Integrand size = 22, antiderivative size = 128 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a^3 (A b-a B)}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a+b x^2}}-\frac {3 a (A b-2 a B) \sqrt {a+b x^2}}{b^5}+\frac {(A b-4 a B) \left (a+b x^2\right )^{3/2}}{3 b^5}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^5} \]
1/3*a^3*(A*b-B*a)/b^5/(b*x^2+a)^(3/2)+1/3*(A*b-4*B*a)*(b*x^2+a)^(3/2)/b^5+ 1/5*B*(b*x^2+a)^(5/2)/b^5-a^2*(3*A*b-4*B*a)/b^5/(b*x^2+a)^(1/2)-3*a*(A*b-2 *B*a)*(b*x^2+a)^(1/2)/b^5
Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.77 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {128 a^4 B+24 a^2 b^2 x^2 \left (-5 A+2 B x^2\right )+b^4 x^6 \left (5 A+3 B x^2\right )-2 a b^3 x^4 \left (15 A+4 B x^2\right )+a^3 \left (-80 A b+192 b B x^2\right )}{15 b^5 \left (a+b x^2\right )^{3/2}} \]
(128*a^4*B + 24*a^2*b^2*x^2*(-5*A + 2*B*x^2) + b^4*x^6*(5*A + 3*B*x^2) - 2 *a*b^3*x^4*(15*A + 4*B*x^2) + a^3*(-80*A*b + 192*b*B*x^2))/(15*b^5*(a + b* x^2)^(3/2))
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (B x^2+A\right )}{\left (b x^2+a\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(a B-A b) a^3}{b^4 \left (b x^2+a\right )^{5/2}}-\frac {(4 a B-3 A b) a^2}{b^4 \left (b x^2+a\right )^{3/2}}+\frac {3 (2 a B-A b) a}{b^4 \sqrt {b x^2+a}}+\frac {B \left (b x^2+a\right )^{3/2}}{b^4}+\frac {(A b-4 a B) \sqrt {b x^2+a}}{b^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 a^3 (A b-a B)}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac {2 a^2 (3 A b-4 a B)}{b^5 \sqrt {a+b x^2}}-\frac {6 a \sqrt {a+b x^2} (A b-2 a B)}{b^5}+\frac {2 \left (a+b x^2\right )^{3/2} (A b-4 a B)}{3 b^5}+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b^5}\right )\) |
((2*a^3*(A*b - a*B))/(3*b^5*(a + b*x^2)^(3/2)) - (2*a^2*(3*A*b - 4*a*B))/( b^5*Sqrt[a + b*x^2]) - (6*a*(A*b - 2*a*B)*Sqrt[a + b*x^2])/b^5 + (2*(A*b - 4*a*B)*(a + b*x^2)^(3/2))/(3*b^5) + (2*B*(a + b*x^2)^(5/2))/(5*b^5))/2
3.6.84.3.1 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 = 2.86 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(-\frac {16 \left (-\frac {x^{6} \left (\frac {3 x^{2} B}{5}+A \right ) b^{4}}{16}+\frac {3 x^{4} \left (\frac {4 x^{2} B}{15}+A \right ) a \,b^{3}}{8}+\frac {3 x^{2} \left (-\frac {2 x^{2} B}{5}+A \right ) a^{2} b^{2}}{2}+a^{3} \left (-\frac {12 x^{2} B}{5}+A \right ) b -\frac {8 B \,a^{4}}{5}\right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{5}}\) | \(87\) |
gosper | \(-\frac {-3 B \,x^{8} b^{4}-5 A \,x^{6} b^{4}+8 B \,x^{6} a \,b^{3}+30 A a \,b^{3} x^{4}-48 B \,a^{2} b^{2} x^{4}+120 A \,a^{2} b^{2} x^{2}-192 B \,a^{3} b \,x^{2}+80 A \,a^{3} b -128 B \,a^{4}}{15 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{5}}\) | \(101\) |
trager | \(-\frac {-3 B \,x^{8} b^{4}-5 A \,x^{6} b^{4}+8 B \,x^{6} a \,b^{3}+30 A a \,b^{3} x^{4}-48 B \,a^{2} b^{2} x^{4}+120 A \,a^{2} b^{2} x^{2}-192 B \,a^{3} b \,x^{2}+80 A \,a^{3} b -128 B \,a^{4}}{15 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{5}}\) | \(101\) |
risch | \(-\frac {\left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+14 B a b \,x^{2}+40 a b A -73 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{15 b^{5}}-\frac {\sqrt {b \,x^{2}+a}\, \left (9 A \,b^{2} x^{2}-12 B a b \,x^{2}+8 a b A -11 a^{2} B \right ) a^{2}}{3 b^{5} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}\) | \(120\) |
default | \(B \left (\frac {x^{8}}{5 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {8 a \left (\frac {x^{6}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )}{b}\right )}{5 b}\right )+A \left (\frac {x^{6}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )}{b}\right )\) | \(190\) |
-16/3/(b*x^2+a)^(3/2)*(-1/16*x^6*(3/5*x^2*B+A)*b^4+3/8*x^4*(4/15*x^2*B+A)* a*b^3+3/2*x^2*(-2/5*x^2*B+A)*a^2*b^2+a^3*(-12/5*x^2*B+A)*b-8/5*B*a^4)/b^5
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.96 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, B b^{4} x^{8} - {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} x^{6} + 128 \, B a^{4} - 80 \, A a^{3} b + 6 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 24 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} \]
1/15*(3*B*b^4*x^8 - (8*B*a*b^3 - 5*A*b^4)*x^6 + 128*B*a^4 - 80*A*a^3*b + 6 *(8*B*a^2*b^2 - 5*A*a*b^3)*x^4 + 24*(8*B*a^3*b - 5*A*a^2*b^2)*x^2)*sqrt(b* x^2 + a)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)
Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (119) = 238\).
