Integrand size = 22, antiderivative size = 126 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {1}{2 a^2 c x^2}-\frac {b^2}{2 a^2 (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c+a d) \log (x)}{a^3 c^2}+\frac {b^2 (2 b c-3 a d) \log \left (a+b x^2\right )}{2 a^3 (b c-a d)^2}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^2} \]
-1/2/a^2/c/x^2-1/2*b^2/a^2/(-a*d+b*c)/(b*x^2+a)-(a*d+2*b*c)*ln(x)/a^3/c^2+ 1/2*b^2*(-3*a*d+2*b*c)*ln(b*x^2+a)/a^3/(-a*d+b*c)^2+1/2*d^3*ln(d*x^2+c)/c^ 2/(-a*d+b*c)^2
Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {1}{2} \left (-\frac {1}{a^2 c x^2}+\frac {b^2}{a^2 (-b c+a d) \left (a+b x^2\right )}-\frac {2 (2 b c+a d) \log (x)}{a^3 c^2}+\frac {b^2 (2 b c-3 a d) \log \left (a+b x^2\right )}{a^3 (b c-a d)^2}+\frac {d^3 \log \left (c+d x^2\right )}{c^2 (b c-a d)^2}\right ) \]
(-(1/(a^2*c*x^2)) + b^2/(a^2*(-(b*c) + a*d)*(a + b*x^2)) - (2*(2*b*c + a*d )*Log[x])/(a^3*c^2) + (b^2*(2*b*c - 3*a*d)*Log[a + b*x^2])/(a^3*(b*c - a*d )^2) + (d^3*Log[c + d*x^2])/(c^2*(b*c - a*d)^2))/2
Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^2+a\right )^2 \left (d x^2+c\right )}dx^2\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d^4}{c^2 (b c-a d)^2 \left (d x^2+c\right )}-\frac {b^3 (3 a d-2 b c)}{a^3 (a d-b c)^2 \left (b x^2+a\right )}+\frac {-2 b c-a d}{a^3 c^2 x^2}-\frac {b^3}{a^2 (a d-b c) \left (b x^2+a\right )^2}+\frac {1}{a^2 c x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {b^2 (2 b c-3 a d) \log \left (a+b x^2\right )}{a^3 (b c-a d)^2}-\frac {\log \left (x^2\right ) (a d+2 b c)}{a^3 c^2}-\frac {b^2}{a^2 \left (a+b x^2\right ) (b c-a d)}-\frac {1}{a^2 c x^2}+\frac {d^3 \log \left (c+d x^2\right )}{c^2 (b c-a d)^2}\right )\) |
(-(1/(a^2*c*x^2)) - b^2/(a^2*(b*c - a*d)*(a + b*x^2)) - ((2*b*c + a*d)*Log [x^2])/(a^3*c^2) + (b^2*(2*b*c - 3*a*d)*Log[a + b*x^2])/(a^3*(b*c - a*d)^2 ) + (d^3*Log[c + d*x^2])/(c^2*(b*c - a*d)^2))/2
3.3.96.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.71 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {1}{2 a^{2} c \,x^{2}}+\frac {\left (-a d -2 b c \right ) \ln \left (x \right )}{a^{3} c^{2}}-\frac {b^{3} \left (\frac {\left (3 a d -2 b c \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{3} \left (a d -b c \right )^{2}}+\frac {d^{3} \ln \left (d \,x^{2}+c \right )}{2 c^{2} \left (a d -b c \right )^{2}}\) | \(120\) |
norman | \(\frac {-\frac {1}{2 a c}+\frac {\left (a b d -2 b^{2} c \right ) b \,x^{4}}{2 c \,a^{3} \left (a d -b c \right )}}{x^{2} \left (b \,x^{2}+a \right )}+\frac {d^{3} \ln \left (d \,x^{2}+c \right )}{2 c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\left (a d +2 b c \right ) \ln \left (x \right )}{a^{3} c^{2}}-\frac {b^{2} \left (3 a d -2 b c \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(162\) |
risch | \(\frac {-\frac {b \left (a d -2 b c \right ) x^{2}}{2 a^{2} c \left (a d -b c \right )}-\frac {1}{2 a c}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {\ln \left (x \right ) d}{a^{2} c^{2}}-\frac {2 \ln \left (x \right ) b}{a^{3} c}+\frac {d^{3} \ln \left (-d \,x^{2}-c \right )}{2 c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {3 b^{2} \ln \left (b \,x^{2}+a \right ) d}{2 a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {b^{3} \ln \left (b \,x^{2}+a \right ) c}{a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(197\) |
parallelrisch | \(-\frac {2 \ln \left (x \right ) x^{4} a^{3} b \,d^{3}-6 \ln \left (x \right ) x^{4} a \,b^{3} c^{2} d +4 \ln \left (x \right ) x^{4} b^{4} c^{3}+3 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{3} c^{2} d -2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{4} c^{3}-\ln \left (d \,x^{2}+c \right ) x^{4} a^{3} b \,d^{3}-a^{2} b^{2} c \,d^{2} x^{4}+3 a \,b^{3} c^{2} d \,x^{4}-2 b^{4} c^{3} x^{4}+2 \ln \left (x \right ) x^{2} a^{4} d^{3}-6 \ln \left (x \right ) x^{2} a^{2} b^{2} c^{2} d +4 \ln \left (x \right ) x^{2} a \,b^{3} c^{3}+3 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b^{2} c^{2} d -2 \ln \left (b \,x^{2}+a \right ) x^{2} a \,b^{3} c^{3}-\ln \left (d \,x^{2}+c \right ) x^{2} a^{4} d^{3}+c \,d^{2} a^{4}-2 a^{3} b \,c^{2} d +a^{2} b^{2} c^{3}}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right ) x^{2} a^{3} c^{2}}\) | \(320\) |
-1/2/a^2/c/x^2+1/a^3/c^2*(-a*d-2*b*c)*ln(x)-1/2*b^3/a^3/(a*d-b*c)^2*((3*a* d-2*b*c)/b*ln(b*x^2+a)-(a*d-b*c)*a/b/(b*x^2+a))+1/2*d^3/c^2/(a*d-b*c)^2*ln (d*x^2+c)
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (118) = 236\).
