Integrand size = 19, antiderivative size = 121 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {(A b-a B) x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a B) x}{35 a^2 b \left (a+b x^2\right )^{5/2}}+\frac {4 (6 A b+a B) x}{105 a^3 b \left (a+b x^2\right )^{3/2}}+\frac {8 (6 A b+a B) x}{105 a^4 b \sqrt {a+b x^2}} \] Output:
1/7*(A*b-B*a)*x/a/b/(b*x^2+a)^(7/2)+1/35*(6*A*b+B*a)*x/a^2/b/(b*x^2+a)^(5/ 2)+4/105*(6*A*b+B*a)*x/a^3/b/(b*x^2+a)^(3/2)+8/105*(6*A*b+B*a)*x/a^4/b/(b* x^2+a)^(1/2)
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {48 A b^3 x^7+8 a b^2 x^5 \left (21 A+B x^2\right )+14 a^2 b x^3 \left (15 A+2 B x^2\right )+35 a^3 \left (3 A x+B x^3\right )}{105 a^4 \left (a+b x^2\right )^{7/2}} \] Input:
Integrate[(A + B*x^2)/(a + b*x^2)^(9/2),x]
Output:
(48*A*b^3*x^7 + 8*a*b^2*x^5*(21*A + B*x^2) + 14*a^2*b*x^3*(15*A + 2*B*x^2) + 35*a^3*(3*A*x + B*x^3))/(105*a^4*(a + b*x^2)^(7/2))
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {298, 209, 209, 208}
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.
\(\displaystyle \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {(a B+6 A b) \int \frac {1}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}+\frac {x (A b-a B)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a B+6 A b) \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}+\frac {x (A b-a B)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a B+6 A b) \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}+\frac {x (A b-a B)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\left (\frac {4 \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right ) (a B+6 A b)}{7 a b}+\frac {x (A b-a B)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
Input:
Int[(A + B*x^2)/(a + b*x^2)^(9/2),x]
Output:
((A*b - a*B)*x)/(7*a*b*(a + b*x^2)^(7/2)) + ((6*A*b + a*B)*(x/(5*a*(a + b* x^2)^(5/2)) + (4*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2] )))/(5*a)))/(7*a*b)
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Time = 0.55 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.59
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {x^{2} B}{3}+A \right ) a^{3}+2 \left (\frac {2 x^{2} B}{15}+A \right ) b \,x^{2} a^{2}+\frac {8 b^{2} \left (\frac {x^{2} B}{21}+A \right ) x^{4} a}{5}+\frac {16 A \,b^{3} x^{6}}{35}\right ) x}{\left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4}}\) | \(71\) |
gosper | \(\frac {x \left (48 A \,b^{3} x^{6}+8 B a \,b^{2} x^{6}+168 a A \,b^{2} x^{4}+28 B \,a^{2} b \,x^{4}+210 a^{2} A b \,x^{2}+35 B \,a^{3} x^{2}+105 a^{3} A \right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4}}\) | \(81\) |
trager | \(\frac {x \left (48 A \,b^{3} x^{6}+8 B a \,b^{2} x^{6}+168 a A \,b^{2} x^{4}+28 B \,a^{2} b \,x^{4}+210 a^{2} A b \,x^{2}+35 B \,a^{3} x^{2}+105 a^{3} A \right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4}}\) | \(81\) |
orering | \(\frac {x \left (48 A \,b^{3} x^{6}+8 B a \,b^{2} x^{6}+168 a A \,b^{2} x^{4}+28 B \,a^{2} b \,x^{4}+210 a^{2} A b \,x^{2}+35 B \,a^{3} x^{2}+105 a^{3} A \right )}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4}}\) | \(81\) |
default | \(A \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )+B \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )\) | \(174\) |
Input:
int((B*x^2+A)/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
Output:
((1/3*x^2*B+A)*a^3+2*(2/15*x^2*B+A)*b*x^2*a^2+8/5*b^2*(1/21*x^2*B+A)*x^4*a +16/35*A*b^3*x^6)/(b*x^2+a)^(7/2)*x/a^4
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x^{7} + 28 \, {\left (B a^{2} b + 6 \, A a b^{2}\right )} x^{5} + 105 \, A a^{3} x + 35 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")
Output:
1/105*(8*(B*a*b^2 + 6*A*b^3)*x^7 + 28*(B*a^2*b + 6*A*a*b^2)*x^5 + 105*A*a^ 3*x + 35*(B*a^3 + 6*A*a^2*b)*x^3)*sqrt(b*x^2 + a)/(a^4*b^4*x^8 + 4*a^5*b^3 *x^6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 + a^8)
Leaf count of result is larger than twice the leaf count of optimal. 1787 vs. \(2 (114) = 228\).
