\(\int \frac {1}{x (a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\) [1037]

Optimal result

Integrand size = 24, antiderivative size = 225 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {d (3 b c+2 a d)}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {b}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {d \left (b^2 c^2+6 a b c d-2 a^2 d^2\right )}{2 a c^2 (b c-a d)^3 \sqrt {c+d x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2 c^{5/2}}+\frac {b^{5/2} (2 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 (b c-a d)^{7/2}} \] Output:

1/6*d*(2*a*d+3*b*c)/a/c/(-a*d+b*c)^2/(d*x^2+c)^(3/2)+1/2*b/a/(-a*d+b*c)/(b 
*x^2+a)/(d*x^2+c)^(3/2)+1/2*d*(-2*a^2*d^2+6*a*b*c*d+b^2*c^2)/a/c^2/(-a*d+b 
*c)^3/(d*x^2+c)^(1/2)-arctanh((d*x^2+c)^(1/2)/c^(1/2))/a^2/c^(5/2)+1/2*b^( 
5/2)*(-7*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)^(7/2)
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {\frac {a \left (3 b^3 c^2 \left (c+d x^2\right )^2-2 a^3 d^3 \left (4 c+3 d x^2\right )+2 a b^2 c d^2 x^2 \left (10 c+9 d x^2\right )+2 a^2 b d^2 \left (10 c^2+5 c d x^2-3 d^2 x^4\right )\right )}{c^2 (b c-a d)^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {3 b^{5/2} (2 b c-7 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}-\frac {6 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{5/2}}}{6 a^2} \] Input:

Integrate[1/(x*(a + b*x^2)^2*(c + d*x^2)^(5/2)),x]
 

Output:

((a*(3*b^3*c^2*(c + d*x^2)^2 - 2*a^3*d^3*(4*c + 3*d*x^2) + 2*a*b^2*c*d^2*x 
^2*(10*c + 9*d*x^2) + 2*a^2*b*d^2*(10*c^2 + 5*c*d*x^2 - 3*d^2*x^4)))/(c^2* 
(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)^(3/2)) + (3*b^(5/2)*(2*b*c - 7*a*d)* 
ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2) 
 - (6*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/c^(5/2))/(6*a^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {354, 114, 27, 169, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {5 b d x^2+2 b c-2 a d}{2 x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2}{a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {5 b d x^2+2 (b c-a d)}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\)

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {\frac {2 (b c-a d)^3 \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{a}-\frac {b^3 c^2 (2 b c-7 a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{a}}{c (b c-a d)}+\frac {2 d \left (-2 a^2 d^2+6 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}+\frac {2 d (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {\frac {4 (b c-a d)^3 \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{a d}-\frac {2 b^3 c^2 (2 b c-7 a d) \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{a d}}{c (b c-a d)}+\frac {2 d \left (-2 a^2 d^2+6 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}+\frac {2 d (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\frac {2 d \left (-2 a^2 d^2+6 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}+\frac {\frac {2 b^{5/2} c^2 (2 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {4 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}}{c (b c-a d)}+\frac {2 d (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\)

Input:

Int[1/(x*(a + b*x^2)^2*(c + d*x^2)^(5/2)),x]
 

Output:

