Integrand size = 19, antiderivative size = 140 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=-\frac {a^3 c \sqrt {c \left (a+b x^2\right )^2}}{2 x^2 \left (a+b x^2\right )}+\frac {3 a b^2 c x^2 \sqrt {c \left (a+b x^2\right )^2}}{2 \left (a+b x^2\right )}+\frac {b^3 c x^4 \sqrt {c \left (a+b x^2\right )^2}}{4 \left (a+b x^2\right )}+\frac {3 a^2 b c \sqrt {c \left (a+b x^2\right )^2} \log (x)}{a+b x^2} \]
-1/2*a^3*c*(c*(b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+3/2*a*b^2*c*x^2*(c*(b*x^2+a )^2)^(1/2)/(b*x^2+a)+1/4*b^3*c*x^4*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+3*a^2*b *c*ln(x)*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.70 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.69 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\frac {1}{16} c \left (\frac {\left (8 a^3+21 a^2 b x^2-24 a b^2 x^4-4 b^3 x^6\right ) \left (a b c+b^2 c x^2-\sqrt {b^2 c} \sqrt {c \left (a+b x^2\right )^2}\right )}{x^2 \left (a \sqrt {b^2 c}+b \sqrt {b^2 c} x^2-b \sqrt {c \left (a+b x^2\right )^2}\right )}+24 a^2 b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {b^2 c} x^2-\sqrt {c \left (a+b x^2\right )^2}}{a \sqrt {c}}\right )-12 a^2 \sqrt {b^2 c} \log \left (x^2 \left (a b c+b^2 c x^2-\sqrt {b^2 c} \sqrt {c \left (a+b x^2\right )^2}\right )\right )\right ) \]
(c*(((8*a^3 + 21*a^2*b*x^2 - 24*a*b^2*x^4 - 4*b^3*x^6)*(a*b*c + b^2*c*x^2 - Sqrt[b^2*c]*Sqrt[c*(a + b*x^2)^2]))/(x^2*(a*Sqrt[b^2*c] + b*Sqrt[b^2*c]* x^2 - b*Sqrt[c*(a + b*x^2)^2])) + 24*a^2*b*Sqrt[c]*ArcTanh[(Sqrt[b^2*c]*x^ 2 - Sqrt[c*(a + b*x^2)^2])/(a*Sqrt[c])] - 12*a^2*Sqrt[b^2*c]*Log[x^2*(a*b* c + b^2*c*x^2 - Sqrt[b^2*c]*Sqrt[c*(a + b*x^2)^2])]))/16
Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.48, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2045, 27, 243, 49, 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.
\(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{a^3 x^3}dx}{a+b x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x^3}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x^4}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \left (\frac {a^3}{x^4}+\frac {3 b a^2}{x^2}+3 b^2 a+b^3 x^2\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \left (-\frac {a^3}{x^2}+3 a^2 b \log \left (x^2\right )+3 a b^2 x^2+\frac {b^3 x^4}{2}\right )}{2 \left (a+b x^2\right )}\) |
(c*Sqrt[c*(a + b*x^2)^2]*(-(a^3/x^2) + 3*a*b^2*x^2 + (b^3*x^4)/2 + 3*a^2*b *Log[x^2]))/(2*(a + b*x^2))
3.3.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.44
method | result | size |
default | \(\frac {{\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (b^{3} x^{6}+6 b^{2} x^{4} a +12 a^{2} b \ln \left (x \right ) x^{2}-2 a^{3}\right )}{4 x^{2} \left (b \,x^{2}+a \right )^{3}}\) | \(61\) |
pseudoelliptic | \(-\frac {c \left (-\frac {b^{3} x^{6}}{2}-3 b^{2} x^{4} a -3 a^{2} b \ln \left (x^{2}\right ) x^{2}+a^{3}\right ) \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{2 \left (b \,x^{2}+a \right ) x^{2}}\) | \(63\) |
risch | \(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, b \left (b \,x^{2}+3 a \right )^{2}}{4 b \,x^{2}+4 a}-\frac {a^{3} c \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{2 x^{2} \left (b \,x^{2}+a \right )}+\frac {3 a^{2} b c \ln \left (x \right ) \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(101\) |
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.54 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\frac {{\left (b^{3} c x^{6} + 6 \, a b^{2} c x^{4} + 12 \, a^{2} b c x^{2} \log \left (x\right ) - 2 \, a^{3} c\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{4 \, {\left (b x^{4} + a x^{2}\right )}} \]
1/4*(b^3*c*x^6 + 6*a*b^2*c*x^4 + 12*a^2*b*c*x^2*log(x) - 2*a^3*c)*sqrt(b^2 *c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^4 + a*x^2)
\[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{3}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.26 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\frac {3}{2} \, \left (-1\right )^{2 \, b^{2} c x^{2} + 2 \, a b c} a^{2} b c^{\frac {3}{2}} \log \left (2 \, b^{2} c x^{2} + 2 \, a b c\right ) - \frac {3}{2} \, \left (-1\right )^{2 \, a b c x^{2} + 2 \, a^{2} c} a^{2} b c^{\frac {3}{2}} \log \left (2 \, a b c + \frac {2 \, a^{2} c}{x^{2}}\right ) + \frac {3}{4} \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} b^{2} c x^{2} + \frac {9}{4} \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} a b c - \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}}}{2 \, x^{2}} \]
3/2*(-1)^(2*b^2*c*x^2 + 2*a*b*c)*a^2*b*c^(3/2)*log(2*b^2*c*x^2 + 2*a*b*c) - 3/2*(-1)^(2*a*b*c*x^2 + 2*a^2*c)*a^2*b*c^(3/2)*log(2*a*b*c + 2*a^2*c/x^2 ) + 3/4*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*b^2*c*x^2 + 9/4*sqrt(b^2*c*x ^4 + 2*a*b*c*x^2 + a^2*c)*a*b*c - 1/2*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3 /2)/x^2
Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\frac {1}{4} \, {\left (b^{3} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a b^{2} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{2} b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {2 \, {\left (3 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{x^{2}}\right )} c^{\frac {3}{2}} \]
1/4*(b^3*x^4*sgn(b*x^2 + a) + 6*a*b^2*x^2*sgn(b*x^2 + a) + 6*a^2*b*log(x^2 )*sgn(b*x^2 + a) - 2*(3*a^2*b*x^2*sgn(b*x^2 + a) + a^3*sgn(b*x^2 + a))/x^2 )*c^(3/2)
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2}}{x^3} \,d x \]