\(\int \frac {1}{x (a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [1028]

Optimal result

Integrand size = 24, antiderivative size = 170 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2 c^{3/2}}+\frac {b^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 (b c-a d)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {a \left (2 a^2 d^2+2 a b d^2 x^2+b^2 c \left (c+d x^2\right )\right )}{c (b c-a d)^2 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {b^{3/2} (2 b c-5 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}}{2 a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {354, 114, 27, 169, 27, 174, 73, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {2 (b c-a d)^2 \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{a}-\frac {b^2 c (2 b c-5 a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{a}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {4 (b c-a d)^2 \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{a d}-\frac {2 b^2 c (2 b c-5 a d) \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{a d}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {2 b^{3/2} c (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {4 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\right )\)

Input:

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

Output:

(b/(a*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) + ((2*d*(b*c + 2*a*d))/(c*( 
b*c - a*d)*Sqrt[c + d*x^2]) + ((-4*(b*c - a*d)^2*ArcTanh[Sqrt[c + d*x^2]/S 
qrt[c]])/(a*Sqrt[c]) + (2*b^(3/2)*c*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[ 
c + d*x^2])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(b*c - a*d)))/(2*a*( 
b*c - a*d)))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.17

method	result	size
pseudoelliptic	\(\frac {-2 \left (-\frac {5 a d}{2}+b c \right ) \left (b \,x^{2}+a \right ) \sqrt {x^{2} d +c}\, c^{\frac {5}{2}} b^{2} \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (-2 c \sqrt {x^{2} d +c}\, \left (b \,x^{2}+a \right ) \left (a d -b c \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )+a \,c^{\frac {3}{2}} \left (b^{2} c \left (x^{2} d +c \right )+2 x^{2} a b \,d^{2}+2 a^{2} d^{2}\right )\right ) \sqrt {\left (a d -b c \right ) b}}{2 \sqrt {\left (a d -b c \right ) b}\, \sqrt {x^{2} d +c}\, c^{\frac {5}{2}} a^{2} \left (b \,x^{2}+a \right ) \left (a d -b c \right )^{2}}\)	\(199\)
default	\(\text {Expression too large to display}\)	\(1968\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (144) = 288\).

Time = 1.57 (sec) , antiderivative size = 1998, normalized size of antiderivative = 11.75 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/8*((2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 
+ (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2)*x^2)*sqrt(b/(b*c - a*d))*l 
og((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d 
^2)*x^2 - 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sq 
rt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(a*b^2 
*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3) 
*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)*sqrt(c)*log(-( 
d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 4*(a*b^2*c^3 + 2*a^3*c*d^2 
 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^2)*sqrt(d*x^2 + c))/(a^3*b^2*c^5 - 2*a^ 
4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c^2*d 
^3)*x^4 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*x^2) 
, 1/8*(8*(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d 
^2 + a^2*b*d^3)*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2) 
*sqrt(-c)*arctan(sqrt(d*x^2 + c)*sqrt(-c)/c) - (2*a*b^2*c^4 - 5*a^2*b*c^3* 
d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a 
^2*b*c^2*d^2)*x^2)*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a* 
b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4*(2*b^2*c^2 - 3*a*b*c*d 
 + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d))) 
/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(a*b^2*c^3 + 2*a^3*c*d^2 + (a*b^2*c^2*d 
+ 2*a^2*b*c*d^2)*x^2)*sqrt(d*x^2 + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {{\left (2 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {{\left (d x^{2} + c\right )} b^{2} c d + 2 \, {\left (d x^{2} + c\right )} a b d^{2} - 2 \, a b c d^{2} + 2 \, a^{2} d^{3}}{2 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{2} + c} b c + \sqrt {d x^{2} + c} a d\right )}} + \frac {\arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c} c} \] Input:

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

Output:

-1/2*(2*b^3*c - 5*a*b^2*d)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/ 
((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(-b^2*c + a*b*d)) + 1/2*((d*x^2 
 + c)*b^2*c*d + 2*(d*x^2 + c)*a*b*d^2 - 2*a*b*c*d^2 + 2*a^2*d^3)/((a*b^2*c 
^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*((d*x^2 + c)^(3/2)*b - sqrt(d*x^2 + c)*b*c 
 + sqrt(d*x^2 + c)*a*d)) + arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a^2*sqrt(-c)* 
c)
 

