Integrand size = 29, antiderivative size = 84 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}-a A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
-1/15*(d*x^2+c)^(3/2)*(2*B*b*c-5*(A*b+B*a)*d-3*b*B*d*x^2)/d^2-a*A*arctanh( (d*x^2+c)^(1/2)/c^(1/2))*c^(1/2)+a*A*(d*x^2+c)^(1/2)
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (-b \left (c+d x^2\right ) \left (2 B c-5 A d-3 B d x^2\right )+5 a d \left (3 A d+B \left (c+d x^2\right )\right )\right )}{15 d^2}-a A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
(Sqrt[c + d*x^2]*(-(b*(c + d*x^2)*(2*B*c - 5*A*d - 3*B*d*x^2)) + 5*a*d*(3* A*d + B*(c + d*x^2))))/(15*d^2) - a*A*Sqrt[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt [c]]
Time = 0.22 (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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx\) |
\(\Big \downarrow \) 435 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right ) \left (B x^2+A\right ) \sqrt {d x^2+c}}{x^2}dx^2\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {1}{2} \left (a A \int \frac {\sqrt {d x^2+c}}{x^2}dx^2-\frac {2 \left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (a A \left (c \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+2 \sqrt {c+d x^2}\right )-\frac {2 \left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (a A \left (\frac {2 c \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{d}+2 \sqrt {c+d x^2}\right )-\frac {2 \left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (a A \left (2 \sqrt {c+d x^2}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )-\frac {2 \left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}\right )\) |
((-2*(c + d*x^2)^(3/2)*(2*b*B*c - 5*(A*b + a*B)*d - 3*b*B*d*x^2))/(15*d^2) + a*A*(2*Sqrt[c + d*x^2] - 2*Sqrt[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]))/2
3.1.51.3.1 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 = 3.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {-A a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\sqrt {d \,x^{2}+c}\, \left (\left (\frac {\left (\frac {3 b \,x^{2}}{5}+a \right ) x^{2} B}{3}+A \left (\frac {b \,x^{2}}{3}+a \right )\right ) d^{2}+\frac {c \left (\left (\frac {b \,x^{2}}{5}+a \right ) B +A b \right ) d}{3}-\frac {2 B b \,c^{2}}{15}\right )}{d^{2}}\) | \(95\) |
default | \(B b \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )+\frac {A b \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}+\frac {B a \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}+A a \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\) | \(111\) |
(-A*a*c^(1/2)*d^2*arctanh((d*x^2+c)^(1/2)/c^(1/2))+(d*x^2+c)^(1/2)*((1/3*( 3/5*b*x^2+a)*x^2*B+A*(1/3*b*x^2+a))*d^2+1/3*c*((1/5*b*x^2+a)*B+A*b)*d-2/15 *B*b*c^2))/d^2
Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.58 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\left [\frac {15 \, A a \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d^{2}}, \frac {15 \, A a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, d^{2}}\right ] \]
[1/30*(15*A*a*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x ^2) + 2*(3*B*b*d^2*x^4 - 2*B*b*c^2 + 15*A*a*d^2 + 5*(B*a + A*b)*c*d + (B*b *c*d + 5*(B*a + A*b)*d^2)*x^2)*sqrt(d*x^2 + c))/d^2, 1/15*(15*A*a*sqrt(-c) *d^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (3*B*b*d^2*x^4 - 2*B*b*c^2 + 15*A* a*d^2 + 5*(B*a + A*b)*c*d + (B*b*c*d + 5*(B*a + A*b)*d^2)*x^2)*sqrt(d*x^2 + c))/d^2]
Time = 9.67 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {\begin {cases} \frac {2 A a c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 A a \sqrt {c + d x^{2}} + \frac {2 B b \left (c + d x^{2}\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {3}{2}} \left (A b d + B a d - B b c\right )}{3 d^{2}} & \text {for}\: d \neq 0 \\A a \sqrt {c} \log {\left (x^{2} \right )} + A b \sqrt {c} x^{2} + B a \sqrt {c} x^{2} + \frac {B b \sqrt {c} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
Piecewise((2*A*a*c*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + 2*A*a*sqrt(c + d*x**2) + 2*B*b*(c + d*x**2)**(5/2)/(5*d**2) + 2*(c + d*x**2)**(3/2)*(A *b*d + B*a*d - B*b*c)/(3*d**2), Ne(d, 0)), (A*a*sqrt(c)*log(x**2) + A*b*sq rt(c)*x**2 + B*a*sqrt(c)*x**2 + B*b*sqrt(c)*x**4/2, True))/2
Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B b x^{2}}{5 \, d} - A a \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \sqrt {d x^{2} + c} A a - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c}{15 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B a}{3 \, d} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} A b}{3 \, d} \]
1/5*(d*x^2 + c)^(3/2)*B*b*x^2/d - A*a*sqrt(c)*arcsinh(c/(sqrt(c*d)*abs(x)) ) + sqrt(d*x^2 + c)*A*a - 2/15*(d*x^2 + c)^(3/2)*B*b*c/d^2 + 1/3*(d*x^2 + c)^(3/2)*B*a/d + 1/3*(d*x^2 + c)^(3/2)*A*b/d
Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {A a c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} B b d^{8} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c d^{8} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B a d^{9} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} A b d^{9} + 15 \, \sqrt {d x^{2} + c} A a d^{10}}{15 \, d^{10}} \]
A*a*c*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/15*(3*(d*x^2 + c)^(5/2 )*B*b*d^8 - 5*(d*x^2 + c)^(3/2)*B*b*c*d^8 + 5*(d*x^2 + c)^(3/2)*B*a*d^9 + 5*(d*x^2 + c)^(3/2)*A*b*d^9 + 15*sqrt(d*x^2 + c)*A*a*d^10)/d^10
Time = 5.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {B\,b\,x^4}{5}+\frac {B\,c\,\left (5\,a\,d-2\,b\,c\right )}{15\,d^2}+\frac {B\,x^2\,\left (5\,a\,d+b\,c\right )}{15\,d}\right )+A\,a\,\sqrt {d\,x^2+c}-A\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )+\frac {A\,b\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d} \]