Integrand size = 19, antiderivative size = 140 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=-\frac {a^3 c \sqrt {c \left (a+b x^2\right )^2}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a^2 b c \sqrt {c \left (a+b x^2\right )^2}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 c x^2 \sqrt {c \left (a+b x^2\right )^2}}{2 \left (a+b x^2\right )}+\frac {3 a b^2 c \sqrt {c \left (a+b x^2\right )^2} \log (x)}{a+b x^2} \] Output:
-1/4*a^3*c*(c*(b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-3/2*a^2*b*c*(c*(b*x^2+a)^2) ^(1/2)/x^2/(b*x^2+a)+b^3*c*x^2*(c*(b*x^2+a)^2)^(1/2)/(2*b*x^2+2*a)+3*a*b^2 *c*(c*(b*x^2+a)^2)^(1/2)*ln(x)/(b*x^2+a)
Time = 0.66 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.69 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\frac {1}{4} c \left (-\frac {c \left (-a^3-6 a^2 b x^2+a b^2 x^4+2 b^3 x^6\right ) \left (a \sqrt {b^2 c}+b \sqrt {b^2 c} x^2-b \sqrt {c \left (a+b x^2\right )^2}\right )}{x^4 \left (a b c+b^2 c x^2-\sqrt {b^2 c} \sqrt {c \left (a+b x^2\right )^2}\right )}+6 a b^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {b^2 c} x^2-\sqrt {c \left (a+b x^2\right )^2}}{a \sqrt {c}}\right )-3 a b \sqrt {b^2 c} \log \left (x^2 \left (a b c+b^2 c x^2-\sqrt {b^2 c} \sqrt {c \left (a+b x^2\right )^2}\right )\right )\right ) \] Input:
Integrate[(c*(a + b*x^2)^2)^(3/2)/x^5,x]
Output:
(c*(-((c*(-a^3 - 6*a^2*b*x^2 + a*b^2*x^4 + 2*b^3*x^6)*(a*Sqrt[b^2*c] + b*S qrt[b^2*c]*x^2 - b*Sqrt[c*(a + b*x^2)^2]))/(x^4*(a*b*c + b^2*c*x^2 - Sqrt[ b^2*c]*Sqrt[c*(a + b*x^2)^2]))) + 6*a*b^2*Sqrt[c]*ArcTanh[(Sqrt[b^2*c]*x^2 - Sqrt[c*(a + b*x^2)^2])/(a*Sqrt[c])] - 3*a*b*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])]))/4
Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.47, 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.
\(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{a^3 x^5}dx}{a+b x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x^5}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x^6}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \left (\frac {a^3}{x^6}+\frac {3 b a^2}{x^4}+\frac {3 b^2 a}{x^2}+b^3\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \left (-\frac {a^3}{2 x^4}-\frac {3 a^2 b}{x^2}+3 a b^2 \log \left (x^2\right )+b^3 x^2\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[(c*(a + b*x^2)^2)^(3/2)/x^5,x]
Output:
(c*Sqrt[c*(a + b*x^2)^2]*(-1/2*a^3/x^4 - (3*a^2*b)/x^2 + b^3*x^2 + 3*a*b^2 *Log[x^2]))/(2*(a + b*x^2))
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 = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.44
method | result | size |
default | \(\frac {{\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (2 b^{3} x^{6}+12 a \,b^{2} \ln \left (x \right ) x^{4}-6 a^{2} b \,x^{2}-a^{3}\right )}{4 x^{4} \left (b \,x^{2}+a \right )^{3}}\) | \(62\) |
pseudoelliptic | \(-\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, \left (-6 a \,b^{2} \ln \left (x^{2}\right ) x^{4}-2 b^{3} x^{6}+6 a^{2} b \,x^{2}+a^{3}\right )}{4 \left (b \,x^{2}+a \right ) x^{4}}\) | \(63\) |
risch | \(\frac {b^{3} c \,x^{2} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3}{2} a^{2} b \,x^{2}-\frac {1}{4} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{4}}+\frac {3 a \,b^{2} c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{b \,x^{2}+a}\) | \(106\) |
Input:
