Integrand size = 13, antiderivative size = 96 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {6 b^2 x}{a^5}-\frac {b x^3}{a^4}+\frac {x^5}{5 a^3}-\frac {b^4 x}{4 a^5 \left (b+a x^2\right )^2}+\frac {17 b^3 x}{8 a^5 \left (b+a x^2\right )}-\frac {63 b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{11/2}} \] Output:
6*b^2*x/a^5-b*x^3/a^4+1/5*x^5/a^3-1/4*b^4*x/a^5/(a*x^2+b)^2+17/8*b^3*x/a^5 /(a*x^2+b)-63/8*b^(5/2)*arctan(a^(1/2)*x/b^(1/2))/a^(11/2)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {315 b^4 x+525 a b^3 x^3+168 a^2 b^2 x^5-24 a^3 b x^7+8 a^4 x^9}{40 a^5 \left (b+a x^2\right )^2}-\frac {63 b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{11/2}} \] Input:
Integrate[x^4/(a + b/x^2)^3,x]
Output:
(315*b^4*x + 525*a*b^3*x^3 + 168*a^2*b^2*x^5 - 24*a^3*b*x^7 + 8*a^4*x^9)/( 40*a^5*(b + a*x^2)^2) - (63*b^(5/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(11/ 2))
Time = 0.37 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {795, 252, 252, 254, 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.
\(\displaystyle \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^{10}}{\left (a x^2+b\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {9 \int \frac {x^8}{\left (a x^2+b\right )^2}dx}{4 a}-\frac {x^9}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {9 \left (\frac {7 \int \frac {x^6}{a x^2+b}dx}{2 a}-\frac {x^7}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^9}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {9 \left (\frac {7 \int \left (\frac {x^4}{a}-\frac {b x^2}{a^2}+\frac {b^2}{a^3}-\frac {b^3}{a^3 \left (a x^2+b\right )}\right )dx}{2 a}-\frac {x^7}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^9}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {b^{5/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{a^{7/2}}+\frac {b^2 x}{a^3}-\frac {b x^3}{3 a^2}+\frac {x^5}{5 a}\right )}{2 a}-\frac {x^7}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^9}{4 a \left (a x^2+b\right )^2}\) |
Input:
Int[x^4/(a + b/x^2)^3,x]
Output:
-1/4*x^9/(a*(b + a*x^2)^2) + (9*(-1/2*x^7/(a*(b + a*x^2)) + (7*((b^2*x)/a^ 3 - (b*x^3)/(3*a^2) + x^5/(5*a) - (b^(5/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/a^ (7/2)))/(2*a)))/(4*a)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {\frac {1}{5} a^{2} x^{5}-a b \,x^{3}+6 b^{2} x}{a^{5}}-\frac {b^{3} \left (\frac {-\frac {17}{8} a \,x^{3}-\frac {15}{8} b x}{\left (a \,x^{2}+b \right )^{2}}+\frac {63 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}\) | \(74\) |
risch | \(\frac {x^{5}}{5 a^{3}}-\frac {b \,x^{3}}{a^{4}}+\frac {6 b^{2} x}{a^{5}}+\frac {\frac {17}{8} a \,x^{3} b^{3}+\frac {15}{8} b^{4} x}{a^{5} \left (a \,x^{2}+b \right )^{2}}+\frac {63 \sqrt {-a b}\, b^{2} \ln \left (-\sqrt {-a b}\, x -b \right )}{16 a^{6}}-\frac {63 \sqrt {-a b}\, b^{2} \ln \left (\sqrt {-a b}\, x -b \right )}{16 a^{6}}\) | \(112\) |
Input:
int(x^4/(a+b/x^2)^3,x,method=_RETURNVERBOSE)
Output:
1/a^5*(1/5*a^2*x^5-a*b*x^3+6*b^2*x)-1/a^5*b^3*((-17/8*a*x^3-15/8*b*x)/(a*x ^2+b)^2+63/8/(a*b)^(1/2)*arctan(a*x/(a*b)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.67 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\left [\frac {16 \, a^{4} x^{9} - 48 \, a^{3} b x^{7} + 336 \, a^{2} b^{2} x^{5} + 1050 \, a b^{3} x^{3} + 630 \, b^{4} x + 315 \, {\left (a^{2} b^{2} x^{4} + 2 \, a b^{3} x^{2} + b^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right )}{80 \, {\left (a^{7} x^{4} + 2 \, a^{6} b x^{2} + a^{5} b^{2}\right )}}, \frac {8 \, a^{4} x^{9} - 24 \, a^{3} b x^{7} + 168 \, a^{2} b^{2} x^{5} + 525 \, a b^{3} x^{3} + 315 \, b^{4} x - 315 \, {\left (a^{2} b^{2} x^{4} + 2 \, a b^{3} x^{2} + b^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right )}{40 \, {\left (a^{7} x^{4} + 2 \, a^{6} b x^{2} + a^{5} b^{2}\right )}}\right ] \] Input:
