\(\int \frac {(\frac {c}{(a+b x^2)^2})^{3/2}}{x^4} \, dx\) [38]

Optimal result

Integrand size = 19, antiderivative size = 179 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=\frac {11 b^2 c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{8 a^4}+\frac {b^2 c x \sqrt {\frac {c}{\left (a+b x^2\right )^2}}}{4 a^3 \left (a+b x^2\right )}-\frac {c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right )}{3 a^3 x^3}+\frac {3 b c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right )}{a^4 x}+\frac {35 b^{3/2} c \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}} \] Output:

11/8*b^2*c*x*(c/(b*x^2+a)^2)^(1/2)/a^4+1/4*b^2*c*x*(c/(b*x^2+a)^2)^(1/2)/a 
^3/(b*x^2+a)-1/3*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)/a^3/x^3+3*b*c*(c/(b*x^2 
+a)^2)^(1/2)*(b*x^2+a)/a^4/x+35/8*b^(3/2)*c*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a 
)*arctan(b^(1/2)*x/a^(1/2))/a^(9/2)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.59 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=-\frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \left (a+b x^2\right ) \left (\sqrt {a} \left (8 a^3-56 a^2 b x^2-175 a b^2 x^4-105 b^3 x^6\right )-105 b^{3/2} x^3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{24 a^{9/2} x^3} \] Input:

Integrate[(c/(a + b*x^2)^2)^(3/2)/x^4,x]
 

Output:

-1/24*((c/(a + b*x^2)^2)^(3/2)*(a + b*x^2)*(Sqrt[a]*(8*a^3 - 56*a^2*b*x^2 
- 175*a*b^2*x^4 - 105*b^3*x^6) - 105*b^(3/2)*x^3*(a + b*x^2)^2*ArcTan[(Sqr 
t[b]*x)/Sqrt[a]]))/(a^(9/2)*x^3)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2045, 27, 253, 253, 264, 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.

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

\(\Big \downarrow \) 2045

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 253

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {7 \int \frac {1}{x^4 \left (b x^2+a\right )^2}dx}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 253

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {7 \left (\frac {5 \int \frac {1}{x^4 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 264

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {7 \left (\frac {5 \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 264

\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle c \left (a+b x^2\right ) \left (\frac {7 \left (\frac {5 \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}}\)

Input:

Int[(c/(a + b*x^2)^2)^(3/2)/x^4,x]
 

Output:

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

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {\left (-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{7}-105 \sqrt {a b}\, b^{3} x^{6}-210 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{5}-175 \sqrt {a b}\, a \,b^{2} x^{4}-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{3}-56 \sqrt {a b}\, a^{2} b \,x^{2}+8 \sqrt {a b}\, a^{3}\right ) \left (b \,x^{2}+a \right ) {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}}{24 x^{3} \sqrt {a b}\, a^{4}}\)	\(141\)
risch	\(\frac {c \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \left (\frac {35 b^{3} x^{6}}{8 a^{4}}+\frac {175 b^{2} x^{4}}{24 a^{3}}+\frac {7 b \,x^{2}}{3 a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{2}+a \right ) x^{3}}+\frac {35 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{16 a^{5}}-\frac {35 c \left (b \,x^{2}+a \right ) \sqrt {\frac {c}{\left (b \,x^{2}+a \right )^{2}}}\, \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{16 a^{5}}\)	\(158\)