Time = 24.52 (sec) , antiderivative size = 1787, normalized size of antiderivative = 14.77 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \] Input:
integrate((B*x**2+A)/(b*x**2+a)**(9/2),x)
Output:
A*(35*a**14*x/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt (1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2 )*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a ) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12* sqrt(1 + b*x**2/a)) + 175*a**13*b*x**3/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b* *4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 371*a**12*b**2*x**5/(35*a** (37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525* a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b** 5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 429*a**11*b**3*x**7/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x* *2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a **(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b *x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6 *x**12*sqrt(1 + b*x**2/a)) + 286*a**10*b**4*x**9/(35*a**(37/2)*sqrt(1 + b* x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x** 4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525...
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {16 \, A x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {B x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, B x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, B x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")
Output:
16/35*A*x/(sqrt(b*x^2 + a)*a^4) + 8/35*A*x/((b*x^2 + a)^(3/2)*a^3) + 6/35* A*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*A*x/((b*x^2 + a)^(7/2)*a) - 1/7*B*x/((b* x^2 + a)^(7/2)*b) + 8/105*B*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*B*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*B*x/((b*x^2 + a)^(5/2)*a*b)
Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (4 \, x^{2} {\left (\frac {2 \, {\left (B a b^{5} + 6 \, A b^{6}\right )} x^{2}}{a^{4} b^{3}} + \frac {7 \, {\left (B a^{2} b^{4} + 6 \, A a b^{5}\right )}}{a^{4} b^{3}}\right )} + \frac {35 \, {\left (B a^{3} b^{3} + 6 \, A a^{2} b^{4}\right )}}{a^{4} b^{3}}\right )} x^{2} + \frac {105 \, A}{a}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="giac")
Output:
1/105*((4*x^2*(2*(B*a*b^5 + 6*A*b^6)*x^2/(a^4*b^3) + 7*(B*a^2*b^4 + 6*A*a* b^5)/(a^4*b^3)) + 35*(B*a^3*b^3 + 6*A*a^2*b^4)/(a^4*b^3))*x^2 + 105*A/a)*x /(b*x^2 + a)^(7/2)
Time = 0.49 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {x\,\left (\frac {A}{7\,a}-\frac {B}{7\,b}\right )}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {x\,\left (6\,A\,b+B\,a\right )}{35\,a^2\,b\,{\left (b\,x^2+a\right )}^{5/2}}+\frac {x\,\left (24\,A\,b+4\,B\,a\right )}{105\,a^3\,b\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {x\,\left (48\,A\,b+8\,B\,a\right )}{105\,a^4\,b\,\sqrt {b\,x^2+a}} \] Input:
int((A + B*x^2)/(a + b*x^2)^(9/2),x)
Output:
(x*(A/(7*a) - B/(7*b)))/(a + b*x^2)^(7/2) + (x*(6*A*b + B*a))/(35*a^2*b*(a + b*x^2)^(5/2)) + (x*(24*A*b + 4*B*a))/(105*a^3*b*(a + b*x^2)^(3/2)) + (x *(48*A*b + 8*B*a))/(105*a^4*b*(a + b*x^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b x +20 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}-8 \sqrt {b}\, a^{3}-24 \sqrt {b}\, a^{2} b \,x^{2}-24 \sqrt {b}\, a \,b^{2} x^{4}-8 \sqrt {b}\, b^{3} x^{6}}{15 a^{3} b \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int((B*x^2+A)/(b*x^2+a)^(9/2),x)
Output:
(15*sqrt(a + b*x**2)*a**2*b*x + 20*sqrt(a + b*x**2)*a*b**2*x**3 + 8*sqrt(a + b*x**2)*b**3*x**5 - 8*sqrt(b)*a**3 - 24*sqrt(b)*a**2*b*x**2 - 24*sqrt(b )*a*b**2*x**4 - 8*sqrt(b)*b**3*x**6)/(15*a**3*b*(a**3 + 3*a**2*b*x**2 + 3* a*b**2*x**4 + b**3*x**6))