\(\int \frac {x^5 (a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\) [907]

Optimal result

Integrand size = 24, antiderivative size = 152 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {c^2 (b c-a d)^2}{3 d^5 \left (c+d x^2\right )^{3/2}}+\frac {2 c (b c-a d) (2 b c-a d)}{d^5 \sqrt {c+d x^2}}+\frac {\left (6 b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{d^5}-\frac {2 b (2 b c-a d) \left (c+d x^2\right )^{3/2}}{3 d^5}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {5 a^2 d^2 \left (8 c^2+12 c d x^2+3 d^2 x^4\right )+10 a b d \left (-16 c^3-24 c^2 d x^2-6 c d^2 x^4+d^3 x^6\right )+b^2 \left (128 c^4+192 c^3 d x^2+48 c^2 d^2 x^4-8 c d^3 x^6+3 d^4 x^8\right )}{15 d^5 \left (c+d x^2\right )^{3/2}} \] Input:

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

Output:

(5*a^2*d^2*(8*c^2 + 12*c*d*x^2 + 3*d^2*x^4) + 10*a*b*d*(-16*c^3 - 24*c^2*d 
*x^2 - 6*c*d^2*x^4 + d^3*x^6) + b^2*(128*c^4 + 192*c^3*d*x^2 + 48*c^2*d^2* 
x^4 - 8*c*d^3*x^6 + 3*d^4*x^8))/(15*d^5*(c + d*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {354, 99, 2009}

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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-2*c^2*(b*c - a*d)^2)/(3*d^5*(c + d*x^2)^(3/2)) + (4*c*(b*c - a*d)*(2*b* 
c - a*d))/(d^5*Sqrt[c + d*x^2]) + (2*(6*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Sqr 
t[c + d*x^2])/d^5 - (4*b*(2*b*c - a*d)*(c + d*x^2)^(3/2))/(3*d^5) + (2*b^2 
*(c + d*x^2)^(5/2))/(5*d^5))/2
 

Defintions of rubi rules used

rule 99

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

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {\left (3 b^{2} x^{8}+10 a b \,x^{6}+15 a^{2} x^{4}\right ) d^{4}+60 c \,x^{2} \left (-\frac {2}{15} b^{2} x^{4}-a b \,x^{2}+a^{2}\right ) d^{3}+40 c^{2} \left (\frac {6}{5} b^{2} x^{4}-6 a b \,x^{2}+a^{2}\right ) d^{2}-160 \left (-\frac {6 b \,x^{2}}{5}+a \right ) c^{3} b d +128 b^{2} c^{4}}{15 \left (x^{2} d +c \right )^{\frac {3}{2}} d^{5}}\)	\(122\)
risch	\(\frac {\left (3 b^{2} d^{2} x^{4}+10 x^{2} a b \,d^{2}-14 x^{2} b^{2} c d +15 a^{2} d^{2}-80 a b c d +73 b^{2} c^{2}\right ) \sqrt {x^{2} d +c}}{15 d^{5}}+\frac {\left (a d -b c \right ) c \left (6 a \,d^{2} x^{2}-12 b c d \,x^{2}+5 a c d -11 b \,c^{2}\right ) \sqrt {x^{2} d +c}}{3 d^{5} \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right )}\)	\(142\)
gosper	\(\frac {3 b^{2} x^{8} d^{4}+10 a b \,d^{4} x^{6}-8 c \,x^{6} d^{3} b^{2}+15 a^{2} d^{4} x^{4}-60 a b c \,d^{3} x^{4}+48 b^{2} c^{2} d^{2} x^{4}+60 a^{2} c \,d^{3} x^{2}-240 c^{2} x^{2} d^{2} a b +192 b^{2} c^{3} d \,x^{2}+40 a^{2} c^{2} d^{2}-160 a b \,c^{3} d +128 b^{2} c^{4}}{15 \left (x^{2} d +c \right )^{\frac {3}{2}} d^{5}}\)	\(149\)
trager	\(\frac {3 b^{2} x^{8} d^{4}+10 a b \,d^{4} x^{6}-8 c \,x^{6} d^{3} b^{2}+15 a^{2} d^{4} x^{4}-60 a b c \,d^{3} x^{4}+48 b^{2} c^{2} d^{2} x^{4}+60 a^{2} c \,d^{3} x^{2}-240 c^{2} x^{2} d^{2} a b +192 b^{2} c^{3} d \,x^{2}+40 a^{2} c^{2} d^{2}-160 a b \,c^{3} d +128 b^{2} c^{4}}{15 \left (x^{2} d +c \right )^{\frac {3}{2}} d^{5}}\)	\(149\)
orering	\(\frac {3 b^{2} x^{8} d^{4}+10 a b \,d^{4} x^{6}-8 c \,x^{6} d^{3} b^{2}+15 a^{2} d^{4} x^{4}-60 a b c \,d^{3} x^{4}+48 b^{2} c^{2} d^{2} x^{4}+60 a^{2} c \,d^{3} x^{2}-240 c^{2} x^{2} d^{2} a b +192 b^{2} c^{3} d \,x^{2}+40 a^{2} c^{2} d^{2}-160 a b \,c^{3} d +128 b^{2} c^{4}}{15 \left (x^{2} d +c \right )^{\frac {3}{2}} d^{5}}\)	\(149\)
default	\(a^{2} \left (\frac {x^{4}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {4 c \left (-\frac {x^{2}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {2 c}{3 d^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}\right )}{d}\right )+b^{2} \left (\frac {x^{8}}{5 d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {8 c \left (\frac {x^{6}}{3 d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {2 c \left (\frac {x^{4}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {4 c \left (-\frac {x^{2}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {2 c}{3 d^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}\right )}{d}\right )}{d}\right )}{5 d}\right )+2 a b \left (\frac {x^{6}}{3 d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {2 c \left (\frac {x^{4}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {4 c \left (-\frac {x^{2}}{d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {2 c}{3 d^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}\right )}{d}\right )}{d}\right )\)	\(254\)

