Integrand size = 19, antiderivative size = 192 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\frac {1}{3} a^3 c \sqrt {c \left (a+b x^2\right )^3}+\frac {a^4 c \sqrt {c \left (a+b x^2\right )^3}}{a+b x^2}+\frac {1}{5} a^2 c \left (a+b x^2\right ) \sqrt {c \left (a+b x^2\right )^3}+\frac {1}{7} a c \left (a+b x^2\right )^2 \sqrt {c \left (a+b x^2\right )^3}+\frac {1}{9} c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^3}-\frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \text {arctanh}\left (\sqrt {1+\frac {b x^2}{a}}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \]
1/3*a^3*c*(c*(b*x^2+a)^3)^(1/2)+a^4*c*(c*(b*x^2+a)^3)^(1/2)/(b*x^2+a)+1/5* a^2*c*(b*x^2+a)*(c*(b*x^2+a)^3)^(1/2)+1/7*a*c*(b*x^2+a)^2*(c*(b*x^2+a)^3)^ (1/2)+1/9*c*(b*x^2+a)^3*(c*(b*x^2+a)^3)^(1/2)-a^3*c*arctanh((1+b*x^2/a)^(1 /2))*(c*(b*x^2+a)^3)^(1/2)/(1+b*x^2/a)^(3/2)
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.58 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\frac {\left (c \left (a+b x^2\right )^3\right )^{3/2} \left (\sqrt {a+b x^2} \left (563 a^4+506 a^3 b x^2+408 a^2 b^2 x^4+185 a b^3 x^6+35 b^4 x^8\right )-315 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )}{315 \left (a+b x^2\right )^{9/2}} \]
((c*(a + b*x^2)^3)^(3/2)*(Sqrt[a + b*x^2]*(563*a^4 + 506*a^3*b*x^2 + 408*a ^2*b^2*x^4 + 185*a*b^3*x^6 + 35*b^4*x^8) - 315*a^(9/2)*ArcTanh[Sqrt[a + b* x^2]/Sqrt[a]]))/(315*(a + b*x^2)^(9/2))
Time = 0.25 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.74, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {2045, 243, 60, 60, 60, 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.
| \(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx\) |
| \(\Big \downarrow \) 2045 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}dx}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\int \frac {\left (\frac {b x^2}{a}+1\right )^{7/2}}{x^2}dx^2+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\int \frac {\left (\frac {b x^2}{a}+1\right )^{5/2}}{x^2}dx^2+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\int \frac {\left (\frac {b x^2}{a}+1\right )^{3/2}}{x^2}dx^2+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\int \frac {\sqrt {\frac {b x^2}{a}+1}}{x^2}dx^2+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\int \frac {1}{x^2 \sqrt {\frac {b x^2}{a}+1}}dx^2+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {2 a \int \frac {1}{\frac {a x^4}{b}-\frac {a}{b}}d\sqrt {\frac {b x^2}{a}+1}}{b}+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a^3 c \left (-2 \text {arctanh}\left (\sqrt {\frac {b x^2}{a}+1}\right )+\frac {2}{9} \left (\frac {b x^2}{a}+1\right )^{9/2}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right ) \sqrt {c \left (a+b x^2\right )^3}}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
(a^3*c*Sqrt[c*(a + b*x^2)^3]*(2*Sqrt[1 + (b*x^2)/a] + (2*(1 + (b*x^2)/a)^( 3/2))/3 + (2*(1 + (b*x^2)/a)^(5/2))/5 + (2*(1 + (b*x^2)/a)^(7/2))/7 + (2*( 1 + (b*x^2)/a)^(9/2))/9 - 2*ArcTanh[Sqrt[1 + (b*x^2)/a]]))/(2*(1 + (b*x^2) /a)^(3/2))
3.3.40.3.1 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_)^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_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Time = 1.50 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(-\frac {{\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}} \left (-35 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} b^{2} x^{4}-115 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a b \,x^{2}+315 \ln \left (\frac {2 a c +2 \sqrt {a c}\, \sqrt {b c \,x^{2}+a c}}{x}\right ) a^{5} c^{3}+46 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a^{2}-105 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {3}{2}} a^{3} c -315 \sqrt {a c}\, \sqrt {b c \,x^{2}+a c}\, a^{4} c^{2}-189 a^{2} {\left (c \left (b \,x^{2}+a \right )\right )}^{\frac {5}{2}} \sqrt {a c}\right )}{315 \left (b \,x^{2}+a \right )^{3} {\left (c \left (b \,x^{2}+a \right )\right )}^{\frac {3}{2}} c \sqrt {a c}}\) | \(221\) |
