Integrand size = 22, antiderivative size = 76 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=a A \sqrt {a+b x^2}+\frac {1}{3} A \left (a+b x^2\right )^{3/2}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b}-a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
1/3*A*(b*x^2+a)^(3/2)+1/5*B*(b*x^2+a)^(5/2)/b-a^(3/2)*A*arctanh((b*x^2+a)^ (1/2)/a^(1/2))+a*A*(b*x^2+a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=\frac {\sqrt {a+b x^2} \left (20 a A b+3 a^2 B+5 A b^2 x^2+6 a b B x^2+3 b^2 B x^4\right )}{15 b}-a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(20*a*A*b + 3*a^2*B + 5*A*b^2*x^2 + 6*a*b*B*x^2 + 3*b^2*B *x^4))/(15*b) - a^(3/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {354, 90, 60, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2} \left (B x^2+A\right )}{x^2}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (A \int \frac {\left (b x^2+a\right )^{3/2}}{x^2}dx^2+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (A \left (a \int \frac {\sqrt {b x^2+a}}{x^2}dx^2+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (A \left (a \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (A \left (a \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (A \left (a \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2 B \left (a+b x^2\right )^{5/2}}{5 b}\right )\) |
((2*B*(a + b*x^2)^(5/2))/(5*b) + A*((2*(a + b*x^2)^(3/2))/3 + a*(2*Sqrt[a + b*x^2] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])))/2
3.6.27.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 2.80 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {B \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+A \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\) | \(71\) |
pseudoelliptic | \(\frac {-3 a^{\frac {3}{2}} b A \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+4 \left (\frac {x^{2} \left (\frac {3 x^{2} B}{5}+A \right ) b^{2}}{4}+a \left (\frac {3 x^{2} B}{10}+A \right ) b +\frac {3 a^{2} B}{20}\right ) \sqrt {b \,x^{2}+a}}{3 b}\) | \(73\) |
1/5*B*(b*x^2+a)^(5/2)/b+A*(1/3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2)* ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)))
Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.24 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=\left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, B b^{2} x^{4} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, b}, \frac {15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, B b^{2} x^{4} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, b}\right ] \]
[1/30*(15*A*a^(3/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*B*b^2*x^4 + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x^2)*sqrt(b*x^ 2 + a))/b, 1/15*(15*A*sqrt(-a)*a*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*B *b^2*x^4 + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x^2)*sqrt(b*x^2 + a))/ b]
Time = 10.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=\frac {\begin {cases} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x^{2}} + \frac {2 A \left (a + b x^{2}\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x^{2}\right )^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\A a^{\frac {3}{2}} \log {\left (B a^{\frac {3}{2}} x^{2} \right )} + B a^{\frac {3}{2}} x^{2} & \text {otherwise} \end {cases}}{2} \]
Piecewise((2*A*a**2*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + 2*A*a*sqrt( a + b*x**2) + 2*A*(a + b*x**2)**(3/2)/3 + 2*B*(a + b*x**2)**(5/2)/(5*b), N e(b, 0)), (A*a**(3/2)*log(B*a**(3/2)*x**2) + B*a**(3/2)*x**2, True))/2
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=-A a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A + \sqrt {b x^{2} + a} A a + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, b} \]
-A*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a)^(3/2)*A + sqrt( b*x^2 + a)*A*a + 1/5*(b*x^2 + a)^(5/2)*B/b
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=\frac {A a^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{4} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{5} + 15 \, \sqrt {b x^{2} + a} A a b^{5}}{15 \, b^{5}} \]
A*a^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/15*(3*(b*x^2 + a)^(5/2 )*B*b^4 + 5*(b*x^2 + a)^(3/2)*A*b^5 + 15*sqrt(b*x^2 + a)*A*a*b^5)/b^5
Time = 5.50 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5\,b}-A\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+A\,a\,\sqrt {b\,x^2+a} \]