Integrand size = 13, antiderivative size = 67 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{3 b^2 x^3}+\frac {2 c}{b^3 x}+\frac {c^2 x}{2 b^3 \left (b+c x^2\right )}+\frac {5 c^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Output:
-1/3/b^2/x^3+2*c/b^3/x+1/2*c^2*x/b^3/(c*x^2+b)+5/2*c^(3/2)*arctan(c^(1/2)* x/b^(1/2))/b^(7/2)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{3 b^2 x^3}+\frac {2 c}{b^3 x}+\frac {c^2 x}{2 b^3 \left (b+c x^2\right )}+\frac {5 c^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Input:
Integrate[(b*x^2 + c*x^4)^(-2),x]
Output:
-1/3*1/(b^2*x^3) + (2*c)/(b^3*x) + (c^2*x)/(2*b^3*(b + c*x^2)) + (5*c^(3/2 )*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(7/2))
Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1397, 253, 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.
\(\displaystyle \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1397 |
\(\displaystyle \int \frac {1}{x^4 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{x^4 \left (c x^2+b\right )}dx}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {c \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (b+c x^2\right )}\) |
Input:
Int[(b*x^2 + c*x^4)^(-2),x]
Output:
1/(2*b*x^3*(b + c*x^2)) + (5*(-1/3*1/(b*x^3) - (c*(-(1/(b*x)) - (Sqrt[c]*A rcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2)))/b))/(2*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 = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {1}{3 b^{2} x^{3}}+\frac {2 c}{b^{3} x}+\frac {c^{2} \left (\frac {x}{2 c \,x^{2}+2 b}+\frac {5 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{3}}\) | \(55\) |
risch | \(\frac {\frac {5 c^{2} x^{4}}{2 b^{3}}+\frac {5 c \,x^{2}}{3 b^{2}}-\frac {1}{3 b}}{x^{3} \left (c \,x^{2}+b \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+2 c^{3}\right ) x -b^{4} c \textit {\_R} \right )\right )}{4}\) | \(85\) |
Input:
int(1/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/3/b^2/x^3+2*c/b^3/x+c^2/b^3*(1/2*x/(c*x^2+b)+5/2/(b*c)^(1/2)*arctan(c*x /(b*c)^(1/2)))
Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.57 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=\left [\frac {30 \, c^{2} x^{4} + 20 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) - 4 \, b^{2}}{12 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, \frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) - 2 \, b^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \] Input:
integrate(1/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
[1/12*(30*c^2*x^4 + 20*b*c*x^2 + 15*(c^2*x^5 + b*c*x^3)*sqrt(-c/b)*log((c* x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)) - 4*b^2)/(b^3*c*x^5 + b^4*x^3), 1 /6*(15*c^2*x^4 + 10*b*c*x^2 + 15*(c^2*x^5 + b*c*x^3)*sqrt(c/b)*arctan(x*sq rt(c/b)) - 2*b^2)/(b^3*c*x^5 + b^4*x^3)]
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=- \frac {5 \sqrt {- \frac {c^{3}}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}}}{c^{2}} + x \right )}}{4} + \frac {5 \sqrt {- \frac {c^{3}}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}}}{c^{2}} + x \right )}}{4} + \frac {- 2 b^{2} + 10 b c x^{2} + 15 c^{2} x^{4}}{6 b^{4} x^{3} + 6 b^{3} c x^{5}} \] Input:
integrate(1/(c*x**4+b*x**2)**2,x)
Output:
-5*sqrt(-c**3/b**7)*log(-b**4*sqrt(-c**3/b**7)/c**2 + x)/4 + 5*sqrt(-c**3/ b**7)*log(b**4*sqrt(-c**3/b**7)/c**2 + x)/4 + (-2*b**2 + 10*b*c*x**2 + 15* c**2*x**4)/(6*b**4*x**3 + 6*b**3*c*x**5)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=\frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} - 2 \, b^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}} + \frac {5 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} \] Input:
integrate(1/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
1/6*(15*c^2*x^4 + 10*b*c*x^2 - 2*b^2)/(b^3*c*x^5 + b^4*x^3) + 5/2*c^2*arct an(c*x/sqrt(b*c))/(sqrt(b*c)*b^3)
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=\frac {5 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} + \frac {c^{2} x}{2 \, {\left (c x^{2} + b\right )} b^{3}} + \frac {6 \, c x^{2} - b}{3 \, b^{3} x^{3}} \] Input:
integrate(1/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
5/2*c^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) + 1/2*c^2*x/((c*x^2 + b)*b^3 ) + 1/3*(6*c*x^2 - b)/(b^3*x^3)
Time = 18.48 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {5\,c\,x^2}{3\,b^2}-\frac {1}{3\,b}+\frac {5\,c^2\,x^4}{2\,b^3}}{c\,x^5+b\,x^3}+\frac {5\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{7/2}} \] Input:
int(1/(b*x^2 + c*x^4)^2,x)
Output:
((5*c*x^2)/(3*b^2) - 1/(3*b) + (5*c^2*x^4)/(2*b^3))/(b*x^3 + c*x^5) + (5*c ^(3/2)*atan((c^(1/2)*x)/b^(1/2)))/(2*b^(7/2))
Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx=\frac {15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b c \,x^{3}+15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) c^{2} x^{5}-2 b^{3}+10 b^{2} c \,x^{2}+15 b \,c^{2} x^{4}}{6 b^{4} x^{3} \left (c \,x^{2}+b \right )} \] Input:
int(1/(c*x^4+b*x^2)^2,x)
Output:
(15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b*c*x**3 + 15*sqrt(c)*sq rt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*c**2*x**5 - 2*b**3 + 10*b**2*c*x**2 + 15*b*c**2*x**4)/(6*b**4*x**3*(b + c*x**2))