Integrand size = 22, antiderivative size = 127 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {a B-(A b-a C) x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a C) x}{35 a^2 b \left (a+b x^2\right )^{5/2}}+\frac {4 (6 A b+a C) x}{105 a^3 b \left (a+b x^2\right )^{3/2}}+\frac {8 (6 A b+a C) x}{105 a^4 b \sqrt {a+b x^2}} \]
1/7*(-B*a+(A*b-C*a)*x)/a/b/(b*x^2+a)^(7/2)+1/35*(6*A*b+C*a)*x/a^2/b/(b*x^2 +a)^(5/2)+4/105*(6*A*b+C*a)*x/a^3/b/(b*x^2+a)^(3/2)+8/105*(6*A*b+C*a)*x/a^ 4/b/(b*x^2+a)^(1/2)
Time = 0.61 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-15 a^4 B+48 A b^4 x^7+35 a^3 b x \left (3 A+C x^2\right )+8 a b^3 x^5 \left (21 A+C x^2\right )+14 a^2 b^2 x^3 \left (15 A+2 C x^2\right )}{105 a^4 b \left (a+b x^2\right )^{7/2}} \]
(-15*a^4*B + 48*A*b^4*x^7 + 35*a^3*b*x*(3*A + C*x^2) + 8*a*b^3*x^5*(21*A + C*x^2) + 14*a^2*b^2*x^3*(15*A + 2*C*x^2))/(105*a^4*b*(a + b*x^2)^(7/2))
Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2345, 25, 27, 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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {6 A b+a C}{b \left (b x^2+a\right )^{7/2}}dx}{7 a}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {6 A b+a C}{b \left (b x^2+a\right )^{7/2}}dx}{7 a}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a C+6 A b) \int \frac {1}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a C+6 A b) \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a C+6 A b) \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\left (\frac {4 \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right ) (a C+6 A b)}{7 a b}-\frac {a B-x (A b-a C)}{7 a b \left (a+b x^2\right )^{7/2}}\) |
-1/7*(a*B - (A*b - a*C)*x)/(a*b*(a + b*x^2)^(7/2)) + ((6*A*b + a*C)*(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)
3.1.54.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 3.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76
method | result | size |
gosper | \(\frac {48 A \,b^{4} x^{7}+8 C a \,x^{7} b^{3}+168 A a \,b^{3} x^{5}+28 C \,a^{2} x^{5} b^{2}+210 A \,a^{2} b^{2} x^{3}+35 C \,a^{3} x^{3} b +105 A \,a^{3} b x -15 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b}\) | \(96\) |
trager | \(\frac {48 A \,b^{4} x^{7}+8 C a \,x^{7} b^{3}+168 A a \,b^{3} x^{5}+28 C \,a^{2} x^{5} b^{2}+210 A \,a^{2} b^{2} x^{3}+35 C \,a^{3} x^{3} b +105 A \,a^{3} b x -15 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b}\) | \(96\) |
default | \(A \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )+C \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )-\frac {B}{7 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\) | \(189\) |
1/105*(48*A*b^4*x^7+8*C*a*b^3*x^7+168*A*a*b^3*x^5+28*C*a^2*b^2*x^5+210*A*a ^2*b^2*x^3+35*C*a^3*b*x^3+105*A*a^3*b*x-15*B*a^4)/(b*x^2+a)^(7/2)/a^4/b
Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, {\left (C a b^{3} + 6 \, A b^{4}\right )} x^{7} + 105 \, A a^{3} b x + 28 \, {\left (C a^{2} b^{2} + 6 \, A a b^{3}\right )} x^{5} - 15 \, B a^{4} + 35 \, {\left (C a^{3} b + 6 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{4} b^{5} x^{8} + 4 \, a^{5} b^{4} x^{6} + 6 \, a^{6} b^{3} x^{4} + 4 \, a^{7} b^{2} x^{2} + a^{8} b\right )}} \]
1/105*(8*(C*a*b^3 + 6*A*b^4)*x^7 + 105*A*a^3*b*x + 28*(C*a^2*b^2 + 6*A*a*b ^3)*x^5 - 15*B*a^4 + 35*(C*a^3*b + 6*A*a^2*b^2)*x^3)*sqrt(b*x^2 + a)/(a^4* b^5*x^8 + 4*a^5*b^4*x^6 + 6*a^6*b^3*x^4 + 4*a^7*b^2*x^2 + a^8*b)
Leaf count of result is larger than twice the leaf count of optimal. 1266 vs. \(2 (117) = 234\).
Time = 24.19 (sec) , antiderivative size = 1880, normalized size of antiderivative = 14.80 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \]
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 = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {16 \, A x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {C x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, C x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, C x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {B}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} \]
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*C*x/((b* x^2 + a)^(7/2)*b) + 8/105*C*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*C*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*C*x/((b*x^2 + a)^(5/2)*a*b) - 1/7*B/((b*x^2 + a)^ (7/2)*b)
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (4 \, x^{2} {\left (\frac {2 \, {\left (C a b^{5} + 6 \, A b^{6}\right )} x^{2}}{a^{4} b^{3}} + \frac {7 \, {\left (C a^{2} b^{4} + 6 \, A a b^{5}\right )}}{a^{4} b^{3}}\right )} + \frac {35 \, {\left (C a^{3} b^{3} + 6 \, A a^{2} b^{4}\right )}}{a^{4} b^{3}}\right )} x^{2} + \frac {105 \, A}{a}\right )} x - \frac {15 \, B}{b}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \]
1/105*(((4*x^2*(2*(C*a*b^5 + 6*A*b^6)*x^2/(a^4*b^3) + 7*(C*a^2*b^4 + 6*A*a *b^5)/(a^4*b^3)) + 35*(C*a^3*b^3 + 6*A*a^2*b^4)/(a^4*b^3))*x^2 + 105*A/a)* x - 15*B/b)/(b*x^2 + a)^(7/2)
Time = 5.76 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {x\,\left (6\,A\,b+C\,a\right )}{35\,a^2\,b\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {\frac {B}{7\,b}-x\,\left (\frac {A}{7\,a}-\frac {C}{7\,b}\right )}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {x\,\left (24\,A\,b+4\,C\,a\right )}{105\,a^3\,b\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {x\,\left (48\,A\,b+8\,C\,a\right )}{105\,a^4\,b\,\sqrt {b\,x^2+a}} \]