Integrand size = 29, antiderivative size = 84 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=A c \sqrt {a+b x^2}-\frac {\left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (B c+A d)-3 b B d x^2\right )}{15 b^2}-\sqrt {a} A c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
-1/15*(b*x^2+a)^(3/2)*(2*B*a*d-5*b*(A*d+B*c)-3*b*B*d*x^2)/b^2-A*c*arctanh( (b*x^2+a)^(1/2)/a^(1/2))*a^(1/2)+A*c*(b*x^2+a)^(1/2)
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\frac {\sqrt {a+b x^2} \left (-B \left (a+b x^2\right ) \left (-5 b c+2 a d-3 b d x^2\right )+5 A b \left (3 b c+a d+b d x^2\right )\right )}{15 b^2}-\sqrt {a} A c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(-(B*(a + b*x^2)*(-5*b*c + 2*a*d - 3*b*d*x^2)) + 5*A*b*(3 *b*c + a*d + b*d*x^2)))/(15*b^2) - Sqrt[a]*A*c*ArcTanh[Sqrt[a + b*x^2]/Sqr t[a]]
Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {435, 164, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx\) |
\(\Big \downarrow \) 435 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {b x^2+a} \left (B x^2+A\right ) \left (d x^2+c\right )}{x^2}dx^2\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {1}{2} \left (A c \int \frac {\sqrt {b x^2+a}}{x^2}dx^2-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (A d+B c)-3 b B d x^2\right )}{15 b^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (A c \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (A d+B c)-3 b B d x^2\right )}{15 b^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (A c \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (A d+B c)-3 b B d x^2\right )}{15 b^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (A c \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (A d+B c)-3 b B d x^2\right )}{15 b^2}\right )\) |
((-2*(a + b*x^2)^(3/2)*(2*a*B*d - 5*b*(B*c + A*d) - 3*b*B*d*x^2))/(15*b^2) + A*c*(2*Sqrt[a + b*x^2] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))/2
3.1.50.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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_.)*(( e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2) *(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IntegerQ[(m - 1)/2]
Time = 3.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {-3 A \sqrt {a}\, b^{2} c \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\left (\left (x^{2} \left (\frac {3 d \,x^{2}}{5}+c \right ) B +3 \left (\frac {d \,x^{2}}{3}+c \right ) A \right ) b^{2}+\left (\left (\frac {d \,x^{2}}{5}+c \right ) B +A d \right ) a b -\frac {2 B \,a^{2} d}{5}\right ) \sqrt {b \,x^{2}+a}}{3 b^{2}}\) | \(95\) |
default | \(B d \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+\frac {A d \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+\frac {B c \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+A c \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\) | \(111\) |
1/3*(-3*A*a^(1/2)*b^2*c*arctanh((b*x^2+a)^(1/2)/a^(1/2))+((x^2*(3/5*d*x^2+ c)*B+3*(1/3*d*x^2+c)*A)*b^2+((1/5*d*x^2+c)*B+A*d)*a*b-2/5*B*a^2*d)*(b*x^2+ a)^(1/2))/b^2
Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\left [\frac {15 \, A \sqrt {a} b^{2} c \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, B b^{2} d x^{4} + {\left (5 \, B b^{2} c + {\left (B a b + 5 \, A b^{2}\right )} d\right )} x^{2} + 5 \, {\left (B a b + 3 \, A b^{2}\right )} c - {\left (2 \, B a^{2} - 5 \, A a b\right )} d\right )} \sqrt {b x^{2} + a}}{30 \, b^{2}}, \frac {15 \, A \sqrt {-a} b^{2} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, B b^{2} d x^{4} + {\left (5 \, B b^{2} c + {\left (B a b + 5 \, A b^{2}\right )} d\right )} x^{2} + 5 \, {\left (B a b + 3 \, A b^{2}\right )} c - {\left (2 \, B a^{2} - 5 \, A a b\right )} d\right )} \sqrt {b x^{2} + a}}{15 \, b^{2}}\right ] \]
[1/30*(15*A*sqrt(a)*b^2*c*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x ^2) + 2*(3*B*b^2*d*x^4 + (5*B*b^2*c + (B*a*b + 5*A*b^2)*d)*x^2 + 5*(B*a*b + 3*A*b^2)*c - (2*B*a^2 - 5*A*a*b)*d)*sqrt(b*x^2 + a))/b^2, 1/15*(15*A*sqr t(-a)*b^2*c*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*B*b^2*d*x^4 + (5*B*b^2*c + (B*a*b + 5*A*b^2)*d)*x^2 + 5*(B*a*b + 3*A*b^2)*c - (2*B*a^2 - 5*A*a*b)* d)*sqrt(b*x^2 + a))/b^2]
Time = 9.66 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\frac {\begin {cases} \frac {2 A a c \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A c \sqrt {a + b x^{2}} + \frac {2 B d \left (a + b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {2 \left (a + b x^{2}\right )^{\frac {3}{2}} \left (A b d - B a d + B b c\right )}{3 b^{2}} & \text {for}\: b \neq 0 \\A \sqrt {a} c \log {\left (x^{2} \right )} + A \sqrt {a} d x^{2} + B \sqrt {a} c x^{2} + \frac {B \sqrt {a} d x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
Piecewise((2*A*a*c*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + 2*A*c*sqrt(a + b*x**2) + 2*B*d*(a + b*x**2)**(5/2)/(5*b**2) + 2*(a + b*x**2)**(3/2)*(A *b*d - B*a*d + B*b*c)/(3*b**2), Ne(b, 0)), (A*sqrt(a)*c*log(x**2) + A*sqrt (a)*d*x**2 + B*sqrt(a)*c*x**2 + B*sqrt(a)*d*x**4/2, True))/2
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B d x^{2}}{5 \, b} - A \sqrt {a} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \sqrt {b x^{2} + a} A c + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c}{3 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a d}{15 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A d}{3 \, b} \]
1/5*(b*x^2 + a)^(3/2)*B*d*x^2/b - A*sqrt(a)*c*arcsinh(a/(sqrt(a*b)*abs(x)) ) + sqrt(b*x^2 + a)*A*c + 1/3*(b*x^2 + a)^(3/2)*B*c/b - 2/15*(b*x^2 + a)^( 3/2)*B*a*d/b^2 + 1/3*(b*x^2 + a)^(3/2)*A*d/b
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\frac {A a c \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{9} c + 15 \, \sqrt {b x^{2} + a} A b^{10} c + 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{8} d - 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{8} d + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{9} d}{15 \, b^{10}} \]
A*a*c*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/15*(5*(b*x^2 + a)^(3/2 )*B*b^9*c + 15*sqrt(b*x^2 + a)*A*b^10*c + 3*(b*x^2 + a)^(5/2)*B*b^8*d - 5* (b*x^2 + a)^(3/2)*B*a*b^8*d + 5*(b*x^2 + a)^(3/2)*A*b^9*d)/b^10
Time = 5.99 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right ) \left (c+d x^2\right )}{x} \, dx=\sqrt {b\,x^2+a}\,\left (\frac {B\,d\,x^4}{5}-\frac {B\,a\,\left (2\,a\,d-5\,b\,c\right )}{15\,b^2}+\frac {B\,x^2\,\left (a\,d+5\,b\,c\right )}{15\,b}\right )+A\,c\,\sqrt {b\,x^2+a}+\frac {A\,d\,{\left (b\,x^2+a\right )}^{3/2}}{3\,b}-A\,\sqrt {a}\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right ) \]