\(\int \frac {(c \sqrt {a+b x^2})^{3/2}}{x^3} \, dx\) [44]

Optimal result

Integrand size = 21, antiderivative size = 133 \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{2 x^2}+\frac {3 b \left (c \sqrt {a+b x^2}\right )^{3/2} \arctan \left (\sqrt [4]{1+\frac {b x^2}{a}}\right )}{4 a \left (1+\frac {b x^2}{a}\right )^{3/4}}-\frac {3 b \left (c \sqrt {a+b x^2}\right )^{3/2} \text {arctanh}\left (\sqrt [4]{1+\frac {b x^2}{a}}\right )}{4 a \left (1+\frac {b x^2}{a}\right )^{3/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (2 \sqrt [4]{a} \left (a+b x^2\right )^{3/4}-3 b x^2 \arctan \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )+3 b x^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right )}{4 \sqrt [4]{a} x^2 \left (a+b x^2\right )^{3/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.75, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2045, 243, 51, 73, 25, 27, 827, 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.

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

\(\Big \downarrow \) 2045

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

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)/x^2) - (3*b*(-1/2*ArcT 
an[(1 + (b*x^2)/a)^(1/4)] + ArcTanh[(1 + (b*x^2)/a)^(1/4)]/2))/a))/(2*(1 + 
 (b*x^2)/a)^(3/4))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[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 827

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

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

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

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=\int \frac {\left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{x^{3}}\, dx \] 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 = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=\frac {{\left (3 \, c^{4} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\sqrt {b x^{2} + a} c}}{\left (a c^{2}\right )^{\frac {1}{4}}}\right )}{\left (a c^{2}\right )^{\frac {1}{4}}} + \frac {\log \left (\frac {\sqrt {\sqrt {b x^{2} + a} c} - \left (a c^{2}\right )^{\frac {1}{4}}}{\sqrt {\sqrt {b x^{2} + a} c} + \left (a c^{2}\right )^{\frac {1}{4}}}\right )}{\left (a c^{2}\right )^{\frac {1}{4}}}\right )} - \frac {4 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} c^{4}}{{\left (b x^{2} + a\right )} c^{2} - a c^{2}}\right )} b}{8 \, c^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.57 \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=-\frac {1}{16} \, {\left (\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} - \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{a} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right )}{a} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b x^{2}}\right )} b c^{\frac {3}{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2}}{x^3} \,d x \] 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 = 1.08 \[ \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {c}\, c \left (-3 a^{\frac {3}{4}} \mathit {atan} \left (\frac {a^{\frac {5}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}-a^{\frac {7}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}{2 a b \,x^{2}+2 a^{2}}\right ) b \,x^{2}-4 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a -3 a^{\frac {3}{4}} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}\right ) b \,x^{2}+3 a^{\frac {3}{4}} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}\right ) b \,x^{2}\right )}{8 a \,x^{2}} \] Input:

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

Output:

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