Integrand size = 19, antiderivative size = 208 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\frac {105}{64} a^2 b c x \sqrt {c \left (a+b x^2\right )^3}+\frac {315 a^3 b c x \sqrt {c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac {21}{16} a b c x \left (a+b x^2\right ) \sqrt {c \left (a+b x^2\right )^3}+\frac {9}{8} b c x \left (a+b x^2\right )^2 \sqrt {c \left (a+b x^2\right )^3}-\frac {c \left (a+b x^2\right )^3 \sqrt {c \left (a+b x^2\right )^3}}{x}+\frac {315 a^{5/2} \sqrt {b} c \sqrt {c \left (a+b x^2\right )^3} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 \left (1+\frac {b x^2}{a}\right )^{3/2}} \]
105/64*a^2*b*c*x*(c*(b*x^2+a)^3)^(1/2)+315/128*a^3*b*c*x*(c*(b*x^2+a)^3)^( 1/2)/(b*x^2+a)+21/16*a*b*c*x*(b*x^2+a)*(c*(b*x^2+a)^3)^(1/2)+9/8*b*c*x*(b* x^2+a)^2*(c*(b*x^2+a)^3)^(1/2)-c*(b*x^2+a)^3*(c*(b*x^2+a)^3)^(1/2)/x+315/1 28*a^(5/2)*c*arcsinh(x*b^(1/2)/a^(1/2))*b^(1/2)*(c*(b*x^2+a)^3)^(1/2)/(1+b *x^2/a)^(3/2)
Time = 0.05 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.58 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=-\frac {\left (c \left (a+b x^2\right )^3\right )^{3/2} \left (\sqrt {a+b x^2} \left (128 a^4-325 a^3 b x^2-210 a^2 b^2 x^4-88 a b^3 x^6-16 b^4 x^8\right )+315 a^4 \sqrt {b} x \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{128 x \left (a+b x^2\right )^{9/2}} \]
-1/128*((c*(a + b*x^2)^3)^(3/2)*(Sqrt[a + b*x^2]*(128*a^4 - 325*a^3*b*x^2 - 210*a^2*b^2*x^4 - 88*a*b^3*x^6 - 16*b^4*x^8) + 315*a^4*Sqrt[b]*x*Log[-(S qrt[b]*x) + Sqrt[a + b*x^2]]))/(x*(a + b*x^2)^(9/2))
Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2045, 247, 211, 211, 211, 211, 222}
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.
| \(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 2045 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x^2}dx}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \int \left (\frac {b x^2}{a}+1\right )^{7/2}dx}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\frac {7}{8} \int \left (\frac {b x^2}{a}+1\right )^{5/2}dx+\frac {1}{8} x \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\frac {7}{8} \left (\frac {5}{6} \int \left (\frac {b x^2}{a}+1\right )^{3/2}dx+\frac {1}{6} x \left (\frac {b x^2}{a}+1\right )^{5/2}\right )+\frac {1}{8} x \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {\frac {b x^2}{a}+1}dx+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )+\frac {1}{6} x \left (\frac {b x^2}{a}+1\right )^{5/2}\right )+\frac {1}{8} x \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \left (\frac {9 b \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\frac {b x^2}{a}+1}}dx+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )+\frac {1}{6} x \left (\frac {b x^2}{a}+1\right )^{5/2}\right )+\frac {1}{8} x \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {a^3 c \left (\frac {9 b \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {\frac {b x^2}{a}+1}\right )+\frac {1}{4} x \left (\frac {b x^2}{a}+1\right )^{3/2}\right )+\frac {1}{6} x \left (\frac {b x^2}{a}+1\right )^{5/2}\right )+\frac {1}{8} x \left (\frac {b x^2}{a}+1\right )^{7/2}\right )}{a}-\frac {\left (\frac {b x^2}{a}+1\right )^{9/2}}{x}\right ) \sqrt {c \left (a+b x^2\right )^3}}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
(a^3*c*Sqrt[c*(a + b*x^2)^3]*(-((1 + (b*x^2)/a)^(9/2)/x) + (9*b*((x*(1 + ( b*x^2)/a)^(7/2))/8 + (7*((x*(1 + (b*x^2)/a)^(5/2))/6 + (5*((x*(1 + (b*x^2) /a)^(3/2))/4 + (3*((x*Sqrt[1 + (b*x^2)/a])/2 + (Sqrt[a]*ArcSinh[(Sqrt[b]*x )/Sqrt[a]])/(2*Sqrt[b])))/4))/6))/8))/a))/(1 + (b*x^2)/a)^(3/2)
3.3.41.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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 = 2.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.68
| method | result | size |
| risch | \(-\frac {\left (-16 b^{4} x^{8}-88 a \,b^{3} x^{6}-210 a^{2} b^{2} x^{4}-325 a^{3} b \,x^{2}+128 a^{4}\right ) c \sqrt {c \left (b \,x^{2}+a \right )^{3}}}{128 \left (b \,x^{2}+a \right ) x}+\frac {315 b \,a^{4} \ln \left (\frac {b c x}{\sqrt {b c}}+\sqrt {b c \,x^{2}+a c}\right ) c \sqrt {c \left (b \,x^{2}+a \right )^{3}}\, \sqrt {c \left (b \,x^{2}+a \right )}}{128 \sqrt {b c}\, \left (b \,x^{2}+a \right )^{2}}\) | \(141\) |
