Integrand size = 17, antiderivative size = 81 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {7 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
-7/10/b^2/x^5+7/6*c/b^3/x^3-7/2*c^2/b^4/x+1/2/b/x^5/(c*x^2+b)-7/2*c^(5/2)* arctan(x*c^(1/2)/b^(1/2))/b^(9/2)
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{5 b^2 x^5}+\frac {2 c}{3 b^3 x^3}-\frac {3 c^2}{b^4 x}-\frac {c^3 x}{2 b^4 \left (b+c x^2\right )}-\frac {7 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
-1/5*1/(b^2*x^5) + (2*c)/(3*b^3*x^3) - (3*c^2)/(b^4*x) - (c^3*x)/(2*b^4*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(9/2))
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {9, 253, 264, 264, 264, 218}
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.
\(\displaystyle \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^6 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {7 \int \frac {1}{x^6 \left (c x^2+b\right )}dx}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {7 \left (-\frac {c \int \frac {1}{x^4 \left (c x^2+b\right )}dx}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {7 \left (-\frac {c \left (-\frac {c \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {7 \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {7 \left (-\frac {c \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\) |
1/(2*b*x^5*(b + c*x^2)) + (7*(-1/5*1/(b*x^5) - (c*(-1/3*1/(b*x^3) - (c*(-( 1/(b*x)) - (Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2)))/b))/b))/(2*b)
3.3.4.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {1}{5 b^{2} x^{5}}-\frac {3 c^{2}}{b^{4} x}+\frac {2 c}{3 b^{3} x^{3}}-\frac {c^{3} \left (\frac {x}{2 c \,x^{2}+2 b}+\frac {7 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{4}}\) | \(67\) |
risch | \(\frac {-\frac {7 c^{3} x^{6}}{2 b^{4}}-\frac {7 c^{2} x^{4}}{3 b^{3}}+\frac {7 c \,x^{2}}{15 b^{2}}-\frac {1}{5 b}}{x^{5} \left (c \,x^{2}+b \right )}+\frac {7 \sqrt {-b c}\, c^{2} \ln \left (-c x +\sqrt {-b c}\right )}{4 b^{5}}-\frac {7 \sqrt {-b c}\, c^{2} \ln \left (-c x -\sqrt {-b c}\right )}{4 b^{5}}\) | \(106\) |
-1/5/b^2/x^5-3*c^2/b^4/x+2/3*c/b^3/x^3-c^3/b^4*(1/2*x/(c*x^2+b)+7/2/(b*c)^ (1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\left [-\frac {210 \, c^{3} x^{6} + 140 \, b c^{2} x^{4} - 28 \, b^{2} c x^{2} + 12 \, b^{3} - 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{60 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3} + 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \]
[-1/60*(210*c^3*x^6 + 140*b*c^2*x^4 - 28*b^2*c*x^2 + 12*b^3 - 105*(c^3*x^7 + b*c^2*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/ (b^4*c*x^7 + b^5*x^5), -1/30*(105*c^3*x^6 + 70*b*c^2*x^4 - 14*b^2*c*x^2 + 6*b^3 + 105*(c^3*x^7 + b*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c*x^ 7 + b^5*x^5)]
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (- \frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} - \frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (\frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} + \frac {- 6 b^{3} + 14 b^{2} c x^{2} - 70 b c^{2} x^{4} - 105 c^{3} x^{6}}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \]
7*sqrt(-c**5/b**9)*log(-b**5*sqrt(-c**5/b**9)/c**3 + x)/4 - 7*sqrt(-c**5/b **9)*log(b**5*sqrt(-c**5/b**9)/c**3 + x)/4 + (-6*b**3 + 14*b**2*c*x**2 - 7 0*b*c**2*x**4 - 105*c**3*x**6)/(30*b**5*x**5 + 30*b**4*c*x**7)
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3}}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}} - \frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} \]
-1/30*(105*c^3*x^6 + 70*b*c^2*x^4 - 14*b^2*c*x^2 + 6*b^3)/(b^4*c*x^7 + b^5 *x^5) - 7/2*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4)
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} - \frac {c^{3} x}{2 \, {\left (c x^{2} + b\right )} b^{4}} - \frac {45 \, c^{2} x^{4} - 10 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{4} x^{5}} \]
-7/2*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/2*c^3*x/((c*x^2 + b)*b^ 4) - 1/15*(45*c^2*x^4 - 10*b*c*x^2 + 3*b^2)/(b^4*x^5)
Time = 12.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {\frac {1}{5\,b}-\frac {7\,c\,x^2}{15\,b^2}+\frac {7\,c^2\,x^4}{3\,b^3}+\frac {7\,c^3\,x^6}{2\,b^4}}{c\,x^7+b\,x^5}-\frac {7\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{9/2}} \]