Time = 0.50 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.41 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {80 A a^{3} b}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} - \frac {120 A a^{2} b^{2} x^{2}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} - \frac {30 A a b^{3} x^{4}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} + \frac {5 A b^{4} x^{6}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} + \frac {128 B a^{4}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} + \frac {192 B a^{3} b x^{2}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} + \frac {48 B a^{2} b^{2} x^{4}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} - \frac {8 B a b^{3} x^{6}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} + \frac {3 B b^{4} x^{8}}{15 a b^{5} \sqrt {a + b x^{2}} + 15 b^{6} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{8}}{8} + \frac {B x^{10}}{10}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-80*A*a**3*b/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) - 120*A*a**2*b**2*x**2/(15*a*b**5*sqrt(a + b*x**2) + 15*b**6*x* *2*sqrt(a + b*x**2)) - 30*A*a*b**3*x**4/(15*a*b**5*sqrt(a + b*x**2) + 15*b **6*x**2*sqrt(a + b*x**2)) + 5*A*b**4*x**6/(15*a*b**5*sqrt(a + b*x**2) + 1 5*b**6*x**2*sqrt(a + b*x**2)) + 128*B*a**4/(15*a*b**5*sqrt(a + b*x**2) + 1 5*b**6*x**2*sqrt(a + b*x**2)) + 192*B*a**3*b*x**2/(15*a*b**5*sqrt(a + b*x* *2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 48*B*a**2*b**2*x**4/(15*a*b**5*sqrt (a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) - 8*B*a*b**3*x**6/(15*a*b**5 *sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)) + 3*B*b**4*x**8/(15*a*b **5*sqrt(a + b*x**2) + 15*b**6*x**2*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**8 /8 + B*x**10/10)/a**(5/2), True))
Time = 0.19 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.36 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {B x^{8}}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {8 \, B a x^{6}}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {A x^{6}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {16 \, B a^{2} x^{4}}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} - \frac {2 \, A a x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {64 \, B a^{3} x^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} - \frac {8 \, A a^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {128 \, B a^{4}}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{5}} - \frac {16 \, A a^{3}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} \]
1/5*B*x^8/((b*x^2 + a)^(3/2)*b) - 8/15*B*a*x^6/((b*x^2 + a)^(3/2)*b^2) + 1 /3*A*x^6/((b*x^2 + a)^(3/2)*b) + 16/5*B*a^2*x^4/((b*x^2 + a)^(3/2)*b^3) - 2*A*a*x^4/((b*x^2 + a)^(3/2)*b^2) + 64/5*B*a^3*x^2/((b*x^2 + a)^(3/2)*b^4) - 8*A*a^2*x^2/((b*x^2 + a)^(3/2)*b^3) + 128/15*B*a^4/((b*x^2 + a)^(3/2)*b ^5) - 16/3*A*a^3/((b*x^2 + a)^(3/2)*b^4)
Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.10 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {12 \, {\left (b x^{2} + a\right )} B a^{3} - B a^{4} - 9 \, {\left (b x^{2} + a\right )} A a^{2} b + A a^{3} b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{5}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{20} - 20 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{20} + 90 \, \sqrt {b x^{2} + a} B a^{2} b^{20} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{21} - 45 \, \sqrt {b x^{2} + a} A a b^{21}}{15 \, b^{25}} \]
1/3*(12*(b*x^2 + a)*B*a^3 - B*a^4 - 9*(b*x^2 + a)*A*a^2*b + A*a^3*b)/((b*x ^2 + a)^(3/2)*b^5) + 1/15*(3*(b*x^2 + a)^(5/2)*B*b^20 - 20*(b*x^2 + a)^(3/ 2)*B*a*b^20 + 90*sqrt(b*x^2 + a)*B*a^2*b^20 + 5*(b*x^2 + a)^(3/2)*A*b^21 - 45*sqrt(b*x^2 + a)*A*a*b^21)/b^25
Time = 5.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3\,B\,{\left (b\,x^2+a\right )}^4-5\,B\,a^4+90\,B\,a^2\,{\left (b\,x^2+a\right )}^2+5\,A\,b\,{\left (b\,x^2+a\right )}^3-20\,B\,a\,{\left (b\,x^2+a\right )}^3+60\,B\,a^3\,\left (b\,x^2+a\right )+5\,A\,a^3\,b-45\,A\,a\,b\,{\left (b\,x^2+a\right )}^2-45\,A\,a^2\,b\,\left (b\,x^2+a\right )}{15\,b^5\,{\left (b\,x^2+a\right )}^{3/2}} \]