Integrand size = 22, antiderivative size = 99 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {a^2 (A b-a B)}{b^4 \sqrt {a+b x^2}}-\frac {a (2 A b-3 a B) \sqrt {a+b x^2}}{b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^4} \]
1/3*(A*b-3*B*a)*(b*x^2+a)^(3/2)/b^4+1/5*B*(b*x^2+a)^(5/2)/b^4-a^2*(A*b-B*a )/b^4/(b*x^2+a)^(1/2)-a*(2*A*b-3*B*a)*(b*x^2+a)^(1/2)/b^4
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {48 a^3 B-8 a^2 b \left (5 A-3 B x^2\right )+b^3 x^4 \left (5 A+3 B x^2\right )-2 a b^2 x^2 \left (10 A+3 B x^2\right )}{15 b^4 \sqrt {a+b x^2}} \]
(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^2) + b^3*x^4*(5*A + 3*B*x^2) - 2*a*b^2*x^ 2*(10*A + 3*B*x^2))/(15*b^4*Sqrt[a + b*x^2])
Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 86, 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.
\(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (B x^2+A\right )}{\left (b x^2+a\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(a B-A b) a^2}{b^3 \left (b x^2+a\right )^{3/2}}+\frac {(3 a B-2 A b) a}{b^3 \sqrt {b x^2+a}}+\frac {B \left (b x^2+a\right )^{3/2}}{b^3}+\frac {(A b-3 a B) \sqrt {b x^2+a}}{b^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 a^2 (A b-a B)}{b^4 \sqrt {a+b x^2}}+\frac {2 \left (a+b x^2\right )^{3/2} (A b-3 a B)}{3 b^4}-\frac {2 a \sqrt {a+b x^2} (2 A b-3 a B)}{b^4}+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b^4}\right )\) |
((-2*a^2*(A*b - a*B))/(b^4*Sqrt[a + b*x^2]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x^2])/b^4 + (2*(A*b - 3*a*B)*(a + b*x^2)^(3/2))/(3*b^4) + (2*B*(a + b* x^2)^(5/2))/(5*b^4))/2
3.6.70.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.80 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {8 \left (-\frac {x^{4} \left (\frac {3 x^{2} B}{5}+A \right ) b^{3}}{8}+\frac {x^{2} \left (\frac {3 x^{2} B}{10}+A \right ) a \,b^{2}}{2}+a^{2} \left (-\frac {3 x^{2} B}{5}+A \right ) b -\frac {6 a^{3} B}{5}\right )}{3 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(68\) |
gosper | \(-\frac {-3 b^{3} B \,x^{6}-5 A \,b^{3} x^{4}+6 B a \,b^{2} x^{4}+20 a A \,b^{2} x^{2}-24 B \,a^{2} b \,x^{2}+40 a^{2} b A -48 a^{3} B}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(77\) |
trager | \(-\frac {-3 b^{3} B \,x^{6}-5 A \,b^{3} x^{4}+6 B a \,b^{2} x^{4}+20 a A \,b^{2} x^{2}-24 B \,a^{2} b \,x^{2}+40 a^{2} b A -48 a^{3} B}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(77\) |
risch | \(-\frac {\left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+9 B a b \,x^{2}+25 a b A -33 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{15 b^{4}}-\frac {a^{2} \left (A b -B a \right )}{b^{4} \sqrt {b \,x^{2}+a}}\) | \(79\) |
default | \(B \left (\frac {x^{6}}{5 b \sqrt {b \,x^{2}+a}}-\frac {6 a \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )}{5 b}\right )+A \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )\) | \(142\) |
-8/3/(b*x^2+a)^(1/2)*(-1/8*x^4*(3/5*x^2*B+A)*b^3+1/2*x^2*(3/10*x^2*B+A)*a* b^2+a^2*(-3/5*x^2*B+A)*b-6/5*a^3*B)/b^4
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (3 \, B b^{3} x^{6} - {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{4} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]
1/15*(3*B*b^3*x^6 - (6*B*a*b^2 - 5*A*b^3)*x^4 + 48*B*a^3 - 40*A*a^2*b + 4* (6*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(b*x^2 + a)/(b^5*x^2 + a*b^4)
Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.74 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\begin {cases} - \frac {8 A a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 A a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {A x^{4}}{3 b \sqrt {a + b x^{2}}} + \frac {16 B a^{3}}{5 b^{4} \sqrt {a + b x^{2}}} + \frac {8 B a^{2} x^{2}}{5 b^{3} \sqrt {a + b x^{2}}} - \frac {2 B a x^{4}}{5 b^{2} \sqrt {a + b x^{2}}} + \frac {B x^{6}}{5 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{8}}{8}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-8*A*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*A*a*x**2/(3*b**2*sqrt(a + b*x**2)) + A*x**4/(3*b*sqrt(a + b*x**2)) + 16*B*a**3/(5*b**4*sqrt(a + b *x**2)) + 8*B*a**2*x**2/(5*b**3*sqrt(a + b*x**2)) - 2*B*a*x**4/(5*b**2*sqr t(a + b*x**2)) + B*x**6/(5*b*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**6/6 + B* x**8/8)/a**(3/2), True))
Time = 0.19 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.33 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{6}}{5 \, \sqrt {b x^{2} + a} b} - \frac {2 \, B a x^{4}}{5 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {8 \, B a^{2} x^{2}}{5 \, \sqrt {b x^{2} + a} b^{3}} - \frac {4 \, A a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {16 \, B a^{3}}{5 \, \sqrt {b x^{2} + a} b^{4}} - \frac {8 \, A a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} \]
1/5*B*x^6/(sqrt(b*x^2 + a)*b) - 2/5*B*a*x^4/(sqrt(b*x^2 + a)*b^2) + 1/3*A* x^4/(sqrt(b*x^2 + a)*b) + 8/5*B*a^2*x^2/(sqrt(b*x^2 + a)*b^3) - 4/3*A*a*x^ 2/(sqrt(b*x^2 + a)*b^2) + 16/5*B*a^3/(sqrt(b*x^2 + a)*b^4) - 8/3*A*a^2/(sq rt(b*x^2 + a)*b^3)
Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B a^{3} - A a^{2} b}{\sqrt {b x^{2} + a} b^{4}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{16} - 15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{16} + 45 \, \sqrt {b x^{2} + a} B a^{2} b^{16} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{17} - 30 \, \sqrt {b x^{2} + a} A a b^{17}}{15 \, b^{20}} \]
(B*a^3 - A*a^2*b)/(sqrt(b*x^2 + a)*b^4) + 1/15*(3*(b*x^2 + a)^(5/2)*B*b^16 - 15*(b*x^2 + a)^(3/2)*B*a*b^16 + 45*sqrt(b*x^2 + a)*B*a^2*b^16 + 5*(b*x^ 2 + a)^(3/2)*A*b^17 - 30*sqrt(b*x^2 + a)*A*a*b^17)/b^20
Time = 5.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.90 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\frac {B\,{\left (b\,x^2+a\right )}^3}{5}+B\,a^3+\frac {A\,b\,{\left (b\,x^2+a\right )}^2}{3}-B\,a\,{\left (b\,x^2+a\right )}^2+3\,B\,a^2\,\left (b\,x^2+a\right )-A\,a^2\,b-2\,A\,a\,b\,\left (b\,x^2+a\right )}{b^4\,\sqrt {b\,x^2+a}} \]