Integrand size = 13, antiderivative size = 96 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {6 a^2 x}{b^5}-\frac {a x^3}{b^4}+\frac {x^5}{5 b^3}-\frac {a^4 x}{4 b^5 \left (a+b x^2\right )^2}+\frac {17 a^3 x}{8 b^5 \left (a+b x^2\right )}-\frac {63 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \] Output:
6*a^2*x/b^5-a*x^3/b^4+1/5*x^5/b^3-1/4*a^4*x/b^5/(b*x^2+a)^2+17/8*a^3*x/b^5 /(b*x^2+a)-63/8*a^(5/2)*arctan(b^(1/2)*x/a^(1/2))/b^(11/2)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {315 a^4 x+525 a^3 b x^3+168 a^2 b^2 x^5-24 a b^3 x^7+8 b^4 x^9}{40 b^5 \left (a+b x^2\right )^2}-\frac {63 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \] Input:
Integrate[x^10/(a + b*x^2)^3,x]
Output:
(315*a^4*x + 525*a^3*b*x^3 + 168*a^2*b^2*x^5 - 24*a*b^3*x^7 + 8*b^4*x^9)/( 40*b^5*(a + b*x^2)^2) - (63*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(11/ 2))
Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {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.
\(\displaystyle \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {9 \int \frac {x^8}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {9 \left (\frac {7 \int \frac {x^6}{b x^2+a}dx}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {9 \left (\frac {7 \int \left (\frac {x^4}{b}-\frac {a x^2}{b^2}-\frac {a^3}{b^3 \left (b x^2+a\right )}+\frac {a^2}{b^3}\right )dx}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 \left (\frac {7 \left (-\frac {a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 x}{b^3}-\frac {a x^3}{3 b^2}+\frac {x^5}{5 b}\right )}{2 b}-\frac {x^7}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^9}{4 b \left (a+b x^2\right )^2}\) |
Input:
Int[x^10/(a + b*x^2)^3,x]
Output:
-1/4*x^9/(b*(a + b*x^2)^2) + (9*(-1/2*x^7/(b*(a + b*x^2)) + (7*((a^2*x)/b^ 3 - (a*x^3)/(3*b^2) + x^5/(5*b) - (a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^ (7/2)))/(2*b)))/(4*b)
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]
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {\frac {1}{5} b^{2} x^{5}-a b \,x^{3}+6 a^{2} x}{b^{5}}-\frac {a^{3} \left (\frac {-\frac {17}{8} b \,x^{3}-\frac {15}{8} a x}{\left (b \,x^{2}+a \right )^{2}}+\frac {63 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\) | \(74\) |
risch | \(\frac {x^{5}}{5 b^{3}}-\frac {a \,x^{3}}{b^{4}}+\frac {6 a^{2} x}{b^{5}}+\frac {\frac {17}{8} a^{3} b \,x^{3}+\frac {15}{8} a^{4} x}{b^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {63 \sqrt {-a b}\, a^{2} \ln \left (-\sqrt {-a b}\, x -a \right )}{16 b^{6}}-\frac {63 \sqrt {-a b}\, a^{2} \ln \left (\sqrt {-a b}\, x -a \right )}{16 b^{6}}\) | \(112\) |
Input:
int(x^10/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b^5*(1/5*b^2*x^5-a*b*x^3+6*a^2*x)-1/b^5*a^3*((-17/8*b*x^3-15/8*a*x)/(b*x ^2+a)^2+63/8/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.07 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.67 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, b^{4} x^{9} - 48 \, a b^{3} x^{7} + 336 \, a^{2} b^{2} x^{5} + 1050 \, a^{3} b x^{3} + 630 \, a^{4} x + 315 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{80 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {8 \, b^{4} x^{9} - 24 \, a b^{3} x^{7} + 168 \, a^{2} b^{2} x^{5} + 525 \, a^{3} b x^{3} + 315 \, a^{4} x - 315 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{40 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \] Input:
integrate(x^10/(b*x^2+a)^3,x, algorithm="fricas")
Output:
