Integrand size = 13, antiderivative size = 97 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {1}{5 b^3 x^5}+\frac {c}{b^4 x^3}-\frac {6 c^2}{b^5 x}-\frac {c^3 x}{4 b^4 \left (b+c x^2\right )^2}-\frac {15 c^3 x}{8 b^5 \left (b+c x^2\right )}-\frac {63 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \] Output:
-1/5/b^3/x^5+c/b^4/x^3-6*c^2/b^5/x-1/4*c^3*x/b^4/(c*x^2+b)^2-15/8*c^3*x/b^ 5/(c*x^2+b)-63/8*c^(5/2)*arctan(c^(1/2)*x/b^(1/2))/b^(11/2)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {8 b^4-24 b^3 c x^2+168 b^2 c^2 x^4+525 b c^3 x^6+315 c^4 x^8}{40 b^5 x^5 \left (b+c x^2\right )^2}-\frac {63 c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \] Input:
Integrate[(b*x^2 + c*x^4)^(-3),x]
Output:
-1/40*(8*b^4 - 24*b^3*c*x^2 + 168*b^2*c^2*x^4 + 525*b*c^3*x^6 + 315*c^4*x^ 8)/(b^5*x^5*(b + c*x^2)^2) - (63*c^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b ^(11/2))
Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {1397, 253, 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.
\(\displaystyle \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 1397 |
\(\displaystyle \int \frac {1}{x^6 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {9 \int \frac {1}{x^6 \left (c x^2+b\right )^2}dx}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {9 \left (\frac {7 \int \frac {1}{x^6 \left (c x^2+b\right )}dx}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {c \int \frac {1}{x^4 \left (c x^2+b\right )}dx}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {c \left (-\frac {c \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {c \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {1}{5 b x^5}\right )}{2 b}+\frac {1}{2 b x^5 \left (b+c x^2\right )}\right )}{4 b}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}\) |
Input:
Int[(b*x^2 + c*x^4)^(-3),x]
Output:
1/(4*b*x^5*(b + c*x^2)^2) + (9*(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)/Sqr t[b]])/b^(3/2)))/b))/b))/(2*b)))/(4*b)
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]
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[x^(2*p)*(b + c*x^2 )^p, x] /; FreeQ[{b, c}, x] && IntegerQ[p]
Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {1}{5 b^{3} x^{5}}+\frac {c}{b^{4} x^{3}}-\frac {6 c^{2}}{b^{5} x}-\frac {c^{3} \left (\frac {\frac {15}{8} c \,x^{3}+\frac {17}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {63 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}\) | \(75\) |
risch | \(\frac {-\frac {63 c^{4} x^{8}}{8 b^{5}}-\frac {105 c^{3} x^{6}}{8 b^{4}}-\frac {21 c^{2} x^{4}}{5 b^{3}}+\frac {3 c \,x^{2}}{5 b^{2}}-\frac {1}{5 b}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {63 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{11} \textit {\_Z}^{2}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{11}+2 c^{5}\right ) x +b^{6} c^{2} \textit {\_R} \right )\right )}{16}\) | \(108\) |
Input:
int(1/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
-1/5/b^3/x^5+c/b^4/x^3-6*c^2/b^5/x-1/b^5*c^3*((15/8*c*x^3+17/8*b*x)/(c*x^2 +b)^2+63/8/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.72 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=\left [-\frac {630 \, c^{4} x^{8} + 1050 \, b c^{3} x^{6} + 336 \, b^{2} c^{2} x^{4} - 48 \, b^{3} c x^{2} + 16 \, b^{4} - 315 \, {\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} c^{2} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{80 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}, -\frac {315 \, c^{4} x^{8} + 525 \, b c^{3} x^{6} + 168 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} + 8 \, b^{4} + 315 \, {\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} c^{2} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{40 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}\right ] \] Input:
