Integrand size = 19, antiderivative size = 91 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {(A b-a B) x}{5 a b \left (a+b x^2\right )^{5/2}}+\frac {(4 A b+a B) x}{15 a^2 b \left (a+b x^2\right )^{3/2}}+\frac {2 (4 A b+a B) x}{15 a^3 b \sqrt {a+b x^2}} \] Output:
1/5*(A*b-B*a)*x/a/b/(b*x^2+a)^(5/2)+1/15*(4*A*b+B*a)*x/a^2/b/(b*x^2+a)^(3/ 2)+2/15*(4*A*b+B*a)*x/a^3/b/(b*x^2+a)^(1/2)
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {15 a^2 A x+20 a A b x^3+5 a^2 B x^3+8 A b^2 x^5+2 a b B x^5}{15 a^3 \left (a+b x^2\right )^{5/2}} \] Input:
Integrate[(A + B*x^2)/(a + b*x^2)^(7/2),x]
Output:
(15*a^2*A*x + 20*a*A*b*x^3 + 5*a^2*B*x^3 + 8*A*b^2*x^5 + 2*a*b*B*x^5)/(15* a^3*(a + b*x^2)^(5/2))
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {298, 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.
\(\displaystyle \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {(a B+4 A b) \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a b}+\frac {x (A b-a B)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(a B+4 A b) \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a b}+\frac {x (A b-a B)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right ) (a B+4 A b)}{5 a b}+\frac {x (A b-a B)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
Input:
Int[(A + B*x^2)/(a + b*x^2)^(7/2),x]
Output:
((A*b - a*B)*x)/(5*a*b*(a + b*x^2)^(5/2)) + ((4*A*b + a*B)*(x/(3*a*(a + b* x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/(5*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 = 0.51 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {x^{2} B}{3}+A \right ) a^{2}+\frac {4 b \left (\frac {x^{2} B}{10}+A \right ) x^{2} a}{3}+\frac {8 A \,b^{2} x^{4}}{15}\right ) x}{\left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\) | \(52\) |
gosper | \(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\) | \(57\) |
trager | \(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\) | \(57\) |
orering | \(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\) | \(57\) |
default | \(A \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )+B \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )\) | \(132\) |
Input:
int((B*x^2+A)/(b*x^2+a)^(7/2),x,method=_RETURNVERBOSE)
Output:
((1/3*x^2*B+A)*a^2+4/3*b*(1/10*x^2*B+A)*x^2*a+8/15*A*b^2*x^4)/(b*x^2+a)^(5 /2)*x/a^3
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {{\left (2 \, {\left (B a b + 4 \, A b^{2}\right )} x^{5} + 15 \, A a^{2} x + 5 \, {\left (B a^{2} + 4 \, A a b\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="fricas")
Output:
1/15*(2*(B*a*b + 4*A*b^2)*x^5 + 15*A*a^2*x + 5*(B*a^2 + 4*A*a*b)*x^3)*sqrt (b*x^2 + a)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)
Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (83) = 166\).
Time = 9.89 (sec) , antiderivative size = 566, normalized size of antiderivative = 6.22 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=A \left (\frac {15 a^{5} x}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 a^{4} b x^{3}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {28 a^{3} b^{2} x^{5}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {8 a^{2} b^{3} x^{7}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\frac {5 a x^{3}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 30 a^{\frac {7}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{5}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 30 a^{\frac {7}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:
integrate((B*x**2+A)/(b*x**2+a)**(7/2),x)
Output:
A*(15*a**5*x/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b* *3*x**6*sqrt(1 + b*x**2/a)) + 35*a**4*b*x**3/(15*a**(17/2)*sqrt(1 + b*x**2 /a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt (1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a)) + 28*a**3*b**2 *x**5/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x* *2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6 *sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**7/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a))) + B*(5*a*x**3/(15* a**(9/2)*sqrt(1 + b*x**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a **(5/2)*b**2*x**4*sqrt(1 + b*x**2/a)) + 2*b*x**5/(15*a**(9/2)*sqrt(1 + b*x **2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a**(5/2)*b**2*x**4*sqr t(1 + b*x**2/a)))
Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {8 \, A x}{15 \, \sqrt {b x^{2} + a} a^{3}} + \frac {4 \, A x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {A x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a} - \frac {B x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {2 \, B x}{15 \, \sqrt {b x^{2} + a} a^{2} b} + \frac {B x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="maxima")
Output:
8/15*A*x/(sqrt(b*x^2 + a)*a^3) + 4/15*A*x/((b*x^2 + a)^(3/2)*a^2) + 1/5*A* x/((b*x^2 + a)^(5/2)*a) - 1/5*B*x/((b*x^2 + a)^(5/2)*b) + 2/15*B*x/(sqrt(b *x^2 + a)*a^2*b) + 1/15*B*x/((b*x^2 + a)^(3/2)*a*b)
Time = 0.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, {\left (B a b^{3} + 4 \, A b^{4}\right )} x^{2}}{a^{3} b^{2}} + \frac {5 \, {\left (B a^{2} b^{2} + 4 \, A a b^{3}\right )}}{a^{3} b^{2}}\right )} + \frac {15 \, A}{a}\right )} x}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}}} \] Input:
integrate((B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="giac")
Output:
1/15*(x^2*(2*(B*a*b^3 + 4*A*b^4)*x^2/(a^3*b^2) + 5*(B*a^2*b^2 + 4*A*a*b^3) /(a^3*b^2)) + 15*A/a)*x/(b*x^2 + a)^(5/2)
Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {8\,A\,b\,x\,{\left (b\,x^2+a\right )}^2-3\,B\,a^3\,x+2\,B\,a\,x\,{\left (b\,x^2+a\right )}^2+B\,a^2\,x\,\left (b\,x^2+a\right )+3\,A\,a^2\,b\,x+4\,A\,a\,b\,x\,\left (b\,x^2+a\right )}{15\,a^3\,b\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:
int((A + B*x^2)/(a + b*x^2)^(7/2),x)
Output:
(8*A*b*x*(a + b*x^2)^2 - 3*B*a^3*x + 2*B*a*x*(a + b*x^2)^2 + B*a^2*x*(a + b*x^2) + 3*A*a^2*b*x + 4*A*a*b*x*(a + b*x^2))/(15*a^3*b*(a + b*x^2)^(5/2))
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {3 \sqrt {b \,x^{2}+a}\, a b x +2 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}-2 \sqrt {b}\, a^{2}-4 \sqrt {b}\, a b \,x^{2}-2 \sqrt {b}\, b^{2} x^{4}}{3 a^{2} b \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int((B*x^2+A)/(b*x^2+a)^(7/2),x)
Output:
(3*sqrt(a + b*x**2)*a*b*x + 2*sqrt(a + b*x**2)*b**2*x**3 - 2*sqrt(b)*a**2 - 4*sqrt(b)*a*b*x**2 - 2*sqrt(b)*b**2*x**4)/(3*a**2*b*(a**2 + 2*a*b*x**2 + b**2*x**4))