\(\int \frac {1}{x^3 (\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [90]

Optimal result

Integrand size = 26, antiderivative size = 184 \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=-\frac {b c-a d}{a^2 e \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}-\frac {c \left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{2 a^2 e^2 x^2}+\frac {3 \sqrt {c} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{5/2} e^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 3.89 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {-\sqrt {a} \sqrt {c+d x^2} \left (3 b c x^2+a \left (c-2 d x^2\right )\right )+3 \sqrt {c} (b c-a d) x^2 \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} e x^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.61, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2053, 2052, 253, 264, 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.

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(b*c - a*d)*e*(1/(2*a*e*x^2*(a*e - c*x^4)) + (3*(-(1/(a*e*x^2)) + (Sqrt[c] 
*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/( 
a^(3/2)*e^(3/2))))/(2*a*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 253

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 = 0.21 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.32

method	result	size
risch	\(-\frac {c \left (b \,x^{2}+a \right )}{2 a^{2} x^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (d \,x^{2}+c \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {3 c \left (a d -b c \right ) \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right )}{2 \sqrt {a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{2 a^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(242\)
default	\(\frac {\left (b \,x^{2}+a \right ) \left (2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} d \,x^{6}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} b c d \,x^{4}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a \,b^{2} c^{2} x^{4}+4 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b d \,x^{4}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c \,x^{4}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{3} c d \,x^{2}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} b \,c^{2} x^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a^{2} d \,x^{2}-4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a b c \,x^{2}-2 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b c \,x^{2}-2 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a \right )}{4 \sqrt {a c}\, x^{2} a^{3} \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}\)	\(641\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.55 \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{4} + {\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt {\frac {c}{a e}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (a b c d + a^{2} d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} + {\left (a b c^{2} + 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {c}{a e}}}{x^{4}}\right ) + 4 \, {\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} + {\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, {\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{4} + {\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt {-\frac {c}{a e}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {c}{a e}}}{2 \, {\left (b c x^{2} + a c\right )}}\right ) + 2 \, {\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} + {\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, {\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),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. \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),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,[1,0,0]%%%},[6,1,0,0]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1, 
[1,1,1]%%
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.68 \[ \int \frac {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {\sqrt {e}\, \left (-\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} c +2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} d \,x^{2}-3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a b c \,x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a^{2} d \,x^{2}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a b c \,x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a b d \,x^{4}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c -\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) b^{2} c \,x^{4}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a^{2} d \,x^{2}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a b c \,x^{2}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a b d \,x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) b^{2} c \,x^{4}\right )}{2 a^{3} e^{2} x^{2} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

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