\(\int \frac {\sqrt {a+b (c x^2)^{3/2}}}{x^3} \, dx\) [43]

Optimal result

Integrand size = 21, antiderivative size = 298 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^3} \, dx=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{2 x^2}+\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} c \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} c x^2-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}} \] Output:

-1/2*(a+b*(c*x^2)^(3/2))^(1/2)/x^2+1/2*3^(3/4)*(1/2*6^(1/2)+1/2*2^(1/2))*b 
^(2/3)*c*(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))*((a^(2/3)+b^(2/3)*c*x^2-a^(1/3)*b 
^(1/3)*(c*x^2)^(1/2))/((1+3^(1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2) 
*EllipticF(((1-3^(1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2))/((1+3^(1/2))*a^(1/3 
)+b^(1/3)*(c*x^2)^(1/2)),I*3^(1/2)+2*I)/(a^(1/3)*(a^(1/3)+b^(1/3)*(c*x^2)^ 
(1/2))/((1+3^(1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2)/(a+b*(c*x^2)^( 
3/2))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^3} \, dx=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-\frac {1}{2},\frac {1}{3},-\frac {b \left (c x^2\right )^{3/2}}{a}\right )}{2 x^2 \sqrt {1+\frac {b \left (c x^2\right )^{3/2}}{a}}} \] Input:

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

Output:

-1/2*(Sqrt[a + b*(c*x^2)^(3/2)]*Hypergeometric2F1[-2/3, -1/2, 1/3, -((b*(c 
*x^2)^(3/2))/a)])/(x^2*Sqrt[1 + (b*(c*x^2)^(3/2))/a])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {892, 809, 759}

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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 809

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

\(\Big \downarrow \) 759

\(\displaystyle c \left (\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{2 c x^2}\right )\)

Input:

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

Output:

c*(-1/2*Sqrt[a + b*(c*x^2)^(3/2)]/(c*x^2) + (3^(3/4)*Sqrt[2 + Sqrt[3]]*b^( 
2/3)*(a^(1/3) + b^(1/3)*Sqrt[c*x^2])*Sqrt[(a^(2/3) + b^(2/3)*c*x^2 - a^(1/ 
3)*b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]*E 
llipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3 
])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])], -7 - 4*Sqrt[3]])/(2*Sqrt[(a^(1/3)*(a^( 
1/3) + b^(1/3)*Sqrt[c*x^2]))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2]) 
^2]*Sqrt[a + b*(c*x^2)^(3/2)]))
 

Defintions of rubi rules used

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[a, b, c, n, m, p, x]
 

rule 892

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1))   Subst[Int[x^m*(a + b 
*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x 
] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.22 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^3} \, dx=\frac {3 \, \sqrt {\frac {\sqrt {c x^{2}} b c}{x}} x^{2} {\rm weierstrassPInverse}\left (0, -\frac {4 \, \sqrt {c x^{2}} a}{b c^{2} x}, x\right ) - \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{2 \, x^{2}} \] Input:

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

Output:

1/2*(3*sqrt(sqrt(c*x^2)*b*c/x)*x^2*weierstrassPInverse(0, -4*sqrt(c*x^2)*a 
/(b*c^2*x), x) - sqrt(sqrt(c*x^2)*b*c*x^2 + a))/x^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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