\(\int \frac {\sqrt {\frac {e (a+b x^2)}{c+d x^2}}}{x^3} \, dx\) [46]

Optimal result

Integrand size = 26, antiderivative size = 130 \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=-\frac {\left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{2 c x^2}-\frac {(b c-a d) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {a} \sqrt {c} \sqrt {a+b x^2} \left (c+d x^2\right )+(b c-a d) x^2 \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )\right )}{2 \sqrt {a} c^{3/2} x^2 \sqrt {a+b x^2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2053, 2052, 252, 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.

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(b*c - a*d)*e*(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/(2*c*(a*e - c*x^4)) - Arc 
Tanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])]/(2*Sqr 
t[a]*c^(3/2)*Sqrt[e]))
 

Defintions of rubi rules used

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 2052

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S 
ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d)   Subst[Int[x^(q* 
(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* 
x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] 
&& IntegerQ[m]
 

rule 2053

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) 
))^(p_), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( 
(a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, 
 x] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {\left (d \,x^{2}+c \right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{2 c \,x^{2}}+\frac {\left (a d -b c \right ) \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{4 c \sqrt {a c e}\, \left (b \,x^{2}+a \right )}\)	\(162\)
default	\(-\frac {\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-2 d b \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, x^{4} \sqrt {a c}-a^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) d c \,x^{2}+c^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) b a \,x^{2}-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, d a \,x^{2} \sqrt {a c}-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c \,x^{2} \sqrt {a c}+2 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, c^{2} a \,x^{2} \sqrt {a c}}\)	\(326\)

Input:

int((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2/c*(d*x^2+c)/x^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/4*(a*d-b*c)/c/(a*c*e) 
^(1/2)*ln((2*a*c*e+(a*d*e+b*c*e)*x^2+2*(a*c*e)^(1/2)*(b*d*e*x^4+(a*d*e+b*c 
*e)*x^2+a*c*e)^(1/2))/x^2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*(b*x^2 
+a)*e)^(1/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.56 \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=\left [-\frac {{\left (b c - a d\right )} x^{2} \sqrt {\frac {e}{a c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} + {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{a c}}}{x^{4}}\right ) + 4 \, {\left (d x^{2} + c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, c x^{2}}, \frac {{\left (b c - a d\right )} x^{2} \sqrt {-\frac {e}{a c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{a c}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (d x^{2} + c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c x^{2}}\right ] \] Input:

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

Output:

[-1/8*((b*c - a*d)*x^2*sqrt(e/(a*c))*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)* 
e*x^4 + 8*a^2*c^2*e + 8*(a*b*c^2 + a^2*c*d)*e*x^2 + 4*(2*a^2*c^3 + (a*b*c^ 
2*d + a^2*c*d^2)*x^4 + (a*b*c^3 + 3*a^2*c^2*d)*x^2)*sqrt((b*e*x^2 + a*e)/( 
d*x^2 + c))*sqrt(e/(a*c)))/x^4) + 4*(d*x^2 + c)*sqrt((b*e*x^2 + a*e)/(d*x^ 
2 + c)))/(c*x^2), 1/4*((b*c - a*d)*x^2*sqrt(-e/(a*c))*arctan(1/2*((b*c + a 
*d)*x^2 + 2*a*c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(a*c))/(b*e*x^2 
 + a*e)) - 2*(d*x^2 + c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c*x^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (110) = 220\).

Time = 0.17 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=\frac {1}{2} \, {\left (\frac {{\left (b c e - a d e\right )} \arctan \left (-\frac {\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}}{\sqrt {-a c e}}\right )}{\sqrt {-a c e} c} - \frac {{\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )} b c e + {\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )} a d e + 2 \, \sqrt {b d e} a c e}{{\left (a c e - {\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )}^{2}\right )} c}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \] Input:

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

Output:

1/2*((b*c*e - a*d*e)*arctan(-(sqrt(b*d*e)*x^2 - sqrt(b*d*e*x^4 + b*c*e*x^2 
 + a*d*e*x^2 + a*c*e))/sqrt(-a*c*e))/(sqrt(-a*c*e)*c) - ((sqrt(b*d*e)*x^2 
- sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^2 + a*c*e))*b*c*e + (sqrt(b*d*e)*x^ 
2 - sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^2 + a*c*e))*a*d*e + 2*sqrt(b*d*e) 
*a*c*e)/((a*c*e - (sqrt(b*d*e)*x^2 - sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^ 
2 + a*c*e))^2)*c))*sgn(d*x^2 + c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{x^3} \, dx=\frac {\sqrt {e}\, \left (-\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a c +\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a d \,x^{2}-\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) b c \,x^{2}-\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a d \,x^{2}+\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) b c \,x^{2}\right )}{2 a \,c^{2} x^{2}} \] Input:

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

Output:

(sqrt(e)*( - sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*c + sqrt(c)*sqrt(a)*log(s 
qrt(a)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*a*d*x**2 - sqrt(c) 
*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*b*c* 
x**2 - sqrt(c)*sqrt(a)*log(x)*a*d*x**2 + sqrt(c)*sqrt(a)*log(x)*b*c*x**2)) 
/(2*a*c**2*x**2)