Integrand size = 19, antiderivative size = 159 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=-\frac {b c \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{a^3}-\frac {b c \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 a^2 \left (a+b x^2\right )}-\frac {c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right )}{2 a^3 x^2}-\frac {3 b c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \log (x)}{a^4}+\frac {3 b c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^4} \] Output:
-b*c*(c/(b*x^2+a)^2)^(1/2)/a^3-1/4*b*c*(c/(b*x^2+a)^2)^(1/2)/a^2/(b*x^2+a) -1/2*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)/a^3/x^2-3*b*c*(c/(b*x^2+a)^2)^(1/2) *(b*x^2+a)*ln(x)/a^4+3/2*b*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)*ln(b*x^2+a)/a ^4
Time = 1.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=-\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (a \left (2 a^2+9 a b x^2+6 b^2 x^4\right )+12 b x^2 \left (a+b x^2\right )^2 \log (x)-6 b x^2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )\right )}{4 a^4 x^2} \] Input:
Integrate[(c/(a + b*x^2)^2)^(3/2)/x^3,x]
Output:
-1/4*((c/(a + b*x^2)^2)^(3/2)*(a + b*x^2)*(a*(2*a^2 + 9*a*b*x^2 + 6*b^2*x^ 4) + 12*b*x^2*(a + b*x^2)^2*Log[x] - 6*b*x^2*(a + b*x^2)^2*Log[a + b*x^2]) )/(a^4*x^2)
Time = 0.45 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58, 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, 54, 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.
\(\displaystyle \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {a^3}{x^3 \left (b x^2+a\right )^3}dx}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {1}{x^3 \left (b x^2+a\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {1}{x^4 \left (b x^2+a\right )^3}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \left (\frac {3 b^2}{a^4 \left (b x^2+a\right )}+\frac {2 b^2}{a^3 \left (b x^2+a\right )^2}+\frac {b^2}{a^2 \left (b x^2+a\right )^3}-\frac {3 b}{a^4 x^2}+\frac {1}{a^3 x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (-\frac {3 b \log \left (x^2\right )}{a^4}+\frac {3 b \log \left (a+b x^2\right )}{a^4}-\frac {2 b}{a^3 \left (a+b x^2\right )}-\frac {1}{a^3 x^2}-\frac {b}{2 a^2 \left (a+b x^2\right )^2}\right )\) |
Input:
Int[(c/(a + b*x^2)^2)^(3/2)/x^3,x]
Output:
(c*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*(-(1/(a^3*x^2)) - b/(2*a^2*(a + b*x^2 )^2) - (2*b)/(a^3*(a + b*x^2)) - (3*b*Log[x^2])/a^4 + (3*b*Log[a + b*x^2]) /a^4))/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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (-\frac {3 b^{2} x^{4}}{2 a^{3}}-\frac {9 b \,x^{2}}{4 a^{2}}-\frac {1}{2 a}\right )}{\left (b \,x^{2}+a \right ) x^{2}}-\frac {3 b c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (b \,x^{2}+a \right ) \ln \left (x \right )}{a^{4}}+\frac {3 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, b \ln \left (-b \,x^{2}-a \right )}{2 a^{4}}\) | \(122\) |
default | \(-\frac {\left (12 \ln \left (x \right ) b^{3} x^{6}-6 \ln \left (b \,x^{2}+a \right ) b^{3} x^{6}+24 a \,b^{2} \ln \left (x \right ) x^{4}-12 \ln \left (b \,x^{2}+a \right ) a \,b^{2} x^{4}+6 a \,b^{2} x^{4}+12 a^{2} b \ln \left (x \right ) x^{2}-6 \ln \left (b \,x^{2}+a \right ) a^{2} b \,x^{2}+9 a^{2} b \,x^{2}+2 a^{3}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{4 a^{4} x^{2}}\) | \(135\) |
Input:
int((c/(b*x^2+a)^2)^(3/2)/x^3,x,method=_RETURNVERBOSE)
Output:
c/(b*x^2+a)*(c/(b*x^2+a)^2)^(1/2)*(-3/2*b^2/a^3*x^4-9/4*b/a^2*x^2-1/2/a)/x ^2-3*b*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)*ln(x)/a^4+3/2*c*(b*x^2+a)*(c/(b*x ^2+a)^2)^(1/2)/a^4*b*ln(-b*x^2-a)