-1/315*(c*(b*x^2+a)^3)^(3/2)*(-35*(a*c)^(1/2)*(b*c*x^2+a*c)^(5/2)*b^2*x^4- 115*(a*c)^(1/2)*(b*c*x^2+a*c)^(5/2)*a*b*x^2+315*ln(2*((a*c)^(1/2)*(b*c*x^2 +a*c)^(1/2)+a*c)/x)*a^5*c^3+46*(a*c)^(1/2)*(b*c*x^2+a*c)^(5/2)*a^2-105*(a* c)^(1/2)*(b*c*x^2+a*c)^(3/2)*a^3*c-315*(a*c)^(1/2)*(b*c*x^2+a*c)^(1/2)*a^4 *c^2-189*a^2*(c*(b*x^2+a))^(5/2)*(a*c)^(1/2))/(b*x^2+a)^3/(c*(b*x^2+a))^(3 /2)/c/(a*c)^(1/2)
Time = 0.28 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.04 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\left [\frac {315 \, {\left (a^{4} b c x^{2} + a^{5} c\right )} \sqrt {a c} \log \left (-\frac {b^{2} c x^{4} + 3 \, a b c x^{2} + 2 \, a^{2} c - 2 \, \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {a c}}{b x^{4} + a x^{2}}\right ) + 2 \, {\left (35 \, b^{4} c x^{8} + 185 \, a b^{3} c x^{6} + 408 \, a^{2} b^{2} c x^{4} + 506 \, a^{3} b c x^{2} + 563 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{630 \, {\left (b x^{2} + a\right )}}, \frac {315 \, {\left (a^{4} b c x^{2} + a^{5} c\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {-a c}}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) + {\left (35 \, b^{4} c x^{8} + 185 \, a b^{3} c x^{6} + 408 \, a^{2} b^{2} c x^{4} + 506 \, a^{3} b c x^{2} + 563 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{315 \, {\left (b x^{2} + a\right )}}\right ] \]
[1/630*(315*(a^4*b*c*x^2 + a^5*c)*sqrt(a*c)*log(-(b^2*c*x^4 + 3*a*b*c*x^2 + 2*a^2*c - 2*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c)*sqrt (a*c))/(b*x^4 + a*x^2)) + 2*(35*b^4*c*x^8 + 185*a*b^3*c*x^6 + 408*a^2*b^2* c*x^4 + 506*a^3*b*c*x^2 + 563*a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^ 2*b*c*x^2 + a^3*c))/(b*x^2 + a), 1/315*(315*(a^4*b*c*x^2 + a^5*c)*sqrt(-a* c)*arctan(sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c)*sqrt(-a* c)/(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)) + (35*b^4*c*x^8 + 185*a*b^3*c*x^6 + 408*a^2*b^2*c*x^4 + 506*a^3*b*c*x^2 + 563*a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2* c*x^4 + 3*a^2*b*c*x^2 + a^3*c))/(b*x^2 + a)]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{3}\right )^{\frac {3}{2}}}{x}\, dx \]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\int { \frac {\left ({\left (b x^{2} + a\right )}^{3} c\right )^{\frac {3}{2}}}{x} \,d x } \]
Time = 0.49 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\frac {1}{315} \, {\left (\frac {315 \, a^{5} \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {-a c}} + \frac {315 \, \sqrt {b c x^{2} + a c} a^{4} c^{44} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, {\left (b c x^{2} + a c\right )}^{\frac {3}{2}} a^{3} c^{43} \mathrm {sgn}\left (b x^{2} + a\right ) + 63 \, {\left (b c x^{2} + a c\right )}^{\frac {5}{2}} a^{2} c^{42} \mathrm {sgn}\left (b x^{2} + a\right ) + 45 \, {\left (b c x^{2} + a c\right )}^{\frac {7}{2}} a c^{41} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, {\left (b c x^{2} + a c\right )}^{\frac {9}{2}} c^{40} \mathrm {sgn}\left (b x^{2} + a\right )}{c^{45}}\right )} c^{2} \]
1/315*(315*a^5*arctan(sqrt(b*c*x^2 + a*c)/sqrt(-a*c))*sgn(b*x^2 + a)/sqrt( -a*c) + (315*sqrt(b*c*x^2 + a*c)*a^4*c^44*sgn(b*x^2 + a) + 105*(b*c*x^2 + a*c)^(3/2)*a^3*c^43*sgn(b*x^2 + a) + 63*(b*c*x^2 + a*c)^(5/2)*a^2*c^42*sgn (b*x^2 + a) + 45*(b*c*x^2 + a*c)^(7/2)*a*c^41*sgn(b*x^2 + a) + 35*(b*c*x^2 + a*c)^(9/2)*c^40*sgn(b*x^2 + a))/c^45)*c^2
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^3\right )}^{3/2}}{x} \,d x \]