3.4.48 \(\int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx\) [348]

3.4.48.1 Optimal result

Integrand size = 21, antiderivative size = 108 \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=-\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{2 (b+a c) x^2}-\frac {b d \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{2 \sqrt {c} (b+a c)^{3/2}} \]

output

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

3.4.48.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {1}{2} \left (-\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{(b+a c) x^2}-\frac {b d \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {-b-a c}}\right )}{\sqrt {c} (-b-a c)^{3/2}}\right ) \]

input

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

output

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

3.4.48.3 Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2057, 2053, 2052, 215, 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.

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

\(\Big \downarrow \) 2057

\(\displaystyle \int \frac {1}{x^3 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}dx\)

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.4.48.3.1 Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 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]]
 

rule 2057

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* 
((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
 

3.4.48.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(196\) vs. \(2(92)=184\).

Time = 0.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.82

method	result	size
risch	\(-\frac {a d \,x^{2}+a c +b}{2 \left (a c +b \right ) x^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}-\frac {b d \ln \left (\frac {2 a \,c^{2}+2 b c +\left (2 a c d +b d \right ) x^{2}+2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}}{x^{2}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{4 \left (a c +b \right ) \sqrt {a \,c^{2}+b c}\, \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(197\)
default	\(-\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-2 a \,d^{2} \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, x^{4} \sqrt {a \,c^{2}+b c}+\ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}+2 b c}{x^{2}}\right ) a b \,c^{2} d \,x^{2}-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, a c d \,x^{2} \sqrt {a \,c^{2}+b c}+\ln \left (\frac {2 a c d \,x^{2}+b d \,x^{2}+2 a \,c^{2}+2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}+2 b c}{x^{2}}\right ) b^{2} c d \,x^{2}-2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, b d \,x^{2} \sqrt {a \,c^{2}+b c}+2 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} \sqrt {a \,c^{2}+b c}\right )}{4 \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, \left (a c +b \right )^{2} c \,x^{2} \sqrt {a \,c^{2}+b c}}\)	\(452\)

input

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

output

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

3.4.48.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (92) = 184\).

Time = 0.36 (sec) , antiderivative size = 451, normalized size of antiderivative = 4.18 \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=\left [\frac {\sqrt {a c^{2} + b c} b d x^{2} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} - 4 \, {\left ({\left (2 \, a c + b\right )} d^{2} x^{4} + 2 \, a c^{3} + {\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt {a c^{2} + b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left (a c^{3} + {\left (a c^{2} + b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + b^{2} c\right )} x^{2}}, \frac {\sqrt {-a c^{2} - b c} b d x^{2} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {-a c^{2} - b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + {\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) - 2 \, {\left (a c^{3} + {\left (a c^{2} + b c\right )} d x^{2} + b c^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + b^{2} c\right )} x^{2}}\right ] \]

input

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

output

[1/8*(sqrt(a*c^2 + b*c)*b*d*x^2*log(((8*a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 
 8*a^2*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d* 
x^2 - 4*((2*a*c + b)*d^2*x^4 + 2*a*c^3 + (4*a*c^2 + 3*b*c)*d*x^2 + 2*b*c^2 
)*sqrt(a*c^2 + b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/x^4) - 4*(a*c^3 
 + (a*c^2 + b*c)*d*x^2 + b*c^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a 
^2*c^3 + 2*a*b*c^2 + b^2*c)*x^2), 1/4*(sqrt(-a*c^2 - b*c)*b*d*x^2*arctan(1 
/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt(-a*c^2 - b*c)*sqrt((a*d*x^2 
+ a*c + b)/(d*x^2 + c))/(a^2*c^3 + 2*a*b*c^2 + (a^2*c^2 + a*b*c)*d*x^2 + b 
^2*c)) - 2*(a*c^3 + (a*c^2 + b*c)*d*x^2 + b*c^2)*sqrt((a*d*x^2 + a*c + b)/ 
(d*x^2 + c)))/((a^2*c^3 + 2*a*b*c^2 + b^2*c)*x^2)]
 

3.4.48.6 Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {1}{x^{3} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]

input

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

output

Integral(1/(x**3*sqrt((a*c + a*d*x**2 + b)/(c + d*x**2))), x)
 

3.4.48.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.60 \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=-\frac {b d \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{2} + 2 \, a b c + b^{2} - \frac {{\left (a d x^{2} + a c + b\right )} {\left (a c^{2} + b c\right )}}{d x^{2} + c}\right )}} + \frac {b d \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{4 \, \sqrt {{\left (a c + b\right )} c} {\left (a c + b\right )}} \]

input

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

output

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

3.4.48.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (92) = 184\).

Time = 0.40 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.70 \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {\frac {b d \arctan \left (-\frac {\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}}{\sqrt {-a c^{2} - b c}}\right )}{\sqrt {-a c^{2} - b c} {\left (a c + b\right )}} - \frac {2 \, a^{\frac {3}{2}} c^{2} {\left | d \right |} + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a c d + 2 \, \sqrt {a} b c {\left | d \right |} + {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} b d}{{\left (a c^{2} - {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} + b c\right )} {\left (a c + b\right )}}}{2 \, \mathrm {sgn}\left (d x^{2} + c\right )} \]

input

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

output

1/2*(b*d*arctan(-(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 
 + a*c^2 + b*c))/sqrt(-a*c^2 - b*c))/(sqrt(-a*c^2 - b*c)*(a*c + b)) - (2*a 
^(3/2)*c^2*abs(d) + 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b* 
d*x^2 + a*c^2 + b*c))*a*c*d + 2*sqrt(a)*b*c*abs(d) + (sqrt(a*d^2)*x^2 - sq 
rt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*b*d)/((a*c^2 - (sqrt( 
a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2 + b* 
c)*(a*c + b)))/sgn(d*x^2 + c)
 

3.4.48.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {1}{x^3\,\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \]

input

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

output

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