\(\int \frac {(\frac {c}{\sqrt {a+b x^2}})^{3/2}}{x^3} \, dx\) [54]

Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^3} \, dx=-\frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{2 a x^2}+\frac {3 b \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \arctan \left (\sqrt [4]{1+\frac {b x^2}{a}}\right )}{4 a}+\frac {3 b \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \text {arctanh}\left (\sqrt [4]{1+\frac {b x^2}{a}}\right )}{4 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^3} \, dx=\frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )^{3/4} \left (-2 a^{3/4} \sqrt [4]{a+b x^2}+3 b x^2 \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )+3 b x^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right )}{4 a^{7/4} x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.73, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2045, 243, 52, 73, 756, 216, 219}

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.

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

\(\Big \downarrow \) 2045

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (-3 \int \frac {1}{\frac {a x^8}{b}-\frac {a}{b}}d\sqrt [4]{\frac {b x^2}{a}+1}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{x^2}\right )\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

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 [F]

\[\int \frac {{\left (\frac {c}{\sqrt {b \,x^{2}+a}}\right )}^{\frac {3}{2}}}{x^{3}}d x\]

Input:

int((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x)
 

Output:

int((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x)
 

Fricas [F(-1)]

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

integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((c/(b*x**2+a)**(1/2))**(3/2)/x**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.10 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^3} \, dx=-\frac {1}{8} \, b c^{2} {\left (\frac {4 \, \left (\frac {c}{\sqrt {b x^{2} + a}}\right )^{\frac {3}{2}}}{a c^{2} - \frac {a^{2} c^{2}}{b x^{2} + a}} + \frac {3 \, {\left (\frac {2 \, \arctan \left (\frac {\sqrt {a} \sqrt {\frac {c}{\sqrt {b x^{2} + a}}}}{\sqrt {\sqrt {a} c}}\right )}{\sqrt {\sqrt {a} c} \sqrt {a}} + \frac {\log \left (\frac {\sqrt {a} \sqrt {\frac {c}{\sqrt {b x^{2} + a}}} - \sqrt {\sqrt {a} c}}{\sqrt {a} \sqrt {\frac {c}{\sqrt {b x^{2} + a}}} + \sqrt {\sqrt {a} c}}\right )}{\sqrt {\sqrt {a} c} \sqrt {a}}\right )}}{a}\right )} \] Input:

integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.55 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^3} \, dx=\frac {b c^{\frac {3}{2}} {\left (\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{a^{2}} - \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{a^{2}} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a b x^{2}}\right )}}{16 \, \sqrt {\mathrm {sgn}\left (b x^{2} + a\right )}} \] Input:

integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x, algorithm="giac")
 

Output:

1/16*b*c^(3/2)*(6*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4 
) + 2*(b*x^2 + a)^(1/4))/(-a)^(1/4))/a^2 + 6*sqrt(2)*(-a)^(1/4)*arctan(-1/ 
2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^2 + a)^(1/4))/(-a)^(1/4))/a^2 + 3*s 
qrt(2)*(-a)^(1/4)*log(sqrt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + 
a) + sqrt(-a))/a^2 - 3*sqrt(2)*(-a)^(1/4)*log(-sqrt(2)*(b*x^2 + a)^(1/4)*( 
-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a))/a^2 - 8*(b*x^2 + a)^(1/4)/(a*b*x^2 
))/sqrt(sgn(b*x^2 + a))
 

Mupad [F(-1)]

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

int((c/(a + b*x^2)^(1/2))^(3/2)/x^3,x)
 

Output:

int((c/(a + b*x^2)^(1/2))^(3/2)/x^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.01 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {c}\, c \left (-3 a^{\frac {1}{4}} \mathit {atan} \left (\frac {a^{\frac {5}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}-a^{\frac {7}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}{2 a b \,x^{2}+2 a^{2}}\right ) b \,x^{2}-4 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a +3 a^{\frac {1}{4}} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}\right ) b \,x^{2}-3 a^{\frac {1}{4}} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}\right ) b \,x^{2}\right )}{8 a^{2} x^{2}} \] Input:

int((c/(b*x^2+a)^(1/2))^(3/2)/x^3,x)
 

Output:

(sqrt(c)*c*( - 3*a**(1/4)*atan((a**(1/4)*(a + b*x**2)**(3/4)*a - a**(3/4)* 
(a + b*x**2)**(1/4)*a - a**(3/4)*(a + b*x**2)**(1/4)*b*x**2)/(2*a**2 + 2*a 
*b*x**2))*b*x**2 - 4*(a + b*x**2)**(1/4)*a + 3*a**(1/4)*log((a + b*x**2)** 
(1/4) + a**(1/4))*b*x**2 - 3*a**(1/4)*log((a + b*x**2)**(1/4) - a**(1/4))* 
b*x**2))/(8*a**2*x**2)