Integrand size = 22, antiderivative size = 72 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {A b-a B}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {A}{a^2 \sqrt {a+b x^2}}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \] Output:
1/3*(A*b-B*a)/a/b/(b*x^2+a)^(3/2)+A/a^2/(b*x^2+a)^(1/2)-A*arctanh((b*x^2+a )^(1/2)/a^(1/2))/a^(5/2)
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {4 a A b-a^2 B+3 A b^2 x^2}{3 a^2 b \left (a+b x^2\right )^{3/2}}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \] Input:
Integrate[(A + B*x^2)/(x*(a + b*x^2)^(5/2)),x]
Output:
(4*a*A*b - a^2*B + 3*A*b^2*x^2)/(3*a^2*b*(a + b*x^2)^(3/2)) - (A*ArcTanh[S qrt[a + b*x^2]/Sqrt[a]])/a^(5/2)
Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^2 \left (b x^2+a\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (\frac {A \int \frac {1}{x^2 \left (b x^2+a\right )^{3/2}}dx^2}{a}+\frac {2 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {A \left (\frac {\int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{a}+\frac {2}{a \sqrt {a+b x^2}}\right )}{a}+\frac {2 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {A \left (\frac {2 \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a b}+\frac {2}{a \sqrt {a+b x^2}}\right )}{a}+\frac {2 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {A \left (\frac {2}{a \sqrt {a+b x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{a}+\frac {2 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}\right )\) |
Input:
Int[(A + B*x^2)/(x*(a + b*x^2)^(5/2)),x]
Output:
((2*(A*b - a*B))/(3*a*b*(a + b*x^2)^(3/2)) + (A*(2/(a*Sqrt[a + b*x^2]) - ( 2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)))/a)/2
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 = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(-\frac {3 A b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) a^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}+a^{\frac {5}{2}} \left (-3 A \,b^{2} x^{2}-4 a b A +a^{2} B \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b \,a^{\frac {9}{2}}}\) | \(73\) |
default | \(A \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )-\frac {B}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\) | \(80\) |
Input:
int((B*x^2+A)/x/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/3*(3*A*b*arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^2*(b*x^2+a)^(3/2)+a^(5/2)*( -3*A*b^2*x^2-4*A*a*b+B*a^2))/(b*x^2+a)^(3/2)/b/a^(9/2)
Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.39 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}, \frac {3 \, {\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}\right ] \] Input:
integrate((B*x^2+A)/x/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[1/6*(3*(A*b^3*x^4 + 2*A*a*b^2*x^2 + A*a^2*b)*sqrt(a)*log(-(b*x^2 - 2*sqrt (b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*A*a*b^2*x^2 - B*a^3 + 4*A*a^2*b)*sq rt(b*x^2 + a))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b), 1/3*(3*(A*b^3*x^4 + 2*A*a*b^2*x^2 + A*a^2*b)*sqrt(-a)*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (3* A*a*b^2*x^2 - B*a^3 + 4*A*a^2*b)*sqrt(b*x^2 + a))/(a^3*b^3*x^4 + 2*a^4*b^2 *x^2 + a^5*b)]
Time = 9.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {A b}{2 a^{2} \sqrt {a + b x^{2}}} + \frac {A b \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{2 a^{2} \sqrt {- a}} - \frac {- A b + B a}{6 a \left (a + b x^{2}\right )^{\frac {3}{2}}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {A \log {\left (B x^{2} \right )} + B x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((B*x**2+A)/x/(b*x**2+a)**(5/2),x)
Output:
Piecewise((2*(A*b/(2*a**2*sqrt(a + b*x**2)) + A*b*atan(sqrt(a + b*x**2)/sq rt(-a))/(2*a**2*sqrt(-a)) - (-A*b + B*a)/(6*a*(a + b*x**2)**(3/2)))/b, Ne( b, 0)), ((A*log(B*x**2) + B*x**2)/(2*a**(5/2)), True))
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=-\frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {A}{\sqrt {b x^{2} + a} a^{2}} + \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \] Input:
integrate((B*x^2+A)/x/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-A*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + A/(sqrt(b*x^2 + a)*a^2) + 1/3*A /((b*x^2 + a)^(3/2)*a) - 1/3*B/((b*x^2 + a)^(3/2)*b)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {A \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {B a^{2} - 3 \, {\left (b x^{2} + a\right )} A b - A a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} \] Input:
integrate((B*x^2+A)/x/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
A*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) - 1/3*(B*a^2 - 3*(b*x^2 + a)*A*b - A*a*b)/((b*x^2 + a)^(3/2)*a^2*b)
Time = 0.94 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {\frac {A}{3\,a}+\frac {A\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {B}{3\,b\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}} \] Input:
int((A + B*x^2)/(x*(a + b*x^2)^(5/2)),x)
Output:
(A/(3*a) + (A*(a + b*x^2))/a^2)/(a + b*x^2)^(3/2) - B/(3*b*(a + b*x^2)^(3/ 2)) - (A*atanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(5/2)
Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.89 \[ \int \frac {A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {b \,x^{2}+a}\, a +\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a +\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,x^{2}-\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a -\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,x^{2}}{a^{2} \left (b \,x^{2}+a \right )} \] Input:
int((B*x^2+A)/x/(b*x^2+a)^(5/2),x)
Output:
(sqrt(a + b*x**2)*a + sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x) /sqrt(a))*a + sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a) )*b*x**2 - sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a - sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b*x**2)/( a**2*(a + b*x**2))