\(\int \frac {(\frac {c}{a+b x^2})^{3/2}}{x^3} \, dx\) [23]

Optimal result

Integrand size = 19, antiderivative size = 104 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {3 b c \sqrt {\frac {c}{a+b x^2}}}{2 a^2}-\frac {c \sqrt {\frac {c}{a+b x^2}}}{2 a x^2}+\frac {3 b c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}} \text {arctanh}\left (\sqrt {1+\frac {b x^2}{a}}\right )}{2 a^2} \] Output:

-3/2*b*c*(c/(b*x^2+a))^(1/2)/a^2-1/2*c*(c/(b*x^2+a))^(1/2)/a/x^2+3/2*b*c*( 
c/(b*x^2+a))^(1/2)*(1+b*x^2/a)^(1/2)*arctanh((1+b*x^2/a)^(1/2))/a^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.75 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {c \sqrt {\frac {c}{a+b x^2}} \left (\sqrt {a} \left (a+3 b x^2\right )-3 b x^2 \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )}{2 a^{5/2} x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2045, 243, 52, 61, 73, 221}

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.

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

\(\Big \downarrow \) 2045

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \left (-\frac {3 b \left (\frac {2}{\sqrt {\frac {b x^2}{a}+1}}-2 \text {arctanh}\left (\sqrt {\frac {b x^2}{a}+1}\right )\right )}{2 a}-\frac {1}{x^2 \sqrt {\frac {b x^2}{a}+1}}\right ) \sqrt {\frac {c}{a+b x^2}}}{2 a}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

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]
 

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 = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \left (3 a^{\frac {3}{2}} b \,x^{2}-3 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) \sqrt {b \,x^{2}+a}\, a b \,x^{2}+a^{\frac {5}{2}}\right )}{2 a^{\frac {7}{2}} x^{2}}\)	\(79\)
risch	\(-\frac {\left (b \,x^{2}+a \right ) c \sqrt {\frac {c}{b \,x^{2}+a}}}{2 a^{2} x^{2}}-\frac {b \left (-\frac {3 \ln \left (\frac {2 a c +2 \sqrt {a c}\, \sqrt {x^{2} b c +a c}}{x}\right )}{\sqrt {a c}}+\frac {\sqrt {b c \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+2 c \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{c \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {b c \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-2 c \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{c \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\right ) c \sqrt {\frac {c}{b \,x^{2}+a}}\, \sqrt {\left (b \,x^{2}+a \right ) c}}{2 a^{2}}\)	\(233\)

Input:

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

Output:

-1/2*(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(3*a^(3/2)*b*x^2-3*ln(2*(a^(1/2)*(b*x^2 
+a)^(1/2)+a)/x)*(b*x^2+a)^(1/2)*a*b*x^2+a^(5/2))/a^(7/2)/x^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, b c x^{2} \sqrt {\frac {c}{a}} \log \left (-\frac {b c x^{2} + 2 \, a c + 2 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {\frac {c}{b x^{2} + a}} \sqrt {\frac {c}{a}}}{x^{2}}\right ) - 2 \, {\left (3 \, b c x^{2} + a c\right )} \sqrt {\frac {c}{b x^{2} + a}}}{4 \, a^{2} x^{2}}, -\frac {3 \, b c x^{2} \sqrt {-\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} \sqrt {-\frac {c}{a}}}{c}\right ) + {\left (3 \, b c x^{2} + a c\right )} \sqrt {\frac {c}{b x^{2} + a}}}{2 \, a^{2} x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {1}{4} \, b c {\left (\frac {2 \, c \sqrt {\frac {c}{b x^{2} + a}}}{a^{2} c - \frac {a^{3} c}{b x^{2} + a}} + \frac {3 \, c \log \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} - \sqrt {a c}}{a \sqrt {\frac {c}{b x^{2} + a}} + \sqrt {a c}}\right )}{\sqrt {a c} a^{2}} + \frac {4 \, \sqrt {\frac {c}{b x^{2} + a}}}{a^{2}}\right )} \] Input:

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

Output:

-1/4*b*c*(2*c*sqrt(c/(b*x^2 + a))/(a^2*c - a^3*c/(b*x^2 + a)) + 3*c*log((a 
*sqrt(c/(b*x^2 + a)) - sqrt(a*c))/(a*sqrt(c/(b*x^2 + a)) + sqrt(a*c)))/(sq 
rt(a*c)*a^2) + 4*sqrt(c/(b*x^2 + a))/a^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {1}{2} \, c {\left (\frac {3 \, b c \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{2}} + \frac {2 \, a b c^{2} - 3 \, {\left (b c x^{2} + a c\right )} b c}{{\left (\sqrt {b c x^{2} + a c} a c - {\left (b c x^{2} + a c\right )}^{\frac {3}{2}}\right )} a^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {c}\, c \left (-\sqrt {b \,x^{2}+a}\, a^{2}-3 \sqrt {b \,x^{2}+a}\, a b \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}\right )}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

(sqrt(c)*c*( - sqrt(a + b*x**2)*a**2 - 3*sqrt(a + b*x**2)*a*b*x**2 - 3*sqr 
t(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*x**2 - 3*sq 
rt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4 + 3* 
sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*x**2 + 3 
*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4)) 
/(2*a**3*x**2*(a + b*x**2))