Input:

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

Output:

1/15*((3*b^2*x^8+10*a*b*x^6+15*a^2*x^4)*d^4+60*c*x^2*(-2/15*b^2*x^4-a*b*x^ 
2+a^2)*d^3+40*c^2*(6/5*b^2*x^4-6*a*b*x^2+a^2)*d^2-160*(-6/5*b*x^2+a)*c^3*b 
*d+128*b^2*c^4)/(d*x^2+c)^(3/2)/d^5
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, b^{2} d^{4} x^{8} - 2 \, {\left (4 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{6} + 128 \, b^{2} c^{4} - 160 \, a b c^{3} d + 40 \, a^{2} c^{2} d^{2} + 3 \, {\left (16 \, b^{2} c^{2} d^{2} - 20 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} x^{4} + 12 \, {\left (16 \, b^{2} c^{3} d - 20 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (d^{7} x^{4} + 2 \, c d^{6} x^{2} + c^{2} d^{5}\right )}} \] Input:

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

Output:

1/15*(3*b^2*d^4*x^8 - 2*(4*b^2*c*d^3 - 5*a*b*d^4)*x^6 + 128*b^2*c^4 - 160* 
a*b*c^3*d + 40*a^2*c^2*d^2 + 3*(16*b^2*c^2*d^2 - 20*a*b*c*d^3 + 5*a^2*d^4) 
*x^4 + 12*(16*b^2*c^3*d - 20*a*b*c^2*d^2 + 5*a^2*c*d^3)*x^2)*sqrt(d*x^2 + 
c)/(d^7*x^4 + 2*c*d^6*x^2 + c^2*d^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 609 vs. \(2 (141) = 282\).

