Integrand size = 19, antiderivative size = 202 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\frac {3}{2} a^2 b c \sqrt {c \left (a+b x^2\right )^3}+\frac {9 a^3 b c \sqrt {c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac {9}{10} a b c \left (a+b x^2\right ) \sqrt {c \left (a+b x^2\right )^3}+\frac {9}{14} b c \left (a+b x^2\right )^2 \sqrt {c \left (a+b x^2\right )^3}-\frac {c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^3}}{2 x^2}-\frac {9 a^2 b c \sqrt {c \left (a+b x^2\right )^3} \text {arctanh}\left (\sqrt {1+\frac {b x^2}{a}}\right )}{2 \left (1+\frac {b x^2}{a}\right )^{3/2}} \]
3/2*a^2*b*c*(c*(b*x^2+a)^3)^(1/2)+9/2*a^3*b*c*(c*(b*x^2+a)^3)^(1/2)/(b*x^2 +a)+9/10*a*b*c*(b*x^2+a)*(c*(b*x^2+a)^3)^(1/2)+9/14*b*c*(b*x^2+a)^2*(c*(b* x^2+a)^3)^(1/2)-1/2*c*(b*x^2+a)^3*(c*(b*x^2+a)^3)^(1/2)/x^2-9/2*a^2*b*c*ar ctanh((1+b*x^2/a)^(1/2))*(c*(b*x^2+a)^3)^(1/2)/(1+b*x^2/a)^(3/2)
Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.58 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=-\frac {\left (c \left (a+b x^2\right )^3\right )^{3/2} \left (\sqrt {a+b x^2} \left (35 a^4-388 a^3 b x^2-156 a^2 b^2 x^4-58 a b^3 x^6-10 b^4 x^8\right )+315 a^{7/2} b x^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )}{70 x^2 \left (a+b x^2\right )^{9/2}} \]
-1/70*((c*(a + b*x^2)^3)^(3/2)*(Sqrt[a + b*x^2]*(35*a^4 - 388*a^3*b*x^2 - 156*a^2*b^2*x^4 - 58*a*b^3*x^6 - 10*b^4*x^8) + 315*a^(7/2)*b*x^2*ArcTanh[S qrt[a + b*x^2]/Sqrt[a]]))/(x^2*(a + b*x^2)^(9/2))
Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.76, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {2045, 243, 51, 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.
| \(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx\) |
| \(\Big \downarrow \) 2045 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^3}dx}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^4}dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \int \frac {\left (\frac {b x^2}{a}+1\right )^{7/2}}{x^2}dx^2}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\int \frac {\left (\frac {b x^2}{a}+1\right )^{5/2}}{x^2}dx^2+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\int \frac {\left (\frac {b x^2}{a}+1\right )^{3/2}}{x^2}dx^2+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\int \frac {\sqrt {\frac {b x^2}{a}+1}}{x^2}dx^2+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\int \frac {1}{x^2 \sqrt {\frac {b x^2}{a}+1}}dx^2+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\frac {2 a \int \frac {1}{\frac {a x^4}{b}-\frac {a}{b}}d\sqrt {\frac {b x^2}{a}+1}}{b}+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right )}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a^3 c \left (\frac {9 b \left (-2 \text {arctanh}\left (\sqrt {\frac {b x^2}{a}+1}\right )+\frac {2}{7} \left (\frac {b x^2}{a}+1\right )^{7/2}+\frac {2}{5} \left (\frac {b x^2}{a}+1\right )^{5/2}+\frac {2}{3} \left (\frac {b x^2}{a}+1\right )^{3/2}+2 \sqrt {\frac {b x^2}{a}+1}\right )}{2 a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}\right ) \sqrt {c \left (a+b x^2\right )^3}}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
(a^3*c*Sqrt[c*(a + b*x^2)^3]*(-((1 + (b*x^2)/a)^(9/2)/x^2) + (9*b*(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*ArcTanh[Sqrt[1 + (b*x^2)/a]]))/(2*a)))/ (2*(1 + (b*x^2)/a)^(3/2))
3.3.42.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 2.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(-\frac {a^{4} c \sqrt {c \left (b \,x^{2}+a \right )^{3}}}{2 \left (b \,x^{2}+a \right ) x^{2}}+\frac {\left (\frac {b^{4} x^{6} \sqrt {b c \,x^{2}+a c}}{7 c}+\frac {29 b^{3} a \,x^{4} \sqrt {b c \,x^{2}+a c}}{35 c}+\frac {78 b^{2} a^{2} x^{2} \sqrt {b c \,x^{2}+a c}}{35 c}-\frac {156 b \,a^{3} \sqrt {b c \,x^{2}+a c}}{35 c}-\frac {9 b \,a^{4} \ln \left (\frac {2 a c +2 \sqrt {a c}\, \sqrt {b c \,x^{2}+a c}}{x}\right )}{2 \sqrt {a c}}+\frac {10 b \,a^{3} \sqrt {c \left (b \,x^{2}+a \right )}}{c}\right ) c \sqrt {c \left (b \,x^{2}+a \right )^{3}}\, \sqrt {c \left (b \,x^{2}+a \right )}}{\left (b \,x^{2}+a \right )^{2}}\) | \(223\) |