Mupad [B] (verification not implemented)

Time = 3.67 (sec) , antiderivative size = 5227, normalized size of antiderivative = 30.75 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

atanh((240*a^3*b^11*c^11*d^4*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(64*a^12*b^2* 
c*d^13 - 240*a^3*b^11*c^10*d^4 + 2080*a^4*b^10*c^9*d^5 - 7760*a^5*b^9*c^8* 
d^6 + 16384*a^6*b^8*c^7*d^7 - 21584*a^7*b^7*c^6*d^8 + 18400*a^8*b^6*c^5*d^ 
9 - 10160*a^9*b^5*c^4*d^10 + 3520*a^10*b^4*c^3*d^11 - 704*a^11*b^3*c^2*d^1 
2)) - (2080*a^4*b^10*c^10*d^5*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(64*a^12*b^2 
*c*d^13 - 240*a^3*b^11*c^10*d^4 + 2080*a^4*b^10*c^9*d^5 - 7760*a^5*b^9*c^8 
*d^6 + 16384*a^6*b^8*c^7*d^7 - 21584*a^7*b^7*c^6*d^8 + 18400*a^8*b^6*c^5*d 
^9 - 10160*a^9*b^5*c^4*d^10 + 3520*a^10*b^4*c^3*d^11 - 704*a^11*b^3*c^2*d^ 
12)) + (7760*a^5*b^9*c^9*d^6*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(64*a^12*b^2* 
c*d^13 - 240*a^3*b^11*c^10*d^4 + 2080*a^4*b^10*c^9*d^5 - 7760*a^5*b^9*c^8* 
d^6 + 16384*a^6*b^8*c^7*d^7 - 21584*a^7*b^7*c^6*d^8 + 18400*a^8*b^6*c^5*d^ 
9 - 10160*a^9*b^5*c^4*d^10 + 3520*a^10*b^4*c^3*d^11 - 704*a^11*b^3*c^2*d^1 
2)) - (16384*a^6*b^8*c^8*d^7*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(64*a^12*b^2* 
c*d^13 - 240*a^3*b^11*c^10*d^4 + 2080*a^4*b^10*c^9*d^5 - 7760*a^5*b^9*c^8* 
d^6 + 16384*a^6*b^8*c^7*d^7 - 21584*a^7*b^7*c^6*d^8 + 18400*a^8*b^6*c^5*d^ 
9 - 10160*a^9*b^5*c^4*d^10 + 3520*a^10*b^4*c^3*d^11 - 704*a^11*b^3*c^2*d^1 
2)) + (21584*a^7*b^7*c^7*d^8*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(64*a^12*b^2* 
c*d^13 - 240*a^3*b^11*c^10*d^4 + 2080*a^4*b^10*c^9*d^5 - 7760*a^5*b^9*c^8* 
d^6 + 16384*a^6*b^8*c^7*d^7 - 21584*a^7*b^7*c^6*d^8 + 18400*a^8*b^6*c^5*d^ 
9 - 10160*a^9*b^5*c^4*d^10 + 3520*a^10*b^4*c^3*d^11 - 704*a^11*b^3*c^2*...
 

Reduce [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 5685, normalized size of antiderivative = 33.44 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(10*sqrt(d)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d 
 - b*c)*sqrt(a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*s 
qrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**2*c**2*d + 10*sq 
rt(d)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c 
)*sqrt(a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2* 
sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**2*c*d**2*x**2 - 4*sqrt 
(d)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)* 
sqrt(a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sq 
rt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a*b*c**3 + 6*sqrt(d)*sqrt(b 
)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sqrt(a*d - 
 b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt 
(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a*b*c**2*d*x**2 + 10*sqrt(d)*sqrt(b)* 
sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sqrt(a*d - b 
*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt(a 
)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a*b*c*d**2*x**4 - 4*sqrt(d)*sqrt(b)*sqr 
t(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sqrt(a*d - b*c) 
*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*s 
qrt(a*d - b*c) + 2*a*d - b*c)))*b**2*c**3*x**2 - 4*sqrt(d)*sqrt(b)*sqrt(a) 
*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sqrt(a*d - b*c)*ata 
n((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*sq...