Time = 0.49 (sec) , antiderivative size = 609, normalized size of antiderivative = 4.01 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\begin {cases} \frac {40 a^{2} c^{2} d^{2}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {60 a^{2} c d^{3} x^{2}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {15 a^{2} d^{4} x^{4}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} - \frac {160 a b c^{3} d}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} - \frac {240 a b c^{2} d^{2} x^{2}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} - \frac {60 a b c d^{3} x^{4}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {10 a b d^{4} x^{6}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {128 b^{2} c^{4}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {192 b^{2} c^{3} d x^{2}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {48 b^{2} c^{2} d^{2} x^{4}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} - \frac {8 b^{2} c d^{3} x^{6}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} + \frac {3 b^{2} d^{4} x^{8}}{15 c d^{5} \sqrt {c + d x^{2}} + 15 d^{6} x^{2} \sqrt {c + d x^{2}}} & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{6}}{6} + \frac {a b x^{8}}{4} + \frac {b^{2} x^{10}}{10}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((40*a**2*c**2*d**2/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sq 
rt(c + d*x**2)) + 60*a**2*c*d**3*x**2/(15*c*d**5*sqrt(c + d*x**2) + 15*d** 
6*x**2*sqrt(c + d*x**2)) + 15*a**2*d**4*x**4/(15*c*d**5*sqrt(c + d*x**2) + 
 15*d**6*x**2*sqrt(c + d*x**2)) - 160*a*b*c**3*d/(15*c*d**5*sqrt(c + d*x** 
2) + 15*d**6*x**2*sqrt(c + d*x**2)) - 240*a*b*c**2*d**2*x**2/(15*c*d**5*sq 
rt(c + d*x**2) + 15*d**6*x**2*sqrt(c + d*x**2)) - 60*a*b*c*d**3*x**4/(15*c 
*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sqrt(c + d*x**2)) + 10*a*b*d**4*x**6 
/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sqrt(c + d*x**2)) + 128*b**2*c 
**4/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sqrt(c + d*x**2)) + 192*b** 
2*c**3*d*x**2/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sqrt(c + d*x**2)) 
 + 48*b**2*c**2*d**2*x**4/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x**2*sqrt( 
c + d*x**2)) - 8*b**2*c*d**3*x**6/(15*c*d**5*sqrt(c + d*x**2) + 15*d**6*x* 
*2*sqrt(c + d*x**2)) + 3*b**2*d**4*x**8/(15*c*d**5*sqrt(c + d*x**2) + 15*d 
**6*x**2*sqrt(c + d*x**2)), Ne(d, 0)), ((a**2*x**6/6 + a*b*x**8/4 + b**2*x 
**10/10)/c**(5/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.63 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{2} x^{8}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {8 \, b^{2} c x^{6}}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} + \frac {2 \, a b x^{6}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {16 \, b^{2} c^{2} x^{4}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} - \frac {4 \, a b c x^{4}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} + \frac {a^{2} x^{4}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {64 \, b^{2} c^{3} x^{2}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{4}} - \frac {16 \, a b c^{2} x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} + \frac {4 \, a^{2} c x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} + \frac {128 \, b^{2} c^{4}}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{5}} - \frac {32 \, a b c^{3}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{4}} + \frac {8 \, a^{2} c^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} \] Input:

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

Output:

1/5*b^2*x^8/((d*x^2 + c)^(3/2)*d) - 8/15*b^2*c*x^6/((d*x^2 + c)^(3/2)*d^2) 
 + 2/3*a*b*x^6/((d*x^2 + c)^(3/2)*d) + 16/5*b^2*c^2*x^4/((d*x^2 + c)^(3/2) 
*d^3) - 4*a*b*c*x^4/((d*x^2 + c)^(3/2)*d^2) + a^2*x^4/((d*x^2 + c)^(3/2)*d 
) + 64/5*b^2*c^3*x^2/((d*x^2 + c)^(3/2)*d^4) - 16*a*b*c^2*x^2/((d*x^2 + c) 
^(3/2)*d^3) + 4*a^2*c*x^2/((d*x^2 + c)^(3/2)*d^2) + 128/15*b^2*c^4/((d*x^2 
 + c)^(3/2)*d^5) - 32/3*a*b*c^3/((d*x^2 + c)^(3/2)*d^4) + 8/3*a^2*c^2/((d* 
x^2 + c)^(3/2)*d^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.32 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {12 \, {\left (d x^{2} + c\right )} b^{2} c^{3} - b^{2} c^{4} - 18 \, {\left (d x^{2} + c\right )} a b c^{2} d + 2 \, a b c^{3} d + 6 \, {\left (d x^{2} + c\right )} a^{2} c d^{2} - a^{2} c^{2} d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{5}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d^{20} - 20 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c d^{20} + 90 \, \sqrt {d x^{2} + c} b^{2} c^{2} d^{20} + 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{21} - 90 \, \sqrt {d x^{2} + c} a b c d^{21} + 15 \, \sqrt {d x^{2} + c} a^{2} d^{22}}{15 \, d^{25}} \] Input:

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

Output:

1/3*(12*(d*x^2 + c)*b^2*c^3 - b^2*c^4 - 18*(d*x^2 + c)*a*b*c^2*d + 2*a*b*c 
^3*d + 6*(d*x^2 + c)*a^2*c*d^2 - a^2*c^2*d^2)/((d*x^2 + c)^(3/2)*d^5) + 1/ 
15*(3*(d*x^2 + c)^(5/2)*b^2*d^20 - 20*(d*x^2 + c)^(3/2)*b^2*c*d^20 + 90*sq 
rt(d*x^2 + c)*b^2*c^2*d^20 + 10*(d*x^2 + c)^(3/2)*a*b*d^21 - 90*sqrt(d*x^2 
 + c)*a*b*c*d^21 + 15*sqrt(d*x^2 + c)*a^2*d^22)/d^25
 

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.97 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {40\,a^2\,c^2\,d^2+60\,a^2\,c\,d^3\,x^2+15\,a^2\,d^4\,x^4-160\,a\,b\,c^3\,d-240\,a\,b\,c^2\,d^2\,x^2-60\,a\,b\,c\,d^3\,x^4+10\,a\,b\,d^4\,x^6+128\,b^2\,c^4+192\,b^2\,c^3\,d\,x^2+48\,b^2\,c^2\,d^2\,x^4-8\,b^2\,c\,d^3\,x^6+3\,b^2\,d^4\,x^8}{15\,d^5\,{\left (d\,x^2+c\right )}^{3/2}} \] Input:

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

Output:

(128*b^2*c^4 + 40*a^2*c^2*d^2 + 15*a^2*d^4*x^4 + 3*b^2*d^4*x^8 + 60*a^2*c* 
d^3*x^2 + 192*b^2*c^3*d*x^2 - 8*b^2*c*d^3*x^6 - 160*a*b*c^3*d + 48*b^2*c^2 
*d^2*x^4 + 10*a*b*d^4*x^6 - 60*a*b*c*d^3*x^4 - 240*a*b*c^2*d^2*x^2)/(15*d^ 
5*(c + d*x^2)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.10 \[ \int \frac {x^5 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d \,x^{2}+c}\, \left (3 b^{2} d^{4} x^{8}+10 a b \,d^{4} x^{6}-8 b^{2} c \,d^{3} x^{6}+15 a^{2} d^{4} x^{4}-60 a b c \,d^{3} x^{4}+48 b^{2} c^{2} d^{2} x^{4}+60 a^{2} c \,d^{3} x^{2}-240 a b \,c^{2} d^{2} x^{2}+192 b^{2} c^{3} d \,x^{2}+40 a^{2} c^{2} d^{2}-160 a b \,c^{3} d +128 b^{2} c^{4}\right )}{15 d^{5} \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right )} \] Input:

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

Output:

(sqrt(c + d*x**2)*(40*a**2*c**2*d**2 + 60*a**2*c*d**3*x**2 + 15*a**2*d**4* 
x**4 - 160*a*b*c**3*d - 240*a*b*c**2*d**2*x**2 - 60*a*b*c*d**3*x**4 + 10*a 
*b*d**4*x**6 + 128*b**2*c**4 + 192*b**2*c**3*d*x**2 + 48*b**2*c**2*d**2*x* 
*4 - 8*b**2*c*d**3*x**6 + 3*b**2*d**4*x**8))/(15*d**5*(c**2 + 2*c*d*x**2 + 
 d**2*x**4))