Integrand size = 24, antiderivative size = 111 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=a^2 c \sqrt {c+d x^2}+\frac {1}{3} a^2 \left (c+d x^2\right )^{3/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^2}-a^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
1/3*a^2*(d*x^2+c)^(3/2)-1/5*b*(-2*a*d+b*c)*(d*x^2+c)^(5/2)/d^2+1/7*b^2*(d* x^2+c)^(7/2)/d^2-a^2*c^(3/2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))+a^2*c*(d*x^2 +c)^(1/2)
Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (42 a b d \left (c+d x^2\right )^2-3 b^2 \left (2 c-5 d x^2\right ) \left (c+d x^2\right )^2+35 a^2 d^2 \left (4 c+d x^2\right )\right )}{105 d^2}-a^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
(Sqrt[c + d*x^2]*(42*a*b*d*(c + d*x^2)^2 - 3*b^2*(2*c - 5*d*x^2)*(c + d*x^ 2)^2 + 35*a^2*d^2*(4*c + d*x^2)))/(105*d^2) - a^2*c^(3/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]
Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {354, 99, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2 \left (d x^2+c\right )^{3/2}}{x^2}dx^2\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{2} \int \left (\frac {b^2 \left (d x^2+c\right )^{5/2}}{d}-\frac {b (b c-2 a d) \left (d x^2+c\right )^{3/2}}{d}+\frac {a^2 \left (d x^2+c\right )^{3/2}}{x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-2 a^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {2}{3} a^2 \left (c+d x^2\right )^{3/2}+2 a^2 c \sqrt {c+d x^2}-\frac {2 b \left (c+d x^2\right )^{5/2} (b c-2 a d)}{5 d^2}+\frac {2 b^2 \left (c+d x^2\right )^{7/2}}{7 d^2}\right )\) |
(2*a^2*c*Sqrt[c + d*x^2] + (2*a^2*(c + d*x^2)^(3/2))/3 - (2*b*(b*c - 2*a*d )*(c + d*x^2)^(5/2))/(5*d^2) + (2*b^2*(c + d*x^2)^(7/2))/(7*d^2) - 2*a^2*c ^(3/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2
3.7.18.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 2.90 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00
method | result | size |
default | \(b^{2} \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{7 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{35 d^{2}}\right )+a^{2} \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )+\frac {2 a b \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5 d}\) | \(111\) |
pseudoelliptic | \(\frac {-3 a^{2} c^{\frac {3}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+4 \left (\frac {x^{2} \left (\frac {3}{7} b^{2} x^{4}+\frac {6}{5} a b \,x^{2}+a^{2}\right ) d^{3}}{4}+c \left (\frac {6}{35} b^{2} x^{4}+\frac {3}{5} a b \,x^{2}+a^{2}\right ) d^{2}+\frac {3 b \,c^{2} \left (\frac {b \,x^{2}}{14}+a \right ) d}{10}-\frac {3 b^{2} c^{3}}{70}\right ) \sqrt {d \,x^{2}+c}}{3 d^{2}}\) | \(118\) |
b^2*(1/7*x^2*(d*x^2+c)^(5/2)/d-2/35*c/d^2*(d*x^2+c)^(5/2))+a^2*(1/3*(d*x^2 +c)^(3/2)+c*((d*x^2+c)^(1/2)-c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x) ))+2/5*a*b*(d*x^2+c)^(5/2)/d
Time = 0.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.54 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=\left [\frac {105 \, a^{2} c^{\frac {3}{2}} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (15 \, b^{2} d^{3} x^{6} - 6 \, b^{2} c^{3} + 42 \, a b c^{2} d + 140 \, a^{2} c d^{2} + 6 \, {\left (4 \, b^{2} c d^{2} + 7 \, a b d^{3}\right )} x^{4} + {\left (3 \, b^{2} c^{2} d + 84 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{210 \, d^{2}}, \frac {105 \, a^{2} \sqrt {-c} c d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (15 \, b^{2} d^{3} x^{6} - 6 \, b^{2} c^{3} + 42 \, a b c^{2} d + 140 \, a^{2} c d^{2} + 6 \, {\left (4 \, b^{2} c d^{2} + 7 \, a b d^{3}\right )} x^{4} + {\left (3 \, b^{2} c^{2} d + 84 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{105 \, d^{2}}\right ] \]