(b/(a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) + ((2*d*(3*b*c + 2*a*d))/ 
(3*c*(b*c - a*d)*(c + d*x^2)^(3/2)) + ((2*d*(b^2*c^2 + 6*a*b*c*d - 2*a^2*d 
^2))/(c*(b*c - a*d)*Sqrt[c + d*x^2]) + ((-4*(b*c - a*d)^3*ArcTanh[Sqrt[c + 
 d*x^2]/Sqrt[c]])/(a*Sqrt[c]) + (2*b^(5/2)*c^2*(2*b*c - 7*a*d)*ArcTanh[(Sq 
rt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(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 = 1.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(-\frac {-2 \left (b \,x^{2}+a \right ) c^{\frac {9}{2}} \left (-\frac {7 a d}{2}+b c \right ) b^{3} \left (x^{2} d +c \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (2 \left (b \,x^{2}+a \right ) c^{2} \left (x^{2} d +c \right )^{\frac {3}{2}} \left (a d -b c \right )^{3} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )+a \,c^{\frac {5}{2}} \left (c^{4} b^{3}+2 b^{3} c^{3} d \,x^{2}+\frac {20 c^{2} \left (\frac {3}{20} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) b \,d^{2}}{3}-\frac {8 a \left (b \,x^{2}+a \right ) c \left (-\frac {9 b \,x^{2}}{4}+a \right ) d^{3}}{3}-2 a^{2} d^{4} x^{2} \left (b \,x^{2}+a \right )\right )\right ) \sqrt {\left (a d -b c \right ) b}}{2 \sqrt {\left (a d -b c \right ) b}\, \left (x^{2} d +c \right )^{\frac {3}{2}} a^{2} c^{\frac {9}{2}} \left (b \,x^{2}+a \right ) \left (a d -b c \right )^{3}}\)	\(257\)
default	\(\text {Expression too large to display}\)	\(3548\)

Input:

int(1/x/(b*x^2+a)^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(-2*(b*x^2+a)*c^(9/2)*(-7/2*a*d+b*c)*b^3*(d*x^2+c)^(3/2)*arctan((d*x^ 
2+c)^(1/2)*b/((a*d-b*c)*b)^(1/2))+(2*(b*x^2+a)*c^2*(d*x^2+c)^(3/2)*(a*d-b* 
c)^3*arctanh((d*x^2+c)^(1/2)/c^(1/2))+a*c^(5/2)*(c^4*b^3+2*b^3*c^3*d*x^2+2 
0/3*c^2*(3/20*b^2*x^4+a*b*x^2+a^2)*b*d^2-8/3*a*(b*x^2+a)*c*(-9/4*b*x^2+a)* 
d^3-2*a^2*d^4*x^2*(b*x^2+a)))*((a*d-b*c)*b)^(1/2))/((a*d-b*c)*b)^(1/2)/(d* 
x^2+c)^(3/2)/a^2/c^(9/2)/(b*x^2+a)/(a*d-b*c)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 806 vs. \(2 (195) = 390\).

Time = 4.14 (sec) , antiderivative size = 3409, normalized size of antiderivative = 15.15 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x/(b*x**2+a)**2/(d*x**2+c)**(5/2),x)
 

Output:

Integral(1/(x*(a + b*x**2)**2*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {5}{2}} x} \,d x } \] Input:

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^2*(d*x^2 + c)^(5/2)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{3} d}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} - \frac {{\left (2 \, b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {9 \, {\left (d x^{2} + c\right )} b c d^{2} + b c^{2} d^{2} - 3 \, {\left (d x^{2} + c\right )} a d^{3} - a c d^{3}}{3 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {\arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c} c^{2}} \] Input:

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

1/2*sqrt(d*x^2 + c)*b^3*d/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - 
a^4*d^3)*((d*x^2 + c)*b - b*c + a*d)) - 1/2*(2*b^4*c - 7*a*b^3*d)*arctan(s 
qrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3 
*a^4*b*c*d^2 - a^5*d^3)*sqrt(-b^2*c + a*b*d)) + 1/3*(9*(d*x^2 + c)*b*c*d^2 
 + b*c^2*d^2 - 3*(d*x^2 + c)*a*d^3 - a*c*d^3)/((b^3*c^5 - 3*a*b^2*c^4*d + 
3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*(d*x^2 + c)^(3/2)) + arctan(sqrt(d*x^2 + c) 
/sqrt(-c))/(a^2*sqrt(-c)*c^2)
 

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 8467, normalized size of antiderivative = 37.63 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

int(1/(x*(a + b*x^2)^2*(c + d*x^2)^(5/2)),x)
                                                                                    