int((c*(b*x^2+a)^2)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
1/4*(c*(b*x^2+a)^2)^(3/2)*(2*b^3*x^6+12*a*b^2*ln(x)*x^4-6*a^2*b*x^2-a^3)/x ^4/(b*x^2+a)^3
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.55 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\frac {{\left (2 \, b^{3} c x^{6} + 12 \, a b^{2} c x^{4} \log \left (x\right ) - 6 \, a^{2} b c x^{2} - a^{3} c\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{4 \, {\left (b x^{6} + a x^{4}\right )}} \] Input:
integrate((c*(b*x^2+a)^2)^(3/2)/x^5,x, algorithm="fricas")
Output:
1/4*(2*b^3*c*x^6 + 12*a*b^2*c*x^4*log(x) - 6*a^2*b*c*x^2 - a^3*c)*sqrt(b^2 *c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^6 + a*x^4)
\[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{5}}\, dx \] Input:
integrate((c*(b*x**2+a)**2)**(3/2)/x**5,x)
Output:
Integral((c*(a + b*x**2)**2)**(3/2)/x**5, x)
Time = 0.04 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.79 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\frac {3}{2} \, \left (-1\right )^{2 \, b^{2} c x^{2} + 2 \, a b c} a b^{2} c^{\frac {3}{2}} \log \left (2 \, b^{2} c x^{2} + 2 \, a b c\right ) - \frac {3}{2} \, \left (-1\right )^{2 \, a b c x^{2} + 2 \, a^{2} c} a b^{2} c^{\frac {3}{2}} \log \left (2 \, a b c + \frac {2 \, a^{2} c}{x^{2}}\right ) + \frac {3 \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} b^{3} c x^{2}}{4 \, a} + \frac {9}{4} \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} b^{2} c + \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2}}{4 \, a^{2}} - \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b}{4 \, a x^{2}} - \frac {{\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {5}{2}}}{4 \, a^{2} c x^{4}} \] Input:
integrate((c*(b*x^2+a)^2)^(3/2)/x^5,x, algorithm="maxima")
Output:
3/2*(-1)^(2*b^2*c*x^2 + 2*a*b*c)*a*b^2*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*b^2*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^3*c*x^2/a + 9/4*sqrt(b^2*c *x^4 + 2*a*b*c*x^2 + a^2*c)*b^2*c + 1/4*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^ (3/2)*b^2/a^2 - 1/4*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)*b/(a*x^2) - 1/ 4*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(5/2)/(a^2*c*x^4)
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.66 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\frac {1}{4} \, {\left (2 \, b^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a b^{2} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {9 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{x^{4}}\right )} c^{\frac {3}{2}} \] Input:
integrate((c*(b*x^2+a)^2)^(3/2)/x^5,x, algorithm="giac")
Output:
1/4*(2*b^3*x^2*sgn(b*x^2 + a) + 6*a*b^2*log(x^2)*sgn(b*x^2 + a) - (9*a*b^2 *x^4*sgn(b*x^2 + a) + 6*a^2*b*x^2*sgn(b*x^2 + a) + a^3*sgn(b*x^2 + a))/x^4 )*c^(3/2)
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2}}{x^5} \,d x \] Input:
int((c*(a + b*x^2)^2)^(3/2)/x^5,x)
Output:
int((c*(a + b*x^2)^2)^(3/2)/x^5, x)
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.30 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^5} \, dx=\frac {\sqrt {c}\, c \left (12 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{4}-a^{3}-6 a^{2} b \,x^{2}+2 b^{3} x^{6}\right )}{4 x^{4}} \] Input:
int((c*(b*x^2+a)^2)^(3/2)/x^5,x)
Output:
(sqrt(c)*c*(12*log(x)*a*b**2*x**4 - a**3 - 6*a**2*b*x**2 + 2*b**3*x**6))/( 4*x**4)