[1/210*(105*a^2*c^(3/2)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c) /x^2) + 2*(15*b^2*d^3*x^6 - 6*b^2*c^3 + 42*a*b*c^2*d + 140*a^2*c*d^2 + 6*( 4*b^2*c*d^2 + 7*a*b*d^3)*x^4 + (3*b^2*c^2*d + 84*a*b*c*d^2 + 35*a^2*d^3)*x ^2)*sqrt(d*x^2 + c))/d^2, 1/105*(105*a^2*sqrt(-c)*c*d^2*arctan(sqrt(-c)/sq rt(d*x^2 + c)) + (15*b^2*d^3*x^6 - 6*b^2*c^3 + 42*a*b*c^2*d + 140*a^2*c*d^ 2 + 6*(4*b^2*c*d^2 + 7*a*b*d^3)*x^4 + (3*b^2*c^2*d + 84*a*b*c*d^2 + 35*a^2 *d^3)*x^2)*sqrt(d*x^2 + c))/d^2]
Time = 15.91 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=\frac {\begin {cases} \frac {2 a^{2} c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a^{2} c \sqrt {c + d x^{2}} + \frac {2 a^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}{3} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {5}{2}} \cdot \left (2 a b d - b^{2} c\right )}{5 d^{2}} & \text {for}\: d \neq 0 \\a^{2} c^{\frac {3}{2}} \log {\left (x^{2} \right )} + 2 a b c^{\frac {3}{2}} x^{2} + \frac {b^{2} c^{\frac {3}{2}} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
Piecewise((2*a**2*c**2*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + 2*a**2*c *sqrt(c + d*x**2) + 2*a**2*(c + d*x**2)**(3/2)/3 + 2*b**2*(c + d*x**2)**(7 /2)/(7*d**2) + 2*(c + d*x**2)**(5/2)*(2*a*b*d - b**2*c)/(5*d**2), Ne(d, 0) ), (a**2*c**(3/2)*log(x**2) + 2*a*b*c**(3/2)*x**2 + b**2*c**(3/2)*x**4/2, True))/2
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x^{2}}{7 \, d} - a^{2} c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} + \sqrt {d x^{2} + c} a^{2} c - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c}{35 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b}{5 \, d} \]
1/7*(d*x^2 + c)^(5/2)*b^2*x^2/d - a^2*c^(3/2)*arcsinh(c/(sqrt(c*d)*abs(x)) ) + 1/3*(d*x^2 + c)^(3/2)*a^2 + sqrt(d*x^2 + c)*a^2*c - 2/35*(d*x^2 + c)^( 5/2)*b^2*c/d^2 + 2/5*(d*x^2 + c)^(5/2)*a*b/d
Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx=\frac {a^{2} c^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {15 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} d^{12} - 21 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c d^{12} + 42 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d^{13} + 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{14} + 105 \, \sqrt {d x^{2} + c} a^{2} c d^{14}}{105 \, d^{14}} \]
a^2*c^2*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/105*(15*(d*x^2 + c)^ (7/2)*b^2*d^12 - 21*(d*x^2 + c)^(5/2)*b^2*c*d^12 + 42*(d*x^2 + c)^(5/2)*a* b*d^13 + 35*(d*x^2 + c)^(3/2)*a^2*d^14 + 105*sqrt(d*x^2 + c)*a^2*c*d^14)/d ^14
Time = 5.30 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x} \, dx={\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{3\,d^2}-\frac {c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )}{3}\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{5\,d^2}-\frac {b^2\,c}{5\,d^2}\right )\,{\left (d\,x^2+c\right )}^{5/2}+\frac {b^2\,{\left (d\,x^2+c\right )}^{7/2}}{7\,d^2}+c\,\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )+a^2\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \]
(c + d*x^2)^(3/2)*((a*d - b*c)^2/(3*d^2) - (c*((2*b^2*c - 2*a*b*d)/d^2 - ( b^2*c)/d^2))/3) - ((2*b^2*c - 2*a*b*d)/(5*d^2) - (b^2*c)/(5*d^2))*(c + d*x ^2)^(5/2) + a^2*c^(3/2)*atan(((c + d*x^2)^(1/2)*1i)/c^(1/2))*1i + (b^2*(c + d*x^2)^(7/2))/(7*d^2) + c*(c + d*x^2)^(1/2)*((a*d - b*c)^2/d^2 - c*((2*b ^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2))