Input:

int((c/(b*x^2+a)^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(-105*arctan(b*x/(a*b)^(1/2))*b^4*x^7-105*(a*b)^(1/2)*b^3*x^6-210*ar 
ctan(b*x/(a*b)^(1/2))*a*b^3*x^5-175*(a*b)^(1/2)*a*b^2*x^4-105*arctan(b*x/( 
a*b)^(1/2))*a^2*b^2*x^3-56*(a*b)^(1/2)*a^2*b*x^2+8*(a*b)^(1/2)*a^3)*(b*x^2 
+a)*(c/(b*x^2+a)^2)^(3/2)/x^3/(a*b)^(1/2)/a^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.87 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=\left [\frac {105 \, {\left (b^{2} c x^{5} + a b c x^{3}\right )} \sqrt {-\frac {b c}{a}} \log \left (\frac {b c x^{2} - a c + 2 \, {\left (a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b c}{a}} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{b x^{2} + a}\right ) + 2 \, {\left (105 \, b^{3} c x^{6} + 175 \, a b^{2} c x^{4} + 56 \, a^{2} b c x^{2} - 8 \, a^{3} c\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{48 \, {\left (a^{4} b x^{5} + a^{5} x^{3}\right )}}, \frac {105 \, {\left (b^{2} c x^{5} + a b c x^{3}\right )} \sqrt {\frac {b c}{a}} \arctan \left (\frac {{\left (b x^{3} + a x\right )} \sqrt {\frac {b c}{a}} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{c}\right ) + {\left (105 \, b^{3} c x^{6} + 175 \, a b^{2} c x^{4} + 56 \, a^{2} b c x^{2} - 8 \, a^{3} c\right )} \sqrt {\frac {c}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}}{24 \, {\left (a^{4} b x^{5} + a^{5} x^{3}\right )}}\right ] \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(105*(b^2*c*x^5 + a*b*c*x^3)*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*(105*b^3*c*x^6 + 175*a*b^2*c*x^4 + 56*a^2*b*c*x^2 - 8*a^3*c)*sqrt( 
c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^4*b*x^5 + a^5*x^3), 1/24*(105*(b^2*c*x^ 
5 + a*b*c*x^3)*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) + (105*b^3*c*x^6 + 175*a*b^2*c*x^4 + 56*a^2*b*c*x 
^2 - 8*a^3*c)*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^4*b*x^5 + a^5*x^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.56 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=\frac {105 \, b^{3} c^{\frac {3}{2}} x^{6} + 175 \, a b^{2} c^{\frac {3}{2}} x^{4} + 56 \, a^{2} b c^{\frac {3}{2}} x^{2} - 8 \, a^{3} c^{\frac {3}{2}}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {35 \, b^{2} c^{\frac {3}{2}} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^4,x, algorithm="maxima")
 

Output:

1/24*(105*b^3*c^(3/2)*x^6 + 175*a*b^2*c^(3/2)*x^4 + 56*a^2*b*c^(3/2)*x^2 - 
 8*a^3*c^(3/2))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3) + 35/8*b^2*c^(3/2)*a 
rctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.47 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=\frac {1}{24} \, c^{\frac {3}{2}} {\left (\frac {105 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {3 \, {\left (11 \, b^{3} x^{3} + 13 \, a b^{2} x\right )}}{{\left (b x^{2} + a\right )}^{2} a^{4}} + \frac {8 \, {\left (9 \, b x^{2} - a\right )}}{a^{4} x^{3}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:

integrate((c/(b*x^2+a)^2)^(3/2)/x^4,x, algorithm="giac")
 

Output:

1/24*c^(3/2)*(105*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) + 3*(11*b^3*x^ 
3 + 13*a*b^2*x)/((b*x^2 + a)^2*a^4) + 8*(9*b*x^2 - a)/(a^4*x^3))*sgn(b*x^2 
 + a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.78 \[ \int \frac {\left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {c}\, c \left (105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{3}+210 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{5}+105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{7}-8 a^{4}+56 a^{3} b \,x^{2}+175 a^{2} b^{2} x^{4}+105 a \,b^{3} x^{6}\right )}{24 a^{5} x^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(sqrt(c)*c*(105*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*x**3 
+ 210*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*x**5 + 105*sqrt 
(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*x**7 - 8*a**4 + 56*a**3*b*x 
**2 + 175*a**2*b**2*x**4 + 105*a*b**3*x**6))/(24*a**5*x**3*(a**2 + 2*a*b*x 
**2 + b**2*x**4))