\(\int \frac {(\frac {e (a+b x^2)}{c+d x^2})^{3/2}}{x^3} \, dx\) [59]

Optimal result

Integrand size = 26, antiderivative size = 179 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\frac {(b c-a d) e \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{c^2}-\frac {a e \left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{2 c^2 x^2}-\frac {3 \sqrt {a} (b c-a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {a} \sqrt {e}}\right )}{2 c^{5/2}} \] Output:

(-a*d+b*c)*e*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/c^2-1/2*a*e*(d*x^2+c)* 
(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/c^2/x^2-3/2*a^(1/2)*(-a*d+b*c)*e^(3 
/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) 
)/c^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.97 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.82 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {c} \sqrt {a+b x^2} \left (2 b c x^2-a \left (c+3 d x^2\right )\right )-3 \sqrt {a} (b c-a d) x^2 \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )\right )}{2 c^{5/2} x^2 \sqrt {a+b x^2}} \] Input:

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

Output:

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[c]*Sqrt[a + b*x^2]*(2*b*c*x^2 - 
 a*(c + 3*d*x^2)) - 3*Sqrt[a]*(b*c - a*d)*x^2*Sqrt[c + d*x^2]*ArcTanh[(Sqr 
t[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])]))/(2*c^(5/2)*x^2*Sqrt[a + 
 b*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.50 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.69, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2053, 2052, 252, 262, 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.

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

\(\displaystyle e (b c-a d) \int \frac {x^8}{\left (a e-c x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\)

\(\Big \downarrow \) 252

\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \int \frac {x^4}{a e-c x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 c}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \left (\frac {a e \int \frac {1}{a e-c x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{c}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \left (\frac {\sqrt {a} \sqrt {e} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{c^{3/2}}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}\right )}{2 c}\right )\)

Input:

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

Output:

(b*c - a*d)*e*(x^6/(2*c*(a*e - c*x^4)) - (3*(-(Sqrt[(e*(a + b*x^2))/(c + d 
*x^2)]/c) + (Sqrt[a]*Sqrt[e]*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d* 
x^2)])/(Sqrt[a]*Sqrt[e])])/c^(3/2)))/(2*c))
 

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 = 0.20 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.34

method	result	size
risch	\(-\frac {a \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{2 c^{2} x^{2}}-\frac {\left (-\frac {\left (-2 a^{2} d^{2}+4 a b c d -2 b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {3 a \left (a d -b c \right ) \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right )}{2 \sqrt {a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{2 c^{2} \left (b \,x^{2}+a \right )}\)	\(239\)
default	\(-\frac {\left (-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,d^{2} x^{6}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} c \,d^{2} x^{4}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a b \,c^{2} d \,x^{4}-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,d^{2} x^{4}-4 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c d \,x^{4}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} c^{2} d \,x^{2}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a b \,c^{3} x^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a c d \,x^{2}-4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, b \,c^{2} x^{2}+2 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, d \,x^{2}-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a c d \,x^{2}-2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,c^{2} x^{2}+2 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, c \right ) \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}{4 \sqrt {a c}\, x^{2} c^{3} \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}}\)	\(641\)

Input:

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

Output:

-1/2*a/c^2*(d*x^2+c)/x^2*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/2/c^2*(-(-2*a^2 
*d^2+4*a*b*c*d-2*b^2*c^2)*(b*x^2+a)/(a*d-b*c)/(b*d*e*x^4+a*d*e*x^2+b*c*e*x 
^2+a*c*e)^(1/2)-3/2*a*(a*d-b*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)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*(b*x^2+a)*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.96 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {a e}{c}} e x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, c^{2} x^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {a e}{c}} e x^{2} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) + 2 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c^{2} x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2)/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 [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{2,[0,1,0]%%%},[6,0,0]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1 
,1,1]%%%}
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.70 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {e}\, e \left (-\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,c^{2}-3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a c d \,x^{2}+2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b \,c^{2} x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a c d \,x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a \,d^{2} x^{4}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) b \,c^{2} x^{2}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) b c d \,x^{4}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a c d \,x^{2}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a \,d^{2} x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) b \,c^{2} x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) b c d \,x^{4}\right )}{2 c^{3} x^{2} \left (d \,x^{2}+c \right )} \] Input:

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

Output:

(sqrt(e)*e*( - sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*c**2 - 3*sqrt(c + d*x** 
2)*sqrt(a + b*x**2)*a*c*d*x**2 + 2*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b*c** 
2*x**2 + 3*sqrt(c)*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c 
 + d*x**2)*a)*a*c*d*x**2 + 3*sqrt(c)*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2)* 
c + sqrt(c)*sqrt(c + d*x**2)*a)*a*d**2*x**4 - 3*sqrt(c)*sqrt(a)*log(sqrt(a 
)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*b*c**2*x**2 - 3*sqrt(c) 
*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*b*c* 
d*x**4 - 3*sqrt(c)*sqrt(a)*log(x)*a*c*d*x**2 - 3*sqrt(c)*sqrt(a)*log(x)*a* 
d**2*x**4 + 3*sqrt(c)*sqrt(a)*log(x)*b*c**2*x**2 + 3*sqrt(c)*sqrt(a)*log(x 
)*b*c*d*x**4))/(2*c**3*x**2*(c + d*x**2))