Integrand size = 22, antiderivative size = 86 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {3 A b-2 a B}{2 a^2 \sqrt {a+b x^2}}-\frac {A}{2 a x^2 \sqrt {a+b x^2}}+\frac {(3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \]
1/2*(3*A*b-2*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+1/2*(-3*A*b+2*B *a)/a^2/(b*x^2+a)^(1/2)-1/2*A/a/x^2/(b*x^2+a)^(1/2)
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=\frac {-a A-3 A b x^2+2 a B x^2}{2 a^2 x^2 \sqrt {a+b x^2}}+\frac {(3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \]
(-(a*A) - 3*A*b*x^2 + 2*a*B*x^2)/(2*a^2*x^2*Sqrt[a + b*x^2]) + ((3*A*b - 2 *a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(5/2))
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {354, 87, 61, 73, 221}
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.
\(\displaystyle \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^4 \left (b x^2+a\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-2 a B) \int \frac {1}{x^2 \left (b x^2+a\right )^{3/2}}dx^2}{2 a}-\frac {A}{a x^2 \sqrt {a+b x^2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-2 a B) \left (\frac {\int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{a}+\frac {2}{a \sqrt {a+b x^2}}\right )}{2 a}-\frac {A}{a x^2 \sqrt {a+b x^2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-2 a B) \left (\frac {2 \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a b}+\frac {2}{a \sqrt {a+b x^2}}\right )}{2 a}-\frac {A}{a x^2 \sqrt {a+b x^2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-2 a B) \left (\frac {2}{a \sqrt {a+b x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {A}{a x^2 \sqrt {a+b x^2}}\right )\) |
(-(A/(a*x^2*Sqrt[a + b*x^2])) - ((3*A*b - 2*a*B)*(2/(a*Sqrt[a + b*x^2]) - (2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)))/(2*a))/2
3.6.78.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.98 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \sqrt {b \,x^{2}+a}\, x^{2} \left (A b -\frac {2 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{2}+\frac {\left (2 x^{2} B -A \right ) a^{\frac {3}{2}}}{2}-\frac {3 A \sqrt {a}\, b \,x^{2}}{2}}{x^{2} a^{\frac {5}{2}} \sqrt {b \,x^{2}+a}}\) | \(79\) |
risch | \(-\frac {A \sqrt {b \,x^{2}+a}}{2 a^{2} x^{2}}-\frac {-\frac {b A}{\sqrt {b \,x^{2}+a}}+a \left (3 A b -2 B a \right ) \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a^{2}}\) | \(92\) |
default | \(B \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )+A \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )\) | \(114\) |
3/2/(b*x^2+a)^(1/2)/a^(5/2)*((b*x^2+a)^(1/2)*x^2*(A*b-2/3*B*a)*arctanh((b* x^2+a)^(1/2)/a^(1/2))+1/3*(2*B*x^2-A)*a^(3/2)-A*a^(1/2)*b*x^2)/x^2
Time = 0.29 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.70 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, \frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \]
[-1/4*(((2*B*a*b - 3*A*b^2)*x^4 + (2*B*a^2 - 3*A*a*b)*x^2)*sqrt(a)*log(-(b *x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(A*a^2 - (2*B*a^2 - 3*A*a *b)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4 + a^4*x^2), 1/2*(((2*B*a*b - 3*A*b^2) *x^4 + (2*B*a^2 - 3*A*a*b)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (A*a^2 - (2*B*a^2 - 3*A*a*b)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4 + a^4*x^2) ]
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (73) = 146\).
Time = 14.78 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.05 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=A \left (- \frac {1}{2 a \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{2 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {5}{2}}}\right ) + B \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{3} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{2} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{2} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}}\right ) \]
A*(-1/(2*a*sqrt(b)*x**3*sqrt(a/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a /(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(5/2))) + B*(2*a**3 *sqrt(1 + b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**3*log(b*x**2/a)/ (2*a**(9/2) + 2*a**(7/2)*b*x**2) - 2*a**3*log(sqrt(1 + b*x**2/a) + 1)/(2*a **(9/2) + 2*a**(7/2)*b*x**2) + a**2*b*x**2*log(b*x**2/a)/(2*a**(9/2) + 2*a **(7/2)*b*x**2) - 2*a**2*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a} - \frac {3 \, A b}{2 \, \sqrt {b x^{2} + a} a^{2}} - \frac {A}{2 \, \sqrt {b x^{2} + a} a x^{2}} \]
-B*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) + 3/2*A*b*arcsinh(a/(sqrt(a*b)*ab s(x)))/a^(5/2) + B/(sqrt(b*x^2 + a)*a) - 3/2*A*b/(sqrt(b*x^2 + a)*a^2) - 1 /2*A/(sqrt(b*x^2 + a)*a*x^2)
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (2 \, B a - 3 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{2}} + \frac {2 \, {\left (b x^{2} + a\right )} B a - 2 \, B a^{2} - 3 \, {\left (b x^{2} + a\right )} A b + 2 \, A a b}{2 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{2} + a} a\right )} a^{2}} \]
1/2*(2*B*a - 3*A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + 1/2* (2*(b*x^2 + a)*B*a - 2*B*a^2 - 3*(b*x^2 + a)*A*b + 2*A*a*b)/(((b*x^2 + a)^ (3/2) - sqrt(b*x^2 + a)*a)*a^2)
Time = 6.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{3/2}} \, dx=\frac {B}{a\,\sqrt {b\,x^2+a}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {3\,A\,b}{2\,a^2\,\sqrt {b\,x^2+a}}-\frac {A}{2\,a\,x^2\,\sqrt {b\,x^2+a}}+\frac {3\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}} \]