\(\int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) [415]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 b x^3}{3 a (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 (b c+a d) x^2}{a (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}+\frac {8 c^2 (b c+a d) \sqrt {a+b x}}{3 a d (b c-a d)^3 (c+d x)^{3/2}}-\frac {8 c (b c-3 a d) (b c+a d) \sqrt {a+b x}}{3 a d (b c-a d)^4 \sqrt {c+d x}} \] Output:

2/3*b*x^3/a/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/2)+2*(a*d+b*c)*x^2/a/(-a*d 
+b*c)^2/(b*x+a)^(1/2)/(d*x+c)^(3/2)+8/3*c^2*(a*d+b*c)*(b*x+a)^(1/2)/a/d/(- 
a*d+b*c)^3/(d*x+c)^(3/2)-8/3*c*(-3*a*d+b*c)*(a*d+b*c)*(b*x+a)^(1/2)/a/d/(- 
a*d+b*c)^4/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (-c^2 d+\frac {3 b c^2 (c+d x)}{a+b x}+\frac {6 a c d (c+d x)}{a+b x}+\frac {6 a b c (c+d x)^2}{(a+b x)^2}+\frac {3 a^2 d (c+d x)^2}{(a+b x)^2}-\frac {a^2 b (c+d x)^3}{(a+b x)^3}\right )}{3 (b c-a d)^4 (c+d x)^{3/2}} \] Input:

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

Output:

(2*(a + b*x)^(3/2)*(-(c^2*d) + (3*b*c^2*(c + d*x))/(a + b*x) + (6*a*c*d*(c 
 + d*x))/(a + b*x) + (6*a*b*c*(c + d*x)^2)/(a + b*x)^2 + (3*a^2*d*(c + d*x 
)^2)/(a + b*x)^2 - (a^2*b*(c + d*x)^3)/(a + b*x)^3))/(3*(b*c - a*d)^4*(c + 
 d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {100, 27, 87, 55, 48}

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.

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

Input:

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

Output:

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

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 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\frac {4}{3} a^{2} b \,d^{3} x^{3}+8 a \,b^{2} c \,d^{2} x^{3}+\frac {4}{3} b^{3} c^{2} d \,x^{3}+2 a^{3} d^{3} x^{2}+14 a^{2} b c \,d^{2} x^{2}+14 a \,b^{2} c^{2} d \,x^{2}+2 b^{3} c^{3} x^{2}+8 a^{3} c \,d^{2} x +16 a^{2} b \,c^{2} d x +8 a \,b^{2} c^{3} x +\frac {16}{3} a^{3} c^{2} d +\frac {16}{3} a^{2} b \,c^{3}}{\left (a d -b c \right )^{4} \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}\)	\(162\)
gosper	\(\frac {\frac {4}{3} a^{2} b \,d^{3} x^{3}+8 a \,b^{2} c \,d^{2} x^{3}+\frac {4}{3} b^{3} c^{2} d \,x^{3}+2 a^{3} d^{3} x^{2}+14 a^{2} b c \,d^{2} x^{2}+14 a \,b^{2} c^{2} d \,x^{2}+2 b^{3} c^{3} x^{2}+8 a^{3} c \,d^{2} x +16 a^{2} b \,c^{2} d x +8 a \,b^{2} c^{3} x +\frac {16}{3} a^{3} c^{2} d +\frac {16}{3} a^{2} b \,c^{3}}{\left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(203\)
orering	\(\frac {\frac {4}{3} a^{2} b \,d^{3} x^{3}+8 a \,b^{2} c \,d^{2} x^{3}+\frac {4}{3} b^{3} c^{2} d \,x^{3}+2 a^{3} d^{3} x^{2}+14 a^{2} b c \,d^{2} x^{2}+14 a \,b^{2} c^{2} d \,x^{2}+2 b^{3} c^{3} x^{2}+8 a^{3} c \,d^{2} x +16 a^{2} b \,c^{2} d x +8 a \,b^{2} c^{3} x +\frac {16}{3} a^{3} c^{2} d +\frac {16}{3} a^{2} b \,c^{3}}{\left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(203\)

Input:

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

Output:

2/3*(2*a^2*b*d^3*x^3+12*a*b^2*c*d^2*x^3+2*b^3*c^2*d*x^3+3*a^3*d^3*x^2+21*a 
^2*b*c*d^2*x^2+21*a*b^2*c^2*d*x^2+3*b^3*c^3*x^2+12*a^3*c*d^2*x+24*a^2*b*c^ 
2*d*x+12*a*b^2*c^3*x+8*a^3*c^2*d+8*a^2*b*c^3)/(a*d-b*c)^4/(b*x+a)^(3/2)/(d 
*x+c)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 468 vs. \(2 (163) = 326\).

Time = 0.46 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.53 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 \, {\left (8 \, a^{2} b c^{3} + 8 \, a^{3} c^{2} d + 2 \, {\left (b^{3} c^{2} d + 6 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3} + 3 \, {\left (b^{3} c^{3} + 7 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2} + 12 \, {\left (a b^{2} c^{3} + 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} + {\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \, {\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} + {\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \] Input:

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

Output:

2/3*(8*a^2*b*c^3 + 8*a^3*c^2*d + 2*(b^3*c^2*d + 6*a*b^2*c*d^2 + a^2*b*d^3) 
*x^3 + 3*(b^3*c^3 + 7*a*b^2*c^2*d + 7*a^2*b*c*d^2 + a^3*d^3)*x^2 + 12*(a*b 
^2*c^3 + 2*a^2*b*c^2*d + a^3*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^ 
4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^ 
4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + 
 a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2 
*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4 
*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^ 
5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b* 
c^2*d^4 + a^6*c*d^5)*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 773 vs. \(2 (163) = 326\).