| default | \(\frac {{\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}} \left (10 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} b^{2} x^{4}-315 \ln \left (\frac {2 a c +2 \sqrt {a c}\, \sqrt {b c \,x^{2}+a c}}{x}\right ) a^{4} b \,c^{3} x^{2}+42 b a {\left (c \left (b \,x^{2}+a \right )\right )}^{\frac {5}{2}} x^{2} \sqrt {a c}-4 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a b \,x^{2}+105 \left (b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a^{2} b c \,x^{2}+315 \sqrt {b c \,x^{2}+a c}\, \sqrt {a c}\, a^{3} b \,c^{2} x^{2}-35 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a^{2}\right )}{70 \left (b \,x^{2}+a \right )^{3} {\left (c \left (b \,x^{2}+a \right )\right )}^{\frac {3}{2}} c \,x^{2} \sqrt {a c}}\) | \(238\) |
-1/2*a^4/(b*x^2+a)/x^2*c*(c*(b*x^2+a)^3)^(1/2)+(1/7*b^4*x^6/c*(b*c*x^2+a*c )^(1/2)+29/35*b^3*a*x^4/c*(b*c*x^2+a*c)^(1/2)+78/35*b^2*a^2*x^2/c*(b*c*x^2 +a*c)^(1/2)-156/35*b*a^3/c*(b*c*x^2+a*c)^(1/2)-9/2*b*a^4/(a*c)^(1/2)*ln((2 *a*c+2*(a*c)^(1/2)*(b*c*x^2+a*c)^(1/2))/x)+10*b*a^3/c*(c*(b*x^2+a))^(1/2)) *c/(b*x^2+a)^2*(c*(b*x^2+a)^3)^(1/2)*(c*(b*x^2+a))^(1/2)
Time = 0.28 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.03 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\left [\frac {315 \, {\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt {a c} \log \left (-\frac {b^{2} c x^{4} + 3 \, a b c x^{2} + 2 \, a^{2} c - 2 \, \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {a c}}{b x^{4} + a x^{2}}\right ) + 2 \, {\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{140 \, {\left (b x^{4} + a x^{2}\right )}}, \frac {315 \, {\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {-a c}}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) + {\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{70 \, {\left (b x^{4} + a x^{2}\right )}}\right ] \]
[1/140*(315*(a^3*b^2*c*x^4 + a^4*b*c*x^2)*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*(10*b^4*c*x^8 + 58*a*b^3*c*x^6 + 156*a ^2*b^2*c*x^4 + 388*a^3*b*c*x^2 - 35*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^4 + a*x^2), 1/70*(315*(a^3*b^2*c*x^4 + a^4* b*c*x^2)*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)) + (10*b^4*c*x^8 + 5 8*a*b^3*c*x^6 + 156*a^2*b^2*c*x^4 + 388*a^3*b*c*x^2 - 35*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^4 + a*x^2)]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{3}\right )^{\frac {3}{2}}}{x^{3}}\, dx \]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\int { \frac {\left ({\left (b x^{2} + a\right )}^{3} c\right )^{\frac {3}{2}}}{x^{3}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\frac {{\left (\frac {315 \, a^{4} b^{2} c \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {-a c}} - \frac {35 \, \sqrt {b c x^{2} + a c} a^{4} b \mathrm {sgn}\left (b x^{2} + a\right )}{x^{2}} + \frac {2 \, {\left (140 \, \sqrt {b c x^{2} + a c} a^{3} b^{2} c^{21} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, {\left (b c x^{2} + a c\right )}^{\frac {3}{2}} a^{2} b^{2} c^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + 14 \, {\left (b c x^{2} + a c\right )}^{\frac {5}{2}} a b^{2} c^{19} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, {\left (b c x^{2} + a c\right )}^{\frac {7}{2}} b^{2} c^{18} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{c^{21}}\right )} c}{70 \, b} \]
1/70*(315*a^4*b^2*c*arctan(sqrt(b*c*x^2 + a*c)/sqrt(-a*c))*sgn(b*x^2 + a)/ sqrt(-a*c) - 35*sqrt(b*c*x^2 + a*c)*a^4*b*sgn(b*x^2 + a)/x^2 + 2*(140*sqrt (b*c*x^2 + a*c)*a^3*b^2*c^21*sgn(b*x^2 + a) + 35*(b*c*x^2 + a*c)^(3/2)*a^2 *b^2*c^20*sgn(b*x^2 + a) + 14*(b*c*x^2 + a*c)^(5/2)*a*b^2*c^19*sgn(b*x^2 + a) + 5*(b*c*x^2 + a*c)^(7/2)*b^2*c^18*sgn(b*x^2 + a))/c^21)*c/b
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^3\right )}^{3/2}}{x^3} \,d x \]