Integrand size = 22, antiderivative size = 97 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=-\frac {a B-(A c-a C) x}{5 a c \left (a+c x^2\right )^{5/2}}+\frac {(4 A c+a C) x}{15 a^2 c \left (a+c x^2\right )^{3/2}}+\frac {2 (4 A c+a C) x}{15 a^3 c \sqrt {a+c x^2}} \]
1/5*(-a*B+(A*c-C*a)*x)/a/c/(c*x^2+a)^(5/2)+1/15*(4*A*c+C*a)*x/a^2/c/(c*x^2 +a)^(3/2)+2/15*(4*A*c+C*a)*x/a^3/c/(c*x^2+a)^(1/2)
Time = 0.52 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {-3 a^3 B+8 A c^3 x^5+5 a^2 c x \left (3 A+C x^2\right )+2 a c^2 x^3 \left (10 A+C x^2\right )}{15 a^3 c \left (a+c x^2\right )^{5/2}} \]
(-3*a^3*B + 8*A*c^3*x^5 + 5*a^2*c*x*(3*A + C*x^2) + 2*a*c^2*x^3*(10*A + C* x^2))/(15*a^3*c*(a + c*x^2)^(5/2))
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2345, 25, 27, 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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {4 A c+a C}{c \left (c x^2+a\right )^{5/2}}dx}{5 a}-\frac {a B-x (A c-a C)}{5 a c \left (a+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 A c+a C}{c \left (c x^2+a\right )^{5/2}}dx}{5 a}-\frac {a B-x (A c-a C)}{5 a c \left (a+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a C+4 A c) \int \frac {1}{\left (c x^2+a\right )^{5/2}}dx}{5 a c}-\frac {a B-x (A c-a C)}{5 a c \left (a+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a C+4 A c) \left (\frac {2 \int \frac {1}{\left (c x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+c x^2\right )^{3/2}}\right )}{5 a c}-\frac {a B-x (A c-a C)}{5 a c \left (a+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\left (\frac {2 x}{3 a^2 \sqrt {a+c x^2}}+\frac {x}{3 a \left (a+c x^2\right )^{3/2}}\right ) (a C+4 A c)}{5 a c}-\frac {a B-x (A c-a C)}{5 a c \left (a+c x^2\right )^{5/2}}\) |
-1/5*(a*B - (A*c - a*C)*x)/(a*c*(a + c*x^2)^(5/2)) + ((4*A*c + a*C)*(x/(3* a*(a + c*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + c*x^2])))/(5*a*c)
3.2.16.3.1 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 = 0.58 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74
method | result | size |
gosper | \(\frac {8 A \,c^{3} x^{5}+2 a \,c^{2} C \,x^{5}+20 a A \,c^{2} x^{3}+5 C \,a^{2} c \,x^{3}+15 a^{2} A c x -3 B \,a^{3}}{15 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a^{3} c}\) | \(72\) |
trager | \(\frac {8 A \,c^{3} x^{5}+2 a \,c^{2} C \,x^{5}+20 a A \,c^{2} x^{3}+5 C \,a^{2} c \,x^{3}+15 a^{2} A c x -3 B \,a^{3}}{15 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a^{3} c}\) | \(72\) |
default | \(A \left (\frac {x}{5 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {c \,x^{2}+a}}}{a}\right )+C \left (-\frac {x}{4 c \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {c \,x^{2}+a}}}{a}\right )}{4 c}\right )-\frac {B}{5 c \left (c \,x^{2}+a \right )^{\frac {5}{2}}}\) | \(147\) |
1/15*(8*A*c^3*x^5+2*C*a*c^2*x^5+20*A*a*c^2*x^3+5*C*a^2*c*x^3+15*A*a^2*c*x- 3*B*a^3)/(c*x^2+a)^(5/2)/a^3/c
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {{\left (2 \, {\left (C a c^{2} + 4 \, A c^{3}\right )} x^{5} + 15 \, A a^{2} c x - 3 \, B a^{3} + 5 \, {\left (C a^{2} c + 4 \, A a c^{2}\right )} x^{3}\right )} \sqrt {c x^{2} + a}}{15 \, {\left (a^{3} c^{4} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{5} c^{2} x^{2} + a^{6} c\right )}} \]
1/15*(2*(C*a*c^2 + 4*A*c^3)*x^5 + 15*A*a^2*c*x - 3*B*a^3 + 5*(C*a^2*c + 4* A*a*c^2)*x^3)*sqrt(c*x^2 + a)/(a^3*c^4*x^6 + 3*a^4*c^3*x^4 + 3*a^5*c^2*x^2 + a^6*c)
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (87) = 174\).