integrate(1/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
[-1/80*(630*c^4*x^8 + 1050*b*c^3*x^6 + 336*b^2*c^2*x^4 - 48*b^3*c*x^2 + 16 *b^4 - 315*(c^4*x^9 + 2*b*c^3*x^7 + b^2*c^2*x^5)*sqrt(-c/b)*log((c*x^2 - 2 *b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5), -1/40*(315*c^4*x^8 + 525*b*c^3*x^6 + 168*b^2*c^2*x^4 - 24*b^3*c*x^2 + 8*b^ 4 + 315*(c^4*x^9 + 2*b*c^3*x^7 + b^2*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b) ))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)]
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=\frac {63 \sqrt {- \frac {c^{5}}{b^{11}}} \log {\left (- \frac {b^{6} \sqrt {- \frac {c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} - \frac {63 \sqrt {- \frac {c^{5}}{b^{11}}} \log {\left (\frac {b^{6} \sqrt {- \frac {c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} + \frac {- 8 b^{4} + 24 b^{3} c x^{2} - 168 b^{2} c^{2} x^{4} - 525 b c^{3} x^{6} - 315 c^{4} x^{8}}{40 b^{7} x^{5} + 80 b^{6} c x^{7} + 40 b^{5} c^{2} x^{9}} \] Input:
integrate(1/(c*x**4+b*x**2)**3,x)
Output:
63*sqrt(-c**5/b**11)*log(-b**6*sqrt(-c**5/b**11)/c**3 + x)/16 - 63*sqrt(-c **5/b**11)*log(b**6*sqrt(-c**5/b**11)/c**3 + x)/16 + (-8*b**4 + 24*b**3*c* x**2 - 168*b**2*c**2*x**4 - 525*b*c**3*x**6 - 315*c**4*x**8)/(40*b**7*x**5 + 80*b**6*c*x**7 + 40*b**5*c**2*x**9)
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {315 \, c^{4} x^{8} + 525 \, b c^{3} x^{6} + 168 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} + 8 \, b^{4}}{40 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}} - \frac {63 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} \] Input:
integrate(1/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
-1/40*(315*c^4*x^8 + 525*b*c^3*x^6 + 168*b^2*c^2*x^4 - 24*b^3*c*x^2 + 8*b^ 4)/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5) - 63/8*c^3*arctan(c*x/sqrt(b*c))/ (sqrt(b*c)*b^5)
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {63 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} - \frac {15 \, c^{4} x^{3} + 17 \, b c^{3} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{5}} - \frac {30 \, c^{2} x^{4} - 5 \, b c x^{2} + b^{2}}{5 \, b^{5} x^{5}} \] Input:
integrate(1/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
-63/8*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^5) - 1/8*(15*c^4*x^3 + 17*b*c ^3*x)/((c*x^2 + b)^2*b^5) - 1/5*(30*c^2*x^4 - 5*b*c*x^2 + b^2)/(b^5*x^5)
Time = 17.69 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {1}{5\,b}-\frac {3\,c\,x^2}{5\,b^2}+\frac {21\,c^2\,x^4}{5\,b^3}+\frac {105\,c^3\,x^6}{8\,b^4}+\frac {63\,c^4\,x^8}{8\,b^5}}{b^2\,x^5+2\,b\,c\,x^7+c^2\,x^9}-\frac {63\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,b^{11/2}} \] Input:
int(1/(b*x^2 + c*x^4)^3,x)
Output:
- (1/(5*b) - (3*c*x^2)/(5*b^2) + (21*c^2*x^4)/(5*b^3) + (105*c^3*x^6)/(8*b ^4) + (63*c^4*x^8)/(8*b^5))/(b^2*x^5 + c^2*x^9 + 2*b*c*x^7) - (63*c^(5/2)* atan((c^(1/2)*x)/b^(1/2)))/(8*b^(11/2))
Time = 0.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-315 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{5}-630 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b \,c^{3} x^{7}-315 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) c^{4} x^{9}-8 b^{5}+24 b^{4} c \,x^{2}-168 b^{3} c^{2} x^{4}-525 b^{2} c^{3} x^{6}-315 b \,c^{4} x^{8}}{40 b^{6} x^{5} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(1/(c*x^4+b*x^2)^3,x)
Output:
( - 315*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c**2*x**5 - 630 *sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b*c**3*x**7 - 315*sqrt(c)*s qrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*c**4*x**9 - 8*b**5 + 24*b**4*c*x**2 - 168*b**3*c**2*x**4 - 525*b**2*c**3*x**6 - 315*b*c**4*x**8)/(40*b**6*x**5* (b**2 + 2*b*c*x**2 + c**2*x**4))