integrate(x^4/(a+b/x^2)^3,x, algorithm="fricas")
Output:
[1/80*(16*a^4*x^9 - 48*a^3*b*x^7 + 336*a^2*b^2*x^5 + 1050*a*b^3*x^3 + 630* b^4*x + 315*(a^2*b^2*x^4 + 2*a*b^3*x^2 + b^4)*sqrt(-b/a)*log((a*x^2 - 2*a* x*sqrt(-b/a) - b)/(a*x^2 + b)))/(a^7*x^4 + 2*a^6*b*x^2 + a^5*b^2), 1/40*(8 *a^4*x^9 - 24*a^3*b*x^7 + 168*a^2*b^2*x^5 + 525*a*b^3*x^3 + 315*b^4*x - 31 5*(a^2*b^2*x^4 + 2*a*b^3*x^2 + b^4)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^ 7*x^4 + 2*a^6*b*x^2 + a^5*b^2)]
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.50 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {63 \sqrt {- \frac {b^{5}}{a^{11}}} \log {\left (- \frac {a^{5} \sqrt {- \frac {b^{5}}{a^{11}}}}{b^{2}} + x \right )}}{16} - \frac {63 \sqrt {- \frac {b^{5}}{a^{11}}} \log {\left (\frac {a^{5} \sqrt {- \frac {b^{5}}{a^{11}}}}{b^{2}} + x \right )}}{16} + \frac {17 a b^{3} x^{3} + 15 b^{4} x}{8 a^{7} x^{4} + 16 a^{6} b x^{2} + 8 a^{5} b^{2}} + \frac {x^{5}}{5 a^{3}} - \frac {b x^{3}}{a^{4}} + \frac {6 b^{2} x}{a^{5}} \] Input:
integrate(x**4/(a+b/x**2)**3,x)
Output:
63*sqrt(-b**5/a**11)*log(-a**5*sqrt(-b**5/a**11)/b**2 + x)/16 - 63*sqrt(-b **5/a**11)*log(a**5*sqrt(-b**5/a**11)/b**2 + x)/16 + (17*a*b**3*x**3 + 15* b**4*x)/(8*a**7*x**4 + 16*a**6*b*x**2 + 8*a**5*b**2) + x**5/(5*a**3) - b*x **3/a**4 + 6*b**2*x/a**5
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {17 \, a b^{3} x^{3} + 15 \, b^{4} x}{8 \, {\left (a^{7} x^{4} + 2 \, a^{6} b x^{2} + a^{5} b^{2}\right )}} - \frac {63 \, b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} + \frac {a^{2} x^{5} - 5 \, a b x^{3} + 30 \, b^{2} x}{5 \, a^{5}} \] Input:
integrate(x^4/(a+b/x^2)^3,x, algorithm="maxima")
Output:
1/8*(17*a*b^3*x^3 + 15*b^4*x)/(a^7*x^4 + 2*a^6*b*x^2 + a^5*b^2) - 63/8*b^3 *arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/5*(a^2*x^5 - 5*a*b*x^3 + 30*b^2 *x)/a^5
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=-\frac {63 \, b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} + \frac {17 \, a b^{3} x^{3} + 15 \, b^{4} x}{8 \, {\left (a x^{2} + b\right )}^{2} a^{5}} + \frac {a^{12} x^{5} - 5 \, a^{11} b x^{3} + 30 \, a^{10} b^{2} x}{5 \, a^{15}} \] Input:
integrate(x^4/(a+b/x^2)^3,x, algorithm="giac")
Output:
-63/8*b^3*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/8*(17*a*b^3*x^3 + 15*b ^4*x)/((a*x^2 + b)^2*a^5) + 1/5*(a^12*x^5 - 5*a^11*b*x^3 + 30*a^10*b^2*x)/ a^15
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {\frac {15\,b^4\,x}{8}+\frac {17\,a\,b^3\,x^3}{8}}{a^7\,x^4+2\,a^6\,b\,x^2+a^5\,b^2}+\frac {x^5}{5\,a^3}-\frac {b\,x^3}{a^4}+\frac {6\,b^2\,x}{a^5}-\frac {63\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,a^{11/2}} \] Input:
int(x^4/(a + b/x^2)^3,x)
Output:
((15*b^4*x)/8 + (17*a*b^3*x^3)/8)/(a^5*b^2 + a^7*x^4 + 2*a^6*b*x^2) + x^5/ (5*a^3) - (b*x^3)/a^4 + (6*b^2*x)/a^5 - (63*b^(5/2)*atan((a^(1/2)*x)/b^(1/ 2)))/(8*a^(11/2))
Time = 0.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{4}-630 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{2}-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4}+8 a^{5} x^{9}-24 a^{4} b \,x^{7}+168 a^{3} b^{2} x^{5}+525 a^{2} b^{3} x^{3}+315 a \,b^{4} x}{40 a^{6} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right )} \] Input:
int(x^4/(a+b/x^2)^3,x)
Output:
( - 315*sqrt(b)*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**4 - 630 *sqrt(b)*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a*b**3*x**2 - 315*sqrt(b)*s qrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*b**4 + 8*a**5*x**9 - 24*a**4*b*x**7 + 168*a**3*b**2*x**5 + 525*a**2*b**3*x**3 + 315*a*b**4*x)/(40*a**6*(a**2*x* *4 + 2*a*b*x**2 + b**2))