Time = 12.06 (sec) , antiderivative size = 638, normalized size of antiderivative = 6.58 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=A \left (\frac {15 a^{5} x}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {35 a^{4} c x^{3}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {28 a^{3} c^{2} x^{5}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {8 a^{2} c^{3} x^{7}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {1}{5 a^{2} c \sqrt {a + c x^{2}} + 10 a c^{2} x^{2} \sqrt {a + c x^{2}} + 5 c^{3} x^{4} \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{2}}{2 a^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) + C \left (\frac {5 a x^{3}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 30 a^{\frac {7}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {5}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {2 c x^{5}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 30 a^{\frac {7}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {5}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}}}\right ) \]
A*(15*a**5*x/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 15*a**(11/2)*c* *3*x**6*sqrt(1 + c*x**2/a)) + 35*a**4*c*x**3/(15*a**(17/2)*sqrt(1 + c*x**2 /a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*sqrt (1 + c*x**2/a) + 15*a**(11/2)*c**3*x**6*sqrt(1 + c*x**2/a)) + 28*a**3*c**2 *x**5/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x* *2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 15*a**(11/2)*c**3*x**6 *sqrt(1 + c*x**2/a)) + 8*a**2*c**3*x**7/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 15*a**(11/2)*c**3*x**6*sqrt(1 + c*x**2/a))) + B*Piecewise((-1/ (5*a**2*c*sqrt(a + c*x**2) + 10*a*c**2*x**2*sqrt(a + c*x**2) + 5*c**3*x**4 *sqrt(a + c*x**2)), Ne(c, 0)), (x**2/(2*a**(7/2)), True)) + C*(5*a*x**3/(1 5*a**(9/2)*sqrt(1 + c*x**2/a) + 30*a**(7/2)*c*x**2*sqrt(1 + c*x**2/a) + 15 *a**(5/2)*c**2*x**4*sqrt(1 + c*x**2/a)) + 2*c*x**5/(15*a**(9/2)*sqrt(1 + c *x**2/a) + 30*a**(7/2)*c*x**2*sqrt(1 + c*x**2/a) + 15*a**(5/2)*c**2*x**4*s qrt(1 + c*x**2/a)))
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {8 \, A x}{15 \, \sqrt {c x^{2} + a} a^{3}} + \frac {4 \, A x}{15 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {A x}{5 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a} - \frac {C x}{5 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} c} + \frac {2 \, C x}{15 \, \sqrt {c x^{2} + a} a^{2} c} + \frac {C x}{15 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a c} - \frac {B}{5 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} c} \]
8/15*A*x/(sqrt(c*x^2 + a)*a^3) + 4/15*A*x/((c*x^2 + a)^(3/2)*a^2) + 1/5*A* x/((c*x^2 + a)^(5/2)*a) - 1/5*C*x/((c*x^2 + a)^(5/2)*c) + 2/15*C*x/(sqrt(c *x^2 + a)*a^2*c) + 1/15*C*x/((c*x^2 + a)^(3/2)*a*c) - 1/5*B/((c*x^2 + a)^( 5/2)*c)
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, {\left (C a c^{3} + 4 \, A c^{4}\right )} x^{2}}{a^{3} c^{2}} + \frac {5 \, {\left (C a^{2} c^{2} + 4 \, A a c^{3}\right )}}{a^{3} c^{2}}\right )} + \frac {15 \, A}{a}\right )} x - \frac {3 \, B}{c}}{15 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}}} \]
1/15*((x^2*(2*(C*a*c^3 + 4*A*c^4)*x^2/(a^3*c^2) + 5*(C*a^2*c^2 + 4*A*a*c^3 )/(a^3*c^2)) + 15*A/a)*x - 3*B/c)/(c*x^2 + a)^(5/2)
Time = 12.96 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {8\,A\,c\,x\,{\left (c\,x^2+a\right )}^2-3\,C\,a^3\,x-3\,B\,a^3+2\,C\,a\,x\,{\left (c\,x^2+a\right )}^2+C\,a^2\,x\,\left (c\,x^2+a\right )+3\,A\,a^2\,c\,x+4\,A\,a\,c\,x\,\left (c\,x^2+a\right )}{15\,a^3\,c\,{\left (c\,x^2+a\right )}^{5/2}} \]