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.89 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=-\frac {{\left (6 \, a b^{2} c x^{4} + 9 \, a^{2} b c x^{2} + 2 \, a^{3} c - 6 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (b x^{2} + a\right ) + 12 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (x\right )\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{4 \, {\left (a^{4} b x^{4} + a^{5} x^{2}\right )}} \] Input:
integrate((c/(b*x^2+a)^2)^(3/2)/x^3,x, algorithm="fricas")
Output:
-1/4*(6*a*b^2*c*x^4 + 9*a^2*b*c*x^2 + 2*a^3*c - 6*(b^3*c*x^6 + 2*a*b^2*c*x ^4 + a^2*b*c*x^2)*log(b*x^2 + a) + 12*(b^3*c*x^6 + 2*a*b^2*c*x^4 + a^2*b*c *x^2)*log(x))*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))/(a^4*b*x^4 + a^5*x^2)
\[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=\int \frac {\left (\frac {c}{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\right )^{\frac {3}{2}}}{x^{3}}\, dx \] Input:
integrate((c/(b*x**2+a)**2)**(3/2)/x**3,x)
Output:
Integral((c/(a**2 + 2*a*b*x**2 + b**2*x**4))**(3/2)/x**3, x)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=-\frac {6 \, b^{2} c^{\frac {3}{2}} x^{4} + 9 \, a b c^{\frac {3}{2}} x^{2} + 2 \, a^{2} c^{\frac {3}{2}}}{4 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} + \frac {3 \, b c^{\frac {3}{2}} \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {3 \, b c^{\frac {3}{2}} \log \left (x^{2}\right )}{2 \, a^{4}} \] Input:
integrate((c/(b*x^2+a)^2)^(3/2)/x^3,x, algorithm="maxima")
Output:
-1/4*(6*b^2*c^(3/2)*x^4 + 9*a*b*c^(3/2)*x^2 + 2*a^2*c^(3/2))/(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2) + 3/2*b*c^(3/2)*log(b*x^2 + a)/a^4 - 3/2*b*c^(3/2 )*log(x^2)/a^4
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.59 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=-\frac {1}{4} \, c^{\frac {3}{2}} {\left (\frac {6 \, b \log \left (x^{2}\right )}{a^{4}} - \frac {6 \, b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{4}} + \frac {9 \, b^{3} x^{4} + 22 \, a b^{2} x^{2} + 14 \, a^{2} b}{{\left (b x^{2} + a\right )}^{2} a^{4}} - \frac {2 \, {\left (3 \, b x^{2} - a\right )}}{a^{4} x^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate((c/(b*x^2+a)^2)^(3/2)/x^3,x, algorithm="giac")
Output:
-1/4*c^(3/2)*(6*b*log(x^2)/a^4 - 6*b*log(abs(b*x^2 + a))/a^4 + (9*b^3*x^4 + 22*a*b^2*x^2 + 14*a^2*b)/((b*x^2 + a)^2*a^4) - 2*(3*b*x^2 - a)/(a^4*x^2) )*sgn(b*x^2 + a)
Timed out. \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2}}{x^3} \,d x \] Input:
int((c/(a + b*x^2)^2)^(3/2)/x^3,x)
Output:
int((c/(a + b*x^2)^2)^(3/2)/x^3, x)
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {c}\, c \left (6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b \,x^{2}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} x^{4}+6 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} x^{6}-12 \,\mathrm {log}\left (x \right ) a^{2} b \,x^{2}-24 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{4}-12 \,\mathrm {log}\left (x \right ) b^{3} x^{6}-2 a^{3}-6 a^{2} b \,x^{2}+3 b^{3} x^{6}\right )}{4 a^{4} x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int((c/(b*x^2+a)^2)^(3/2)/x^3,x)
Output:
(sqrt(c)*c*(6*log(a + b*x**2)*a**2*b*x**2 + 12*log(a + b*x**2)*a*b**2*x**4 + 6*log(a + b*x**2)*b**3*x**6 - 12*log(x)*a**2*b*x**2 - 24*log(x)*a*b**2* x**4 - 12*log(x)*b**3*x**6 - 2*a**3 - 6*a**2*b*x**2 + 3*b**3*x**6))/(4*a** 4*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4))