[1/80*(16*b^4*x^9 - 48*a*b^3*x^7 + 336*a^2*b^2*x^5 + 1050*a^3*b*x^3 + 630* a^4*x + 315*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*sqrt(-a/b)*log((b*x^2 - 2*b* x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5), 1/40*(8 *b^4*x^9 - 24*a*b^3*x^7 + 168*a^2*b^2*x^5 + 525*a^3*b*x^3 + 315*a^4*x - 31 5*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^ 7*x^4 + 2*a*b^6*x^2 + a^2*b^5)]
Time = 0.25 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.50 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {6 a^{2} x}{b^{5}} - \frac {a x^{3}}{b^{4}} + \frac {63 \sqrt {- \frac {a^{5}}{b^{11}}} \log {\left (x - \frac {b^{5} \sqrt {- \frac {a^{5}}{b^{11}}}}{a^{2}} \right )}}{16} - \frac {63 \sqrt {- \frac {a^{5}}{b^{11}}} \log {\left (x + \frac {b^{5} \sqrt {- \frac {a^{5}}{b^{11}}}}{a^{2}} \right )}}{16} + \frac {15 a^{4} x + 17 a^{3} b x^{3}}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} + \frac {x^{5}}{5 b^{3}} \] Input:
integrate(x**10/(b*x**2+a)**3,x)
Output:
6*a**2*x/b**5 - a*x**3/b**4 + 63*sqrt(-a**5/b**11)*log(x - b**5*sqrt(-a**5 /b**11)/a**2)/16 - 63*sqrt(-a**5/b**11)*log(x + b**5*sqrt(-a**5/b**11)/a** 2)/16 + (15*a**4*x + 17*a**3*b*x**3)/(8*a**2*b**5 + 16*a*b**6*x**2 + 8*b** 7*x**4) + x**5/(5*b**3)
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {17 \, a^{3} b x^{3} + 15 \, a^{4} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} - \frac {63 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {b^{2} x^{5} - 5 \, a b x^{3} + 30 \, a^{2} x}{5 \, b^{5}} \] Input:
integrate(x^10/(b*x^2+a)^3,x, algorithm="maxima")
Output:
1/8*(17*a^3*b*x^3 + 15*a^4*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5) - 63/8*a^3 *arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/5*(b^2*x^5 - 5*a*b*x^3 + 30*a^2 *x)/b^5
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=-\frac {63 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {17 \, a^{3} b x^{3} + 15 \, a^{4} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {b^{12} x^{5} - 5 \, a b^{11} x^{3} + 30 \, a^{2} b^{10} x}{5 \, b^{15}} \] Input:
integrate(x^10/(b*x^2+a)^3,x, algorithm="giac")
Output:
-63/8*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/8*(17*a^3*b*x^3 + 15*a ^4*x)/((b*x^2 + a)^2*b^5) + 1/5*(b^12*x^5 - 5*a*b^11*x^3 + 30*a^2*b^10*x)/ b^15
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {15\,a^4\,x}{8}+\frac {17\,b\,a^3\,x^3}{8}}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}+\frac {x^5}{5\,b^3}-\frac {a\,x^3}{b^4}+\frac {6\,a^2\,x}{b^5}-\frac {63\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,b^{11/2}} \] Input:
int(x^10/(a + b*x^2)^3,x)
Output:
((15*a^4*x)/8 + (17*a^3*b*x^3)/8)/(a^2*b^5 + b^7*x^4 + 2*a*b^6*x^2) + x^5/ (5*b^3) - (a*x^3)/b^4 + (6*a^2*x)/b^5 - (63*a^(5/2)*atan((b^(1/2)*x)/a^(1/ 2)))/(8*b^(11/2))
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51 \[ \int \frac {x^{10}}{\left (a+b x^2\right )^3} \, dx=\frac {-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}-630 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,x^{2}-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{4}+315 a^{4} b x +525 a^{3} b^{2} x^{3}+168 a^{2} b^{3} x^{5}-24 a \,b^{4} x^{7}+8 b^{5} x^{9}}{40 b^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^10/(b*x^2+a)^3,x)
Output:
( - 315*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4 - 630*sqrt(b)*s qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*x**2 - 315*sqrt(b)*sqrt(a)*ata n((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**4 + 315*a**4*b*x + 525*a**3*b**2*x **3 + 168*a**2*b**3*x**5 - 24*a*b**4*x**7 + 8*b**5*x**9)/(40*b**6*(a**2 + 2*a*b*x**2 + b**2*x**4))