {b^{2} x^{2}}{2 a^{2} \left (a d -b c \right ) \left (b \,x^{2}+a \right )}+\frac {\ln \left (x \right )}{a^{2} c}-\frac {d^{2} \ln \left (d \,x^{2}+c \right )}{2 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {b \left (2 a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(126\)
risch	\(-\frac {b}{2 \left (a d -b c \right ) a \left (b \,x^{2}+a \right )}+\frac {\ln \left (x \right )}{a^{2} c}-\frac {d^{2} \ln \left (-d \,x^{2}-c \right )}{2 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {b \ln \left (b \,x^{2}+a \right ) d}{a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {b^{2} \ln \left (b \,x^{2}+a \right ) c}{2 a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(155\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2} a^{2} b \,d^{2}-4 \ln \left (x \right ) x^{2} a \,b^{2} c d +2 \ln \left (x \right ) x^{2} b^{3} c^{2}+2 \ln \left (b \,x^{2}+a \right ) x^{2} a \,b^{2} c d -\ln \left (b \,x^{2}+a \right ) x^{2} b^{3} c^{2}-\ln \left (d \,x^{2}+c \right ) x^{2} a^{2} b \,d^{2}+x^{2} a \,b^{2} c d -x^{2} b^{3} c^{2}+2 \ln \left (x \right ) a^{3} d^{2}-4 \ln \left (x \right ) a^{2} b c d +2 \ln \left (x \right ) a \,b^{2} c^{2}+2 \ln \left (b \,x^{2}+a \right ) a^{2} b c d -\ln \left (b \,x^{2}+a \right ) a \,b^{2} c^{2}-\ln \left (d \,x^{2}+c \right ) a^{3} d^{2}}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) c \left (b \,x^{2}+a \right ) a^{2}}\)	\(241\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (93) = 186\).

Time = 0.69 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.20 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {a b^{2} c^{2} - a^{2} b c d - {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + {\left (b^{3} c^{2} - 2 \, a b^{2} c d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - {\left (a^{2} b d^{2} x^{2} + a^{3} d^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2} + {\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {d^{2} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )}} - \frac {{\left (b^{2} c - 2 \, a b d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}} + \frac {b}{2 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.85 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {d^{3} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )}} - \frac {{\left (b^{3} c - 2 \, a b^{2} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )}} + \frac {b^{3} c x^{2} - 2 \, a b^{2} d x^{2} + 2 \, a b^{2} c - 3 \, a^{2} b d}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} {\left (b x^{2} + a\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {\ln \left (x\right )}{a^2\,c}-\frac {d^2\,\ln \left (d\,x^2+c\right )}{2\,a^2\,c\,d^2-4\,a\,b\,c^2\,d+2\,b^2\,c^3}-\frac {\ln \left (b\,x^2+a\right )\,\left (b^2\,c-2\,a\,b\,d\right )}{2\,a^4\,d^2-4\,a^3\,b\,c\,d+2\,a^2\,b^2\,c^2}-\frac {b}{2\,a\,\left (b\,x^2+a\right )\,\left (a\,d-b\,c\right )} \]

[In]

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

[Out]

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