Integrand size = 15, antiderivative size = 64 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {x^3}{7 a \left (a+b x^2\right )^{7/2}}+\frac {4 x^3}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac {8 x^3}{105 a^3 \left (a+b x^2\right )^{3/2}} \] Output:
1/7*x^3/a/(b*x^2+a)^(7/2)+4/35*x^3/a^2/(b*x^2+a)^(5/2)+8/105*x^3/a^3/(b*x^ 2+a)^(3/2)
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {35 a^2 x^3+28 a b x^5+8 b^2 x^7}{105 a^3 \left (a+b x^2\right )^{7/2}} \] Input:
Integrate[x^2/(a + b*x^2)^(9/2),x]
Output:
(35*a^2*x^3 + 28*a*b*x^5 + 8*b^2*x^7)/(105*a^3*(a + b*x^2)^(7/2))
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {245, 245, 242}
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.
\(\displaystyle \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {4 b \int \frac {x^4}{\left (b x^2+a\right )^{9/2}}dx}{3 a}+\frac {x^3}{3 a \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {4 b \left (\frac {2 b \int \frac {x^6}{\left (b x^2+a\right )^{9/2}}dx}{5 a}+\frac {x^5}{5 a \left (a+b x^2\right )^{7/2}}\right )}{3 a}+\frac {x^3}{3 a \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {4 b \left (\frac {2 b x^7}{35 a^2 \left (a+b x^2\right )^{7/2}}+\frac {x^5}{5 a \left (a+b x^2\right )^{7/2}}\right )}{3 a}+\frac {x^3}{3 a \left (a+b x^2\right )^{7/2}}\) |
Input:
Int[x^2/(a + b*x^2)^(9/2),x]
Output:
x^3/(3*a*(a + b*x^2)^(7/2)) + (4*b*(x^5/(5*a*(a + b*x^2)^(7/2)) + (2*b*x^7 )/(35*a^2*(a + b*x^2)^(7/2))))/(3*a)
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] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Time = 0.35 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(\frac {x^{3} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3}}\) | \(39\) |
trager | \(\frac {x^{3} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3}}\) | \(39\) |
pseudoelliptic | \(\frac {x^{3} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3}}\) | \(39\) |
orering | \(\frac {x^{3} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3}}\) | \(39\) |
default | \(-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\) | \(96\) |
Input:
int(x^2/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/105*x^3*(8*b^2*x^4+28*a*b*x^2+35*a^2)/(b*x^2+a)^(7/2)/a^3
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.28 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, b^{2} x^{7} + 28 \, a b x^{5} + 35 \, a^{2} x^{3}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{3} b^{4} x^{8} + 4 \, a^{4} b^{3} x^{6} + 6 \, a^{5} b^{2} x^{4} + 4 \, a^{6} b x^{2} + a^{7}\right )}} \] Input:
integrate(x^2/(b*x^2+a)^(9/2),x, algorithm="fricas")
Output:
1/105*(8*b^2*x^7 + 28*a*b*x^5 + 35*a^2*x^3)*sqrt(b*x^2 + a)/(a^3*b^4*x^8 + 4*a^4*b^3*x^6 + 6*a^5*b^2*x^4 + 4*a^6*b*x^2 + a^7)
Leaf count of result is larger than twice the leaf count of optimal. 517 vs. \(2 (56) = 112\).
Time = 1.00 (sec) , antiderivative size = 517, normalized size of antiderivative = 8.08 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {35 a^{5} x^{3}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {63 a^{4} b x^{5}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {36 a^{3} b^{2} x^{7}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {8 a^{2} b^{3} x^{9}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:
integrate(x**2/(b*x**2+a)**(9/2),x)
Output:
35*a**5*x**3/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt (1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2 )*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a )) + 63*a**4*b*x**5/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x* *2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a **(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b *x**2/a)) + 36*a**3*b**2*x**7/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**( 17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/ a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8* sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**9/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b **4*x**8*sqrt(1 + b*x**2/a))
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} \] Input:
integrate(x^2/(b*x^2+a)^(9/2),x, algorithm="maxima")
Output:
-1/7*x/((b*x^2 + a)^(7/2)*b) + 8/105*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*x/( (b*x^2 + a)^(3/2)*a^2*b) + 1/35*x/((b*x^2 + a)^(5/2)*a*b)
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.67 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (4 \, x^{2} {\left (\frac {2 \, b^{2} x^{2}}{a^{3}} + \frac {7 \, b}{a^{2}}\right )} + \frac {35}{a}\right )} x^{3}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:
integrate(x^2/(b*x^2+a)^(9/2),x, algorithm="giac")
Output:
1/105*(4*x^2*(2*b^2*x^2/a^3 + 7*b/a^2) + 35/a)*x^3/(b*x^2 + a)^(7/2)
Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {8\,x}{105\,a^3\,b\,\sqrt {b\,x^2+a}}-\frac {x}{7\,b\,{\left (b\,x^2+a\right )}^{7/2}}+\frac {4\,x}{105\,a^2\,b\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {x}{35\,a\,b\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:
int(x^2/(a + b*x^2)^(9/2),x)
Output:
(8*x)/(105*a^3*b*(a + b*x^2)^(1/2)) - x/(7*b*(a + b*x^2)^(7/2)) + (4*x)/(1 05*a^2*b*(a + b*x^2)^(3/2)) + x/(35*a*b*(a + b*x^2)^(5/2))
Time = 0.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.42 \[ \int \frac {x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}+28 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}+8 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}-8 \sqrt {b}\, a^{4}-32 \sqrt {b}\, a^{3} b \,x^{2}-48 \sqrt {b}\, a^{2} b^{2} x^{4}-32 \sqrt {b}\, a \,b^{3} x^{6}-8 \sqrt {b}\, b^{4} x^{8}}{105 a^{3} b^{2} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x^2/(b*x^2+a)^(9/2),x)
Output:
(35*sqrt(a + b*x**2)*a**2*b**2*x**3 + 28*sqrt(a + b*x**2)*a*b**3*x**5 + 8* sqrt(a + b*x**2)*b**4*x**7 - 8*sqrt(b)*a**4 - 32*sqrt(b)*a**3*b*x**2 - 48* sqrt(b)*a**2*b**2*x**4 - 32*sqrt(b)*a*b**3*x**6 - 8*sqrt(b)*b**4*x**8)/(10 5*a**3*b**2*(a**4 + 4*a**3*b*x**2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b** 4*x**8))