\(\int \frac {(c+d x^2)^{5/2}}{x^3 (a+b x^2)^2} \, dx\) [1012]

Optimal result

Integrand size = 24, antiderivative size = 180 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}+\frac {c^{3/2} (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3}-\frac {(b c-a d)^{3/2} (4 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {a \sqrt {c+d x^2} \left (2 b^2 c^2 x^2+a^2 d^2 x^2+a b c \left (c-2 d x^2\right )\right )}{b x^2 \left (a+b x^2\right )}+\frac {(-b c+a d)^{3/2} (4 b c+a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}+c^{3/2} (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3} \] Input:

Integrate[(c + d*x^2)^(5/2)/(x^3*(a + b*x^2)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.07, 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, 166, 25, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*x^2)^(5/2)/(x^3*(a + b*x^2)^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 166

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*((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 - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

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 = 0.89 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {-4 \left (b c +\frac {a d}{4}\right ) \left (b \,x^{2}+a \right ) \sqrt {c}\, x^{2} \left (-a d +b c \right )^{2} \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \left (5 \left (b \,x^{2}+a \right ) c^{2} \left (a d -\frac {4 b c}{5}\right ) b \,x^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )+a \left (2 b^{2} c^{2} x^{2}+a c \left (-2 x^{2} d +c \right ) b +a^{2} d^{2} x^{2}\right ) \sqrt {c}\, \sqrt {x^{2} d +c}\right )}{2 \sqrt {\left (a d -b c \right ) b}\, \sqrt {c}\, a^{3} x^{2} b \left (b \,x^{2}+a \right )}\)	\(194\)
risch	\(\text {Expression too large to display}\)	\(1012\)
default	\(\text {Expression too large to display}\)	\(5452\)

Input:

int((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 1272, normalized size of antiderivative = 7.07 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{x^{3} \left (a + b x^{2}\right )^{2}}\, dx \] Input:

integrate((d*x**2+c)**(5/2)/x**3/(b*x**2+a)**2,x)
 

Output:

Integral((c + d*x**2)**(5/2)/(x**3*(a + b*x**2)**2), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{3}} \,d x } \] Input:

integrate((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)^2*x^3), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.57 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{3} \sqrt {-c}} + \frac {{\left (4 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a^{3} b} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} b^{2} c^{3} d - 2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} + 3 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - \sqrt {d x^{2} + c} a^{2} c d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )}^{2} b - 2 \, {\left (d x^{2} + c\right )} b c + b c^{2} + {\left (d x^{2} + c\right )} a d - a c d\right )} a^{2} b} \] Input:

integrate((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.33 (sec) , antiderivative size = 1152, normalized size of antiderivative = 6.40 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((c + d*x^2)^(5/2)/(x^3*(a + b*x^2)^2),x)
 

Output:

(((c + d*x^2)^(1/2)*(a^2*c*d^3 + 2*b^2*c^3*d - 3*a*b*c^2*d^2))/(2*a^2*b) - 
 (d*(c + d*x^2)^(3/2)*(a^2*d^2 + 2*b^2*c^2 - 2*a*b*c*d))/(2*a^2*b))/((c + 
d*x^2)*(a*d - 2*b*c) + b*(c + d*x^2)^2 + b*c^2 - a*c*d) - (atanh((5*d^9*(c 
 + d*x^2)^(1/2)*(c^3)^(1/2))/(4*((5*c^2*d^9)/4 + (4*b*c^3*d^8)/a - (33*b^2 
*c^4*d^7)/(2*a^2) + (65*b^3*c^5*d^6)/(4*a^3) - (5*b^4*c^6*d^5)/a^4)) + (4* 
c*d^8*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*c^3*d^8 + (5*a*c^2*d^9)/(4*b) - (3 
3*b*c^4*d^7)/(2*a) + (65*b^2*c^5*d^6)/(4*a^2) - (5*b^3*c^6*d^5)/a^3) + (65 
*b^2*c^3*d^6*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*(4*a^2*c^3*d^8 + (65*b^2*c^ 
5*d^6)/4 - (5*b^3*c^6*d^5)/a + (5*a^3*c^2*d^9)/(4*b) - (33*a*b*c^4*d^7)/2) 
) - (5*b^3*c^4*d^5*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*a^3*c^3*d^8 - 5*b^3*c 
^6*d^5 + (65*a*b^2*c^5*d^6)/4 - (33*a^2*b*c^4*d^7)/2 + (5*a^4*c^2*d^9)/(4* 
b)) - (33*b*c^2*d^7*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(2*(4*a*c^3*d^8 - (33*b 
*c^4*d^7)/2 + (65*b^2*c^5*d^6)/(4*a) + (5*a^2*c^2*d^9)/(4*b) - (5*b^3*c^6* 
d^5)/a^2)))*(5*a*d - 4*b*c)*(c^3)^(1/2))/(2*a^3) - (atanh((15*c^3*d^6*(c + 
 d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*a*b^5*c^2*d)^(1 
/2))/(4*((7*a^3*c^2*d^9)/4 + (55*b^3*c^5*d^6)/4 - (41*a*b^2*c^4*d^7)/4 - ( 
a^2*b*c^3*d^8)/2 + (a^4*c*d^10)/(4*b) - (5*b^4*c^6*d^5)/a)) + (9*c^2*d^7*( 
c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*a*b^5*c^2*d) 
^(1/2))/(4*((a^3*c*d^10)/4 - (41*b^3*c^4*d^7)/4 - (a*b^2*c^3*d^8)/2 + (7*a 
^2*b*c^2*d^9)/4 + (55*b^4*c^5*d^6)/(4*a) - (5*b^5*c^6*d^5)/a^2)) + (5*c...
 

Reduce [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 4167, normalized size of antiderivative = 23.15 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x)
 

Output:

(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)*sq 
rt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**3*d**2*x**2 + 6*s 
qrt(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*d*x**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*sq 
rt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**2*b*d**2*x**4 - 8*sqrt(d 
)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sq 
rt(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**2*x**2 + 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**2*c*d*x**4 - 8*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)))*b**3*c**2*x**4 - 2*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 - ...