Time = 1.48 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.40 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + a^{3} b c d^{2}\right )} x^{2} - {\left ({\left (2 \, b^{4} c^{3} - 3 \, a b^{3} c^{2} d\right )} x^{4} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - {\left (a^{3} b d^{3} x^{4} + a^{4} d^{3} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left ({\left (2 \, b^{4} c^{3} - 3 \, a b^{3} c^{2} d + a^{3} b d^{3}\right )} x^{4} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + a^{4} d^{3}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left ({\left (a^{3} b^{3} c^{4} - 2 \, a^{4} b^{2} c^{3} d + a^{5} b c^{2} d^{2}\right )} x^{4} + {\left (a^{4} b^{2} c^{4} - 2 \, a^{5} b c^{3} d + a^{6} c^{2} d^{2}\right )} x^{2}\right )}} \]
-1/2*(a^2*b^2*c^3 - 2*a^3*b*c^2*d + a^4*c*d^2 + (2*a*b^3*c^3 - 3*a^2*b^2*c ^2*d + a^3*b*c*d^2)*x^2 - ((2*b^4*c^3 - 3*a*b^3*c^2*d)*x^4 + (2*a*b^3*c^3 - 3*a^2*b^2*c^2*d)*x^2)*log(b*x^2 + a) - (a^3*b*d^3*x^4 + a^4*d^3*x^2)*log (d*x^2 + c) + 2*((2*b^4*c^3 - 3*a*b^3*c^2*d + a^3*b*d^3)*x^4 + (2*a*b^3*c^ 3 - 3*a^2*b^2*c^2*d + a^4*d^3)*x^2)*log(x))/((a^3*b^3*c^4 - 2*a^4*b^2*c^3* d + a^5*b*c^2*d^2)*x^4 + (a^4*b^2*c^4 - 2*a^5*b*c^3*d + a^6*c^2*d^2)*x^2)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{3} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )}} + \frac {{\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )}} - \frac {a b c - a^{2} d + {\left (2 \, b^{2} c - a b d\right )} x^{2}}{2 \, {\left ({\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{4} + {\left (a^{3} b c^{2} - a^{4} c d\right )} x^{2}\right )}} - \frac {{\left (2 \, b c + a d\right )} \log \left (x^{2}\right )}{2 \, a^{3} c^{2}} \]
1/2*d^3*log(d*x^2 + c)/(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2) + 1/2*(2*b^3* c - 3*a*b^2*d)*log(b*x^2 + a)/(a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2) - 1/2* (a*b*c - a^2*d + (2*b^2*c - a*b*d)*x^2)/((a^2*b^2*c^2 - a^3*b*c*d)*x^4 + ( a^3*b*c^2 - a^4*c*d)*x^2) - 1/2*(2*b*c + a*d)*log(x^2)/(a^3*c^2)
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (118) = 236\).
Time = 0.27 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.04 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{4} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )}} + \frac {{\left (2 \, b^{4} c - 3 \, a b^{3} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{3} b^{3} c^{2} - 2 \, a^{4} b^{2} c d + a^{5} b d^{2}\right )}} + \frac {a^{2} b d^{3} x^{4} - 4 \, b^{3} c^{3} x^{2} + 6 \, a b^{2} c^{2} d x^{2} - 2 \, a^{2} b c d^{2} x^{2} + a^{3} d^{3} x^{2} - 2 \, a b^{2} c^{3} + 4 \, a^{2} b c^{2} d - 2 \, a^{3} c d^{2}}{4 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} {\left (b x^{4} + a x^{2}\right )}} - \frac {{\left (2 \, b c + a d\right )} \log \left (x^{2}\right )}{2 \, a^{3} c^{2}} \]
1/2*d^4*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^4*c - 3*a*b^3*d)*log(abs(b*x^2 + a))/(a^3*b^3*c^2 - 2*a^4*b^2*c*d + a^5*b*d^2) + 1/4*(a^2*b*d^3*x^4 - 4*b^3*c^3*x^2 + 6*a*b^2*c^2*d*x^2 - 2*a ^2*b*c*d^2*x^2 + a^3*d^3*x^2 - 2*a*b^2*c^3 + 4*a^2*b*c^2*d - 2*a^3*c*d^2)/ ((a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*(b*x^4 + a*x^2)) - 1/2*(2*b*c + a*d)*log(x^2)/(a^3*c^2)
Time = 5.97 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {\ln \left (b\,x^2+a\right )\,\left (2\,b^3\,c-3\,a\,b^2\,d\right )}{2\,a^5\,d^2-4\,a^4\,b\,c\,d+2\,a^3\,b^2\,c^2}-\frac {\frac {1}{2\,a\,c}-\frac {x^2\,\left (2\,b^2\,c-a\,b\,d\right )}{2\,a^2\,c\,\left (a\,d-b\,c\right )}}{b\,x^4+a\,x^2}+\frac {d^3\,\ln \left (d\,x^2+c\right )}{2\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}-\frac {\ln \left (x\right )\,\left (a\,d+2\,b\,c\right )}{a^3\,c^2} \]