Time = 0.37 (sec) , antiderivative size = 773, normalized size of antiderivative = 4.18 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

2/3*(sqrt(b*x + a)*(2*(b^9*c^5*d^2*abs(b) - 6*a^2*b^7*c^3*d^4*abs(b) + 8*a 
^3*b^6*c^2*d^5*abs(b) - 3*a^4*b^5*c*d^6*abs(b))*(b*x + a)/(b^9*c^7*d - 7*a 
*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3*d^ 
5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8) + 3*(b^10*c^6*d*ab 
s(b) - 2*a*b^9*c^5*d^2*abs(b) - 2*a^2*b^8*c^4*d^3*abs(b) + 8*a^3*b^7*c^3*d 
^4*abs(b) - 7*a^4*b^6*c^2*d^5*abs(b) + 2*a^5*b^5*c*d^6*abs(b))/(b^9*c^7*d 
- 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c 
^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8))/(b^2*c + (b* 
x + a)*b*d - a*b*d)^(3/2) + 4*(3*sqrt(b*d)*a*b^8*c^3 - 5*sqrt(b*d)*a^2*b^7 
*c^2*d + sqrt(b*d)*a^3*b^6*c*d^2 + sqrt(b*d)*a^4*b^5*d^3 - 6*sqrt(b*d)*(sq 
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^2 + 
 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d 
))^2*a^2*b^5*c*d + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* 
x + a)*b*d - a*b*d))^2*a^3*b^4*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) 
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^4*c)/((b^3*c^3*abs(b) - 3*a*b 
^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*(b^2*c - a*b*d - 
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^3))/b^2
 

Mupad [B] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {4\,x^3\,\left (a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{3\,d\,{\left (a\,d-b\,c\right )}^4}+\frac {x^2\,\left (6\,a^3\,d^3+42\,a^2\,b\,c\,d^2+42\,a\,b^2\,c^2\,d+6\,b^3\,c^3\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,a^2\,c^2\,\left (a\,d+b\,c\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {8\,a\,c\,x\,{\left (a\,d+b\,c\right )}^2}{b\,d^2\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,c^2\,\sqrt {a+b\,x}}{b\,d^2}+\frac {x^2\,\left (a\,d+2\,b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {c\,x\,\left (2\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d^2}} \] Input:

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

Output:

((c + d*x)^(1/2)*((4*x^3*(a^2*d^2 + b^2*c^2 + 6*a*b*c*d))/(3*d*(a*d - b*c) 
^4) + (x^2*(6*a^3*d^3 + 6*b^3*c^3 + 42*a*b^2*c^2*d + 42*a^2*b*c*d^2))/(3*b 
*d^2*(a*d - b*c)^4) + (16*a^2*c^2*(a*d + b*c))/(3*b*d^2*(a*d - b*c)^4) + ( 
8*a*c*x*(a*d + b*c)^2)/(b*d^2*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a*c 
^2*(a + b*x)^(1/2))/(b*d^2) + (x^2*(a*d + 2*b*c)*(a + b*x)^(1/2))/(b*d) + 
(c*x*(2*a*d + b*c)*(a + b*x)^(1/2))/(b*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.51 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*( - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*c**2*d**2 - 12*sqrt(d)*sqrt(b) 
*sqrt(a + b*x)*a**3*c*d**3*x - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*d**4*x 
**2 - 4*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*c**3*d - 14*sqrt(d)*sqrt(b)*s 
qrt(a + b*x)*a**2*b*c**2*d**2*x - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b* 
c*d**3*x**2 - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*d**4*x**3 - 6*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*a*b**2*c**4 - 16*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b* 
*2*c**3*d*x - 14*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c**2*d**2*x**2 - 4*s 
qrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c*d**3*x**3 - 6*sqrt(d)*sqrt(b)*sqrt(a 
 + b*x)*b**3*c**4*x - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*c**3*d*x**2 - 
6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*c**2*d**2*x**3 + 8*sqrt(c + d*x)*a**3 
*b*c**2*d**2 + 12*sqrt(c + d*x)*a**3*b*c*d**3*x + 3*sqrt(c + d*x)*a**3*b*d 
**4*x**2 + 8*sqrt(c + d*x)*a**2*b**2*c**3*d + 24*sqrt(c + d*x)*a**2*b**2*c 
**2*d**2*x + 21*sqrt(c + d*x)*a**2*b**2*c*d**3*x**2 + 2*sqrt(c + d*x)*a**2 
*b**2*d**4*x**3 + 12*sqrt(c + d*x)*a*b**3*c**3*d*x + 21*sqrt(c + d*x)*a*b* 
*3*c**2*d**2*x**2 + 12*sqrt(c + d*x)*a*b**3*c*d**3*x**3 + 3*sqrt(c + d*x)* 
b**4*c**3*d*x**2 + 2*sqrt(c + d*x)*b**4*c**2*d**2*x**3))/(3*sqrt(a + b*x)* 
b*d*(a**5*c**2*d**4 + 2*a**5*c*d**5*x + a**5*d**6*x**2 - 4*a**4*b*c**3*d** 
3 - 7*a**4*b*c**2*d**4*x - 2*a**4*b*c*d**5*x**2 + a**4*b*d**6*x**3 + 6*a** 
3*b**2*c**4*d**2 + 8*a**3*b**2*c**3*d**3*x - 2*a**3*b**2*c**2*d**4*x**2 - 
4*a**3*b**2*c*d**5*x**3 - 4*a**2*b**3*c**5*d - 2*a**2*b**3*c**4*d**2*x ...