Integrand size = 22, antiderivative size = 77 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {A}{a x \left (a+b x^2\right )^{3/2}}-\frac {(4 A b-a B) x}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac {2 (4 A b-a B) x}{3 a^3 \sqrt {a+b x^2}} \] Output:
-A/a/x/(b*x^2+a)^(3/2)-1/3*(4*A*b-B*a)*x/a^2/(b*x^2+a)^(3/2)-2/3*(4*A*b-B* a)*x/a^3/(b*x^2+a)^(1/2)
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 a^2 A-12 a A b x^2+3 a^2 B x^2-8 A b^2 x^4+2 a b B x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \] Input:
Integrate[(A + B*x^2)/(x^2*(a + b*x^2)^(5/2)),x]
Output:
(-3*a^2*A - 12*a*A*b*x^2 + 3*a^2*B*x^2 - 8*A*b^2*x^4 + 2*a*b*B*x^4)/(3*a^3 *x*(a + b*x^2)^(3/2))
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {359, 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.
\(\displaystyle \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(4 A b-a B) \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{a}-\frac {A}{a x \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -\frac {(4 A b-a B) \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{a}-\frac {A}{a x \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle -\frac {\left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right ) (4 A b-a B)}{a}-\frac {A}{a x \left (a+b x^2\right )^{3/2}}\) |
Input:
Int[(A + B*x^2)/(x^2*(a + b*x^2)^(5/2)),x]
Output:
-(A/(a*x*(a + b*x^2)^(3/2))) - ((4*A*b - a*B)*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/a
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Time = 0.42 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(-\frac {\left (-x^{2} B +A \right ) a^{2}+4 \left (-\frac {x^{2} B}{6}+A \right ) b \,x^{2} a +\frac {8 A \,b^{2} x^{4}}{3}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} x \,a^{3}}\) | \(55\) |
gosper | \(-\frac {8 A \,b^{2} x^{4}-2 B a b \,x^{4}+12 a A b \,x^{2}-3 B \,a^{2} x^{2}+3 a^{2} A}{3 x \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}\) | \(59\) |
trager | \(-\frac {8 A \,b^{2} x^{4}-2 B a b \,x^{4}+12 a A b \,x^{2}-3 B \,a^{2} x^{2}+3 a^{2} A}{3 x \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}\) | \(59\) |
orering | \(-\frac {8 A \,b^{2} x^{4}-2 B a b \,x^{4}+12 a A b \,x^{2}-3 B \,a^{2} x^{2}+3 a^{2} A}{3 x \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}\) | \(59\) |
risch | \(-\frac {A \sqrt {b \,x^{2}+a}}{a^{3} x}-\frac {\sqrt {b \,x^{2}+a}\, x \left (5 A \,b^{2} x^{2}-2 B a b \,x^{2}+6 a b A -3 a^{2} B \right )}{3 a^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}\) | \(84\) |
default | \(B \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+A \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{a}\right )\) | \(92\) |
Input:
int((B*x^2+A)/x^2/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/(b*x^2+a)^(3/2)*((-B*x^2+A)*a^2+4*(-1/6*x^2*B+A)*b*x^2*a+8/3*A*b^2*x^4) /x/a^3
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, {\left (B a b - 4 \, A b^{2}\right )} x^{4} - 3 \, A a^{2} + 3 \, {\left (B a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} \] Input:
integrate((B*x^2+A)/x^2/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
1/3*(2*(B*a*b - 4*A*b^2)*x^4 - 3*A*a^2 + 3*(B*a^2 - 4*A*a*b)*x^2)*sqrt(b*x ^2 + a)/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x)
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (71) = 142\).
Time = 6.46 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.44 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=A \left (- \frac {3 a^{2} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {12 a b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {8 b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}}\right ) + B \left (\frac {3 a x}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{3}}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:
integrate((B*x**2+A)/x**2/(b*x**2+a)**(5/2),x)
Output:
A*(-3*a**2*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 12*a*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4) - 8*b**(13/2)*x**4*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4)) + B*(3*a*x/(3*a **(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)) + 2*b*x **3/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)) )
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {2 \, B x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {8 \, A b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, A b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} \] Input:
integrate((B*x^2+A)/x^2/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
2/3*B*x/(sqrt(b*x^2 + a)*a^2) + 1/3*B*x/((b*x^2 + a)^(3/2)*a) - 8/3*A*b*x/ (sqrt(b*x^2 + a)*a^3) - 4/3*A*b*x/((b*x^2 + a)^(3/2)*a^2) - A/((b*x^2 + a) ^(3/2)*a*x)
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {x {\left (\frac {{\left (2 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{2}}{a^{5} b} + \frac {3 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )}}{a^{5} b}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \] Input:
integrate((B*x^2+A)/x^2/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
1/3*x*((2*B*a^3*b^2 - 5*A*a^2*b^3)*x^2/(a^5*b) + 3*(B*a^4*b - 2*A*a^3*b^2) /(a^5*b))/(b*x^2 + a)^(3/2) + 2*A*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^ 2 - a)*a^2)
Time = 0.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {A\,a^2-8\,A\,{\left (b\,x^2+a\right )}^2+B\,a^2\,x^2+4\,A\,a\,\left (b\,x^2+a\right )+2\,B\,a\,x^2\,\left (b\,x^2+a\right )}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}} \] Input:
int((A + B*x^2)/(x^2*(a + b*x^2)^(5/2)),x)
Output:
(A*a^2 - 8*A*(a + b*x^2)^2 + B*a^2*x^2 + 4*A*a*(a + b*x^2) + 2*B*a*x^2*(a + b*x^2))/(3*a^3*x*(a + b*x^2)^(3/2))
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{2}+a}\, a -2 \sqrt {b \,x^{2}+a}\, b \,x^{2}-2 \sqrt {b}\, a x -2 \sqrt {b}\, b \,x^{3}}{a^{2} x \left (b \,x^{2}+a \right )} \] Input:
int((B*x^2+A)/x^2/(b*x^2+a)^(5/2),x)
Output:
( - sqrt(a + b*x**2)*a - 2*sqrt(a + b*x**2)*b*x**2 - 2*sqrt(b)*a*x - 2*sqr t(b)*b*x**3)/(a**2*x*(a + b*x**2))