\(\int \frac {(\frac {c}{(a+b x^2)^2})^{3/2}}{x^2} \, dx\) [37]

Optimal result

Integrand size = 19, antiderivative size = 141 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=-\frac {7 b c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{8 a^3}-\frac {b c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 a^2 \left (a+b x^2\right )}-\frac {c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right )}{a^3 x}-\frac {15 \sqrt {b} c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}} \] Output:

-7/8*b*c*x*(c/(b*x^2+a)^2)^(1/2)/a^3-1/4*b*c*x*(c/(b*x^2+a)^2)^(1/2)/a^2/( 
b*x^2+a)-c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)/a^3/x-15/8*b^(1/2)*c*(c/(b*x^2+ 
a)^2)^(1/2)*(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.65 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=-\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (\sqrt {a} \left (8 a^2+25 a b x^2+15 b^2 x^4\right )+15 \sqrt {b} x \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{8 a^{7/2} x} \] Input:

Integrate[(c/(a + b*x^2)^2)^(3/2)/x^2,x]
 

Output:

-1/8*((c/(a + b*x^2)^2)^(3/2)*(a + b*x^2)*(Sqrt[a]*(8*a^2 + 25*a*b*x^2 + 1 
5*b^2*x^4) + 15*Sqrt[b]*x*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]))/(a^( 
7/2)*x)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2045, 27, 253, 253, 264, 218}

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.

\(\displaystyle \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx\)

\(\Big \downarrow \) 2045

\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {a^3}{x^2 \left (b x^2+a\right )^3}dx}{a^3}\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {1}{x^2 \left (b x^2+a\right )^3}dx\)

\(\Big \downarrow \) 253

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {5 \int \frac {1}{x^2 \left (b x^2+a\right )^2}dx}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 253

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 264

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {5 \left (\frac {3 \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle c \left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}}\)

Input:

Int[(c/(a + b*x^2)^2)^(3/2)/x^2,x]
 

Output:

c*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*(1/(4*a*x*(a + b*x^2)^2) + (5*(1/(2*a* 
x*(a + b*x^2)) + (3*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^ 
(3/2)))/(2*a)))/(4*a))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 2045

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (-\frac {15 b^{2} x^{4}}{8 a^{3}}-\frac {25 b \,x^{2}}{8 a^{2}}-\frac {1}{a}\right )}{\left (b \,x^{2}+a \right ) x}+\frac {15 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (3 a^{7} \textit {\_R}^{2}+2 b \right ) x +a^{4} \textit {\_R} \right )\right )}{16}\)	\(114\)
default	\(-\frac {\left (15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{3} x^{5}+30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} x^{3}+15 \sqrt {a b}\, b^{2} x^{4}+15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b x +25 \sqrt {a b}\, a b \,x^{2}+8 \sqrt {a b}\, a^{2}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{8 x \sqrt {a b}\, a^{3}}\)	\(121\)

Input:

int((c/(b*x^2+a)^2)^(3/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

c/(b*x^2+a)*(c/(b*x^2+a)^2)^(1/2)*(-15/8*b^2/a^3*x^4-25/8*b/a^2*x^2-1/a)/x 
+15/16*c*(b*x^2+a)*(c/(b*x^2+a)^2)^(1/2)*sum(_R*ln((3*_R^2*a^7+2*b)*x+a^4* 
_R),_R=RootOf(_Z^2*a^7+b))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.10 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=\left [\frac {15 \, {\left (b c x^{3} + a c x\right )} \sqrt {-\frac {b c}{a}} \log \left (\frac {b c x^{2} - a c - 2 \, {\left (a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b c}{a}} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{b x^{2} + a}\right ) - 2 \, {\left (15 \, b^{2} c x^{4} + 25 \, a b c x^{2} + 8 \, a^{2} c\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{16 \, {\left (a^{3} b x^{3} + a^{4} x\right )}}, -\frac {15 \, {\left (b c x^{3} + a c x\right )} \sqrt {\frac {b c}{a}} \arctan \left (\frac {{\left (b x^{3} + a x\right )} \sqrt {\frac {b c}{a}} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{c}\right ) + {\left (15 \, b^{2} c x^{4} + 25 \, a b c x^{2} + 8 \, a^{2} c\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{8 \, {\left (a^{3} b x^{3} + a^{4} x\right )}}\right ] \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^2,x, algorithm="fricas")
 

Output:

[1/16*(15*(b*c*x^3 + a*c*x)*sqrt(-b*c/a)*log((b*c*x^2 - a*c - 2*(a*b*x^3 + 
 a^2*x)*sqrt(-b*c/a)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b*x^2 + a)) - 2 
*(15*b^2*c*x^4 + 25*a*b*c*x^2 + 8*a^2*c)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2 
)))/(a^3*b*x^3 + a^4*x), -1/8*(15*(b*c*x^3 + a*c*x)*sqrt(b*c/a)*arctan((b* 
x^3 + a*x)*sqrt(b*c/a)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))/c) + (15*b^2*c* 
x^4 + 25*a*b*c*x^2 + 8*a^2*c)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^3*b* 
x^3 + a^4*x)]
 

Sympy [F]

\[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=\int \frac {\left (\frac {c}{a^{2} + 2 a b x^{2} + b^{2} x^{4}}\right )^{\frac {3}{2}}}{x^{2}}\, dx \] Input:

integrate((c/(b*x**2+a)**2)**(3/2)/x**2,x)
 

Output:

Integral((c/(a**2 + 2*a*b*x**2 + b**2*x**4))**(3/2)/x**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.59 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=-\frac {15 \, b^{2} c^{\frac {3}{2}} x^{4} + 25 \, a b c^{\frac {3}{2}} x^{2} + 8 \, a^{2} c^{\frac {3}{2}}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b c^{\frac {3}{2}} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^2,x, algorithm="maxima")
 

Output:

-1/8*(15*b^2*c^(3/2)*x^4 + 25*a*b*c^(3/2)*x^2 + 8*a^2*c^(3/2))/(a^3*b^2*x^ 
5 + 2*a^4*b*x^3 + a^5*x) - 15/8*b*c^(3/2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b) 
*a^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=-\frac {1}{8} \, c^{\frac {3}{2}} {\left (\frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {7 \, b^{2} x^{3} + 9 \, a b x}{{\left (b x^{2} + a\right )}^{2} a^{3}} + \frac {8}{a^{3} x}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^2,x, algorithm="giac")
 

Output:

-1/8*c^(3/2)*(15*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + (7*b^2*x^3 + 9* 
a*b*x)/((b*x^2 + a)^2*a^3) + 8/(a^3*x))*sgn(b*x^2 + a)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=\int \frac {{\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2}}{x^2} \,d x \] Input:

int((c/(a + b*x^2)^2)^(3/2)/x^2,x)
 

Output:

int((c/(a + b*x^2)^2)^(3/2)/x^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.88 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {c}\, c \left (-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{5}-8 a^{3}-25 a^{2} b \,x^{2}-15 a \,b^{2} x^{4}\right )}{8 a^{4} x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((c/(b*x^2+a)^2)^(3/2)/x^2,x)
 

Output:

(sqrt(c)*c*( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*x - 3 
0*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**3 - 15*sqrt(b)*sqrt 
(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*x**5 - 8*a**3 - 25*a**2*b*x**2 - 15 
*a*b**2*x**4))/(8*a**4*x*(a**2 + 2*a*b*x**2 + b**2*x**4))