\(\int \frac {(c+d x^2)^{5/2}}{x (a+b x^2)^2} \, dx\) [1010]

Optimal result

Integrand size = 24, antiderivative size = 160 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx=-\frac {d (b c-3 a d) \sqrt {c+d x^2}}{2 a b^2}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{2 a b \left (a+b x^2\right )}-\frac {c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}+\frac {(b c-a d)^{3/2} (2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {\frac {a \sqrt {c+d x^2} \left (b^2 c^2+3 a^2 d^2+2 a b d \left (-c+d x^2\right )\right )}{b^2 \left (a+b x^2\right )}-\frac {(-b c+a d)^{3/2} (2 b c+3 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{5/2}}-2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {354, 109, 27, 171, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(((b*c - a*d)*(c + d*x^2)^(3/2))/(a*b*(a + b*x^2)) + ((-2*d*(b*c - 3*a*d)* 
Sqrt[c + d*x^2])/b + ((-4*b^2*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/a 
+ (2*(b*c - a*d)^(3/2)*(2*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/S 
qrt[b*c - a*d]])/(a*Sqrt[b]))/b)/(2*a*b))/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 109

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

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && 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 = 0.81 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \left (b \,x^{2}+a \right ) \left (a d +\frac {2 b c}{3}\right ) \left (a d -b c \right )^{2} \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{2}+\frac {3 \left (-\frac {2 b^{2} c^{\frac {5}{2}} \left (b \,x^{2}+a \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )}{3}+a \sqrt {x^{2} d +c}\, \left (\frac {b^{2} c^{2}}{3}-\frac {2 a d \left (-x^{2} d +c \right ) b}{3}+a^{2} d^{2}\right )\right ) \sqrt {\left (a d -b c \right ) b}}{2}}{b^{2} a^{2} \left (b \,x^{2}+a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(165\)
default	\(\text {Expression too large to display}\)	\(5354\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 1138, normalized size of antiderivative = 7.11 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.29 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {\sqrt {d x^{2} + c} d^{2}}{b^{2}} - \frac {{\left (2 \, b^{3} c^{3} - a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a^{2} b^{2}} + \frac {\sqrt {d x^{2} + c} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} a b c d^{2} + \sqrt {d x^{2} + c} a^{2} d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} a b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 1321, normalized size of antiderivative = 8.26 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(d^2*(c + d*x^2)^(1/2))/b^2 + (atan((a^2*d^8*(c + d*x^2)^(1/2)*(c^5)^(1/2) 
*9i)/(2*((9*a^2*c^3*d^8)/2 + 5*b^2*c^5*d^6 + (10*b^3*c^6*d^5)/a - (15*b^4* 
c^7*d^4)/(2*a^2) - 12*a*b*c^4*d^7)) + (c^2*d^6*(c + d*x^2)^(1/2)*(c^5)^(1/ 
2)*5i)/(5*c^5*d^6 - (12*a*c^4*d^7)/b + (10*b*c^6*d^5)/a - (15*b^2*c^7*d^4) 
/(2*a^2) + (9*a^2*c^3*d^8)/(2*b^2)) + (c^3*d^5*(c + d*x^2)^(1/2)*(c^5)^(1/ 
2)*10i)/(10*c^6*d^5 + (5*a*c^5*d^6)/b - (15*b*c^7*d^4)/(2*a) - (12*a^2*c^4 
*d^7)/b^2 + (9*a^3*c^3*d^8)/(2*b^3)) - (a*c*d^7*(c + d*x^2)^(1/2)*(c^5)^(1 
/2)*12i)/(5*b*c^5*d^6 - 12*a*c^4*d^7 + (10*b^2*c^6*d^5)/a + (9*a^2*c^3*d^8 
)/(2*b) - (15*b^3*c^7*d^4)/(2*a^2)) - (b*c^4*d^4*(c + d*x^2)^(1/2)*(c^5)^( 
1/2)*15i)/(2*(10*a*c^6*d^5 - (15*b*c^7*d^4)/2 + (5*a^2*c^5*d^6)/b - (12*a^ 
3*c^4*d^7)/b^2 + (9*a^4*c^3*d^8)/(2*b^3))))*(c^5)^(1/2)*1i)/a^2 + ((c + d* 
x^2)^(1/2)*(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))/(2*a*(b^3*(c + d*x^2) - b^ 
3*c + a*b^2*d)) - (atan((c^4*d^5*(c + d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 
+ 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*35i)/(4*(9*a^3*b*c^3*d^8 - (25*b^ 
4*c^6*d^5)/4 - (85*a*b^3*c^5*d^6)/4 - (81*a^4*c^2*d^9)/4 + (27*a^5*c*d^10) 
/(4*b) + (49*a^2*b^2*c^4*d^7)/2 + (15*b^5*c^7*d^4)/(2*a))) - (c^3*d^6*(c + 
 d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1 
/2)*45i)/(4*((27*a^4*c*d^10)/4 - (85*b^4*c^5*d^6)/4 + (49*a*b^3*c^4*d^7)/2 
 - (81*a^3*b*c^2*d^9)/4 + 9*a^2*b^2*c^3*d^8 - (25*b^5*c^6*d^5)/(4*a) + (15 
*b^6*c^7*d^4)/(2*a^2))) + (c^5*d^4*(c + d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5...
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 3938, normalized size of antiderivative = 24.61 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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