| default | \(-\frac {{\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}} \left (-16 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} \sqrt {b c}\, b^{2} x^{4}-56 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} \sqrt {b c}\, a b \,x^{2}-210 \left (b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b c}\, a^{2} b c \,x^{2}-315 \sqrt {b c \,x^{2}+a c}\, \sqrt {b c}\, a^{3} b \,c^{2} x^{2}-315 \ln \left (\frac {b c x +\sqrt {b c \,x^{2}+a c}\, \sqrt {b c}}{\sqrt {b c}}\right ) a^{4} b \,c^{3} x +128 \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} \sqrt {b c}\, a^{2}\right )}{128 \left (b \,x^{2}+a \right )^{3} {\left (c \left (b \,x^{2}+a \right )\right )}^{\frac {3}{2}} c \sqrt {b c}\, x}\) | \(215\) |
-1/128/(b*x^2+a)*(-16*b^4*x^8-88*a*b^3*x^6-210*a^2*b^2*x^4-325*a^3*b*x^2+1 28*a^4)/x*c*(c*(b*x^2+a)^3)^(1/2)+315/128*b*a^4*ln(b*c*x/(b*c)^(1/2)+(b*c* x^2+a*c)^(1/2))/(b*c)^(1/2)*c/(b*x^2+a)^2*(c*(b*x^2+a)^3)^(1/2)*(c*(b*x^2+ a))^(1/2)
Time = 0.30 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.90 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\left [\frac {315 \, {\left (a^{4} b c x^{3} + a^{5} c x\right )} \sqrt {b c} \log \left (-\frac {2 \, b^{2} c x^{4} + 3 \, a b c x^{2} + a^{2} c + 2 \, \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {b c} x}{b x^{2} + a}\right ) + 2 \, {\left (16 \, b^{4} c x^{8} + 88 \, a b^{3} c x^{6} + 210 \, a^{2} b^{2} c x^{4} + 325 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{256 \, {\left (b x^{3} + a x\right )}}, -\frac {315 \, {\left (a^{4} b c x^{3} + a^{5} c x\right )} \sqrt {-b c} \arctan \left (\frac {\sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {-b c} x}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) - {\left (16 \, b^{4} c x^{8} + 88 \, a b^{3} c x^{6} + 210 \, a^{2} b^{2} c x^{4} + 325 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{128 \, {\left (b x^{3} + a x\right )}}\right ] \]
[1/256*(315*(a^4*b*c*x^3 + a^5*c*x)*sqrt(b*c)*log(-(2*b^2*c*x^4 + 3*a*b*c* x^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)*sq rt(b*c)*x)/(b*x^2 + a)) + 2*(16*b^4*c*x^8 + 88*a*b^3*c*x^6 + 210*a^2*b^2*c *x^4 + 325*a^3*b*c*x^2 - 128*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^3 + a*x), -1/128*(315*(a^4*b*c*x^3 + a^5*c*x)*sqrt (-b*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 (-b*c)*x/(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)) - (16*b^4*c*x^8 + 88*a*b^3*c*x ^6 + 210*a^2*b^2*c*x^4 + 325*a^3*b*c*x^2 - 128*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^3 + a*x)]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{3}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]
\[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\int { \frac {\left ({\left (b x^{2} + a\right )}^{3} c\right )^{\frac {3}{2}}}{x^{2}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\frac {1}{256} \, {\left (\frac {512 \, \sqrt {b c} a^{5} c \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2} - a c} - 315 \, \sqrt {b c} a^{4} \log \left ({\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, {\left (325 \, a^{3} b \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, {\left (105 \, a^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, {\left (2 \, b^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 11 \, a b^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )} x^{2}\right )} x^{2}\right )} \sqrt {b c x^{2} + a c} x\right )} c \]
1/256*(512*sqrt(b*c)*a^5*c*sgn(b*x^2 + a)/((sqrt(b*c)*x - sqrt(b*c*x^2 + a *c))^2 - a*c) - 315*sqrt(b*c)*a^4*log((sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^ 2)*sgn(b*x^2 + a) + 2*(325*a^3*b*sgn(b*x^2 + a) + 2*(105*a^2*b^2*sgn(b*x^2 + a) + 4*(2*b^4*x^2*sgn(b*x^2 + a) + 11*a*b^3*sgn(b*x^2 + a))*x^2)*x^2)*s qrt(b*c*x^2 + a*c)*x)*c
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^3\right )}^{3/2}}{x^2} \,d x \]