                                                                                    
 

Output:

((d^2*(c + d*x^2)*(3*a*d - 8*b*c))/(3*(b*c^2 - a*c*d)^2) - d^2/(3*(b*c^2 - 
 a*c*d)) + (d*(c + d*x^2)^2*(b^3*c^2 - 2*a^2*b*d^2 + 6*a*b^2*c*d))/(2*a*c* 
(b*c^2 - a*c*d)*(a*d - b*c)^2))/(b*(c + d*x^2)^(5/2) + (c + d*x^2)^(3/2)*( 
a*d - b*c)) - atanh((560*a^3*b^16*c^19*d^4*(c + d*x^2)^(1/2))/((c^5)^(1/2) 
*(560*a^3*b^16*c^17*d^4 - 7280*a^4*b^15*c^16*d^5 + 42560*a^5*b^14*c^15*d^6 
 - 149184*a^6*b^13*c^14*d^7 + 351904*a^7*b^12*c^13*d^8 - 593440*a^8*b^11*c 
^12*d^9 + 741120*a^9*b^10*c^11*d^10 - 699840*a^10*b^9*c^10*d^11 + 505008*a 
^11*b^8*c^9*d^12 - 278768*a^12*b^7*c^8*d^13 + 116480*a^13*b^6*c^7*d^14 - 3 
5840*a^14*b^5*c^6*d^15 + 7680*a^15*b^4*c^5*d^16 - 1024*a^16*b^3*c^4*d^17 + 
 64*a^17*b^2*c^3*d^18)) - (7280*a^4*b^15*c^18*d^5*(c + d*x^2)^(1/2))/((c^5 
)^(1/2)*(560*a^3*b^16*c^17*d^4 - 7280*a^4*b^15*c^16*d^5 + 42560*a^5*b^14*c 
^15*d^6 - 149184*a^6*b^13*c^14*d^7 + 351904*a^7*b^12*c^13*d^8 - 593440*a^8 
*b^11*c^12*d^9 + 741120*a^9*b^10*c^11*d^10 - 699840*a^10*b^9*c^10*d^11 + 5 
05008*a^11*b^8*c^9*d^12 - 278768*a^12*b^7*c^8*d^13 + 116480*a^13*b^6*c^7*d 
^14 - 35840*a^14*b^5*c^6*d^15 + 7680*a^15*b^4*c^5*d^16 - 1024*a^16*b^3*c^4 
*d^17 + 64*a^17*b^2*c^3*d^18)) + (42560*a^5*b^14*c^17*d^6*(c + d*x^2)^(1/2 
))/((c^5)^(1/2)*(560*a^3*b^16*c^17*d^4 - 7280*a^4*b^15*c^16*d^5 + 42560*a^ 
5*b^14*c^15*d^6 - 149184*a^6*b^13*c^14*d^7 + 351904*a^7*b^12*c^13*d^8 - 59 
3440*a^8*b^11*c^12*d^9 + 741120*a^9*b^10*c^11*d^10 - 699840*a^10*b^9*c^10* 
d^11 + 505008*a^11*b^8*c^9*d^12 - 278768*a^12*b^7*c^8*d^13 + 116480*a^1...
 

Reduce [B] (verification not implemented)

Time = 2.12 (sec) , antiderivative size = 8814, normalized size of antiderivative = 39.17 \[ \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/x/(b*x^2+a)^2/(d*x^2+c)^(5/2),x)
 

Output:

( - 42*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**2*b*c**4*d - 
84*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)*sq 
rt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**2*b*c**3*d**2*x** 
2 - 42*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**2*b*c**2*d**3 
*x**4 + 12*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)/(sq 
rt(b)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a*b**2*c**5 
- 18*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**2*c**4*d*x**2 
 - 72*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**2*c**3*d**2* 
